Chapter 5 Mechanics II

What happens when a force acts at the pivot?

The torque is zero

What is translational equilibrium?

Vector sum of all the forces acting on the body is equal to zero. (Moving at constant velocity) Fnet = 0

A string is tied around a rock of mass .2kg, then the rock is whirled at a constant speed v in a horizontal circle of radius .4m. If sin theta = .4, and cos theta = .9, what is v?

Vertical tension FT (sin theta) = mg FT = mg / sin theta FT = (.2)(10) / .4 = 5 N Horizontal motion FT cos theta = mv^2/r v^2 = (FT)(r) cos theta / m v^2 = (5N)(.4m)(.9) / .2 v = 3 m/s

In the figure below, three blocks hang below a massless meter stick. Block m, hangs from the 20 cm mark. Block 2 hangs from the 70 cm mark, and block 3 hangs from the 80 cm mark. If m1 = 2 kg, m2 = 5 kg, and m3 = 3kg, at what mark on the meter stick should a string be attached so that this system would hand horizontally.

X(cm) = 63 cm

Is a car moving at constant v in translational equilibrium?

Yes

Is an apple sitting still in translational equilibirum?

Yes

Is terminal velocity an example of translational equilibrium?

Yes

Give an example of rotational equilibrium

Propeller at constant angular velocity

What is the rotational analog of the torque equation?

T = Ia T = torque I = rotational inertia a = acceleration (Normal equation, T = rF)

In uniform circular motion, a always points to the _______

center

The smaller the rotational inertia, the _______ it is to rotate

easier

If nothing is angularly accelerating, what is the net torque?

0

In the figure below, a block of mass 40 kg is held in place by two ropes exerting equal tension forces. If cos theta = 2/3, what is the tension in each rope?

2FT cos theta = mg 2FT (2/3) = (40)(10) 2FT (2/3) = 400 2/3 FT = 200 FT = 300N

An elevator is designed to carry a maximum weight of 9800N (including its own weight), and to move upward at a speed of 5 m/s after an initial period of acceleration. What is the relationship between the maximum tension in the elevator cable and the maximum weight of the elevator while the elevator is accelerating upward? A. The tension is greater than 9800N B. The tension is less than 9800 N C. The tension qualms 9800 N D. It cannot be determined from the information given

A

An object is traveling in a circular path. If the velocity of the object is doubled without changing the path, the force required to maintain the object's motion is: A. quadrupled. B. unchanged. C. doubled. D. halved.

A

A spinning ice skater can increase his rotation rate by bringing his arms in closer to his body while in the spin. Which quantities does this action change? His mass His moment of inertia His translational velocity A. II only B. III only C. II and III only D. I, II, and III

A; Moving one's arms doesn't change one's mass, so choice D is eliminated. Translational velocity v is a measure of the movement of an object's center of mass. A skater spinning in place has zero translational velocity whether he is spinning slowly or quickly. More generally, the rate of spinning has no bearing on the translational velocity of a skater, so that the speed of the skater during a jump is unrelated to his angular speed in rotating while jumping. This eliminates choices B and C. The moment of inertia I is the rotational analog of mass found in the rotational form of Newton's second law of motion τ = Iα (where α is the angular acceleration of an object). The value of Idepends upon the mass of an object and the distribution of that mass around the axis of rotation, so that the closer the mass is to the rotation axis (as when arms are pulled in close to the body), the smaller the value of I.

A car makes a left turn (radius of curvature of 20 m) on a flat road at a constant speed 10 m/s. If the car has a mass of 1000 kg, what minimum static friction coefficient is required between the tires and the road to maintain the turn? A. 0.25 B. 0.5 C. 1 D. Cannot be determined from the information given.

B;

Centrifugal force is an apparent outward force during circular motion. It has been described as a reaction force according to Newton's third law. Which of the following statements is most likely to be correct regarding centrifugal force A. Centrifugal force exists only for uniform circular motion, not uniform circular motion B. Centrifugal force exists only when tension or a normal force provides centripetal acceleration C. Centrifugal force always acts antiparallel to the centripetal vector D. Centrifugal force is a result of repulsive electrostatic interactions.

C

A 1000 kg gondola is operated between two towers 340 m apart. When the gondola is exactly between the towers, it is 100 m below their height. What is the tension in the cable at this point? A. 5 kN B. 8 kN C. 10 kN D. 20 kN

C; 2T = mg 2T = 10000 T = 5000 N (for one side) Multiply by 2 for the total Answer is 10 kN, or 10,000 N

When spinning a coin on a flat surface, two equal forces with opposite directions are applied to the opposite sides of a coin. Which of the following is true about the coin after it leaves the hand? A. The coin does not rotate because equal but opposite forces cancel each other out. B. The coin does not rotate because equal but opposite torques cancel each other out. C. The coin rotates and rotational acceleration is zero D. The coin rotates and rotational acceleration is equal to the nonzero net torque divided by the moment of inertia.

C; Rotational acceleration is 0

When rapidly turning a corner on a flat road, a cyclist leans into the center of the turn. The frame of the bike is nearly parallel to which vector A. The force of gravity on the bike and the rider B. The normal force on the pair C. The centripetal force D. The sum of the normal and frictional forces

D; Gravity acts downward, eliminating A Normal force is perpendicular to the surface, eliminating B The centripetal force will be toward the center of the turn, which will be horizontal on a flat road, eliminating The friction is the source of that horizontal centripetal force and the normal force is up. If you add them, the resultant will be similar to the angle at which the bike leans.

Centripetal force equation

Fc = ma(c) = mv^2/r

Two blocks are in static equilibrium, shown below. If block A has a mass of 15kg and the coefficient of static friction between block A and the surface is .2, what is the maximum mass of block B?

Fs = mg usFN = mg us(m)(g) = m(g) (.2)(15)(10) = 10m 30 = 10m m = 3kg

What is torque?

Measure of a force's effectiveness at making an object spin or rotate.

What is uniform circular motion?

Motion in a circular path at a constant speed.

What is the center of mass?

The center of mass of an object is the point on the object that moves in the same way that a point particle would move.

Consider two spheres of the same size and mass, except one is solid and the other is hollow. If we wanted to rotate each of them around an axis through their centers, which one would be easier to rotate?

The one with the smaller rotational inertia (the solid ball)

A fully unraveled yo-yo is swung around in a vertical circle. The yo-yo is 100 g on a string that is 100 cm long. A yo-yo guru manages to make his yo-yo travel three complete circles in 1 second. Determine the centripetal acceleration of the yo-yo. A. 28 m/s^2 B. 32 m/s^2 C. 284 m/s^2 D. 320 m/s^2

a = v^2/r v = 2pir /t v = (2)(3)(1) / .33 v = 18 m/s a = v^2 / r a = 18^2 / 1 a = 320 m/s^2 (actual answer = 324 m/s^2)

The larger the rotational inertia, the _______ difficult it is to rotate

more

In uniform circular motion, v and a are ________ to each other.

perpendicular

What is tension?

pulling or stretching force directed axially through a body

when is torque maximized?

sin 90 = 1

What is rotational inertia?

the property of an object to resist changes in its rotational state of motion

Centripetal velocity equation

v = 2πr/T

An ammonia molecule (NH3) contains three hydrogens that are positioned at the vertices of an equilateral triangle. The nitrogen atom lies 38 pm (1 pm = 10^-12 m) directly above the center of this triangle. If the N:H mass ratio is 14:1, how far below the N atom is the center of mass of the molecule?

7 pm

An object is traveling in a circular path. If the radius of the circular path is doubled without changing the speed of the object, the force required to maintain the object's motion is: A. halved. B. unchanged. C. doubled. D. quadrupled.

A;

The Singapore Flyer is the world's tallest Ferris wheel at a height of 165 m. It makes a complete revolution in about 40 minutes. What is the centripetal acceleration acting on one of the passenger capsules while the wheel is operating? A. 1/1800 m/s2 B. 1/900 m/s2 C. 1/180 m/s2 D. 1.2 m/s2

A;

A bear of mass 4M sits at the right end of a weightless plank of length L. The plank can rock on a fulcrum L / 6 from its right end. How far from the right end of the plank must a chimpanzee of mass M stand to balance the plank? A. 5L / 6 B. 2L / 3 C. L / 6 D. L

A; A bear of mass 4M sits at the right end of a weightless plank of length L. The plank can rock on a fulcrum L / 6 from its right end. A chimpanzee of mass M must stand 5L / 6 from the right end of the plank to balance the plank. There are a few ways of approaching these questions. Some people have an intuitive sense that if the bear is four times the mass of the chimp, then the chimp must be four times farther away. Balancing torques should also lead easily to this conclusion. Be careful with frame of reference: the chimp must be 4L / 6 = 2L / 3 from the fulcrum, which could make this answer choice catch your eye. The correct answer, 5L / 6, includes the L / 6 from the end of the plank to the fulcrum. L / 6 would be correct only if measuring from the left end of the plank.

A businesswoman is rushing out of a hotel through a revolving door with a force of 80 N applied at the edge of the 3 m wide door. Where is the pivot point and what is the maximum torque? A. Pivot is the door hinge, and torque is 240 N∙m B. Pivot is the door hinge, and torque is 30 N∙m C. Pivot is the point where she is pushing, and torque is 30 N∙m D. Pivot is the point where she is pushing, and torque is 240 N∙m

A; A businesswoman is rushing out of a hotel through a revolving door with a force of 80 N applied at the edge of the 3 m wide door. The pivot is the door hinge, and torque is 240 N·m. This is a two-by-two question, where two decisions must be made to get to the right answer. The pivot point is the hinge of the door. The torque is given by the following formula: τ = rF sinθ = (3)(80)(sin 90°) = 240 N·m.

A housefly of mass m is sitting on the horizontal blade of a ceiling fan and has a tangential speed of v as the fan spins. The coefficient of static friction between the fly and the blade is μs and the acceleration due to gravity is g. As the fly walks toward the center of the fan, which of the following is true? A. The force of static friction decreases and the fly stays on the fan. B. The force of static friction decreases and the fly slides off the fan. C. The force of static friction increases and the fly stays on the fan. Your Answer D. The force of static friction increases and the fly slides off the fan.

A; A housefly of mass m is sitting on the horizontal blade of a ceiling fan and has a tangential speed of v as the fan spins. The coefficient of static friction between the fly and the blade is μs and the acceleration due to gravity is g. As the fly walks toward the center of the fan, the force of static friction decreases and the fly stays on the fan. The centripetal force on the fly is the force of static friction so FCentripetal = FStatic Friction and mv2/r = FStatic Friction where r is the distance from the fly to the center of rotation of the fan. As the fly walks toward the center of the fan, r decreases. However, v also decreases since v is equal to r times the angular speed (which is constant at all points of a rotating object). The centripetal force therefore decreases. Since static friction has a maximum value but not a minimum value, it can decrease accordingly, and the fly will remain on the fan blade. (Note that mass in this case remains constant, and acceleration due to gravity is constant, eliminating those two answer choices).

A uniform bar is lying on a flat table. In addition to its weight and the normal force exerted by the table (which exactly balances the bar's weight), exactly two other forces, F1 and F2, act on the rod. If the net force acting on the rod is zero, then: A. the net torque will be zero if F1 and F2 are applied at the same point on the rod. B. the rod cannot accelerate translationally or rotationally. C. the rod can accelerate translationally if F1 and F2 are not applied at the same point on the rod. D. the net torque on the rod must also be zero.

A; A uniform bar is lying on a flat table. In addition to its weight and the normal force exerted by the table (which exactly balances the bar's weight), exactly two other forces, F1 and F2, act on the rod. If the net force acting on the rod is zero, then the net torque will be zero if F1 and F2 are applied at the same point on the rod. "The net torque on the rod must also be zero" is false: Fnet = 0 does not automatically imply that net= 0. Since the net torque need not be zero, there could be rotational acceleration, so "the rod cannot accelerate translationally or rotationally" can be eliminated. And "the rod can accelerate translationally if F1 and F2 are not applied at the same point on the rod" is false because we are told in the question that the net force on the rod is zero; this does automatically imply that the translational acceleration must be zero (regardless of where the forces are applied). The correct answer must be "the net torque will be zero if F1 and F2 are applied at the same point on the rod". (You might find it instructive to draw a quick sketch and try placing two forces F1 and F2, with the same magnitude and exactly opposite directions [because F1 + F2 = 0] at the same point on an object, and get a nonzero net torque. It can't be done.)

A uniform rod of mass M sticks out from a vertical wall and points toward the floor. If the smaller angle it makes with the wall is θ, and its far end is attached to the ceiling by a string parallel to the wall, find the tension in the supporting string. A. Mg/2 B. Mg sin θ C. (Mg sin θ)/2 D. Mg

A; A uniform rod of mass M sticks out from a vertical wall and points toward the floor. If the smaller angle it makes with the wall is θ, and its far end is attached to the ceiling by a string parallel to the wall, the tension in the supporting string is Mg/2. Since the system is in static equilibrium, the net torque must be zero. Let the length of the rod be L; then the counterclockwise torque (due to the tension T in the string) is LT sin θ. The clockwise torque is due to the weight of the rod, and the weight vector acts at the rod's center of mass, which is halfway (L/2) down the rod. Therefore, the clockwise torque is (L/2)Mg sin θ. Since the net torque is 0, the counterclockwise torque must balance the clockwise torque: This gives T = Mg/2.

A homogenous rectangular sheet of metal lies on a flat table and is able to rotate around an axis through its center, perpendicular to the table. Four forces, all of the same magnitude, are exerted on the sheet as shown below. Which of the following statements is true? A. The net force is zero, but the net torque is not B. The net torque is zero, but the net force is not C. Neither the net force nor the net torque is zero D. Both the net force and the net torque equal zero

A; Forces cancel out, but each level arm creates a CCW torque.

A car of mass m drives around a circular banked track of radius R at constant speed v, and maintains a constant height h. The angle of inclination of the bank is θ, and v is the track's critical speed, meaning that friction is not necessary to keep the car moving in uniform circular motion. What is the relation between the magnitudes of the normal force exerted by the track on the car and the weight of the car? A. FN > mg B. FN = mg C. FN < mg D. Cannot be uniquely determined using the information given.

A; If the car is moving in a circular path of constant radius, this means it is not sliding up or down the bank of the road. For that to be true in the effective absence of friction, the vertical component of the normal force must be balancing out the weight of the car: FN cosθ = mg. Since cosθ < 1, this means that FN > mg. The horizontal component of the normal force, FN sinθ, is what provides the centripetal acceleration toward the center of the curved track. (Note: the fact that the magnitude of the normal force is greater than the weight is precisely the opposite relation one sees in an inclined plane problem, where FN = mgcosθ.)

At the beach, Kate collects sea shells of mass 100 g and places them in a bucket. She swings the bucket in a large circle (radius 0.4 m), vertical with the ground, so that the bucket swings over her head. When the bucket is at the top of the circle, upside down, the tangential velocity of the shells is 4 m/s. The sea shells do not fall out of the bucket. What is the net force on the sea shells inside the bucket at this point? A. The net force is the same as the centripetal force. B. The net force is the same as the force of gravity. C. The net force is zero because the force of gravity cancels the centripetal force and the shells do not fall out of the bucket. D. The net force is the centripetal force plus the force of gravity since both are acting downward when the bucket is at the top of the circle.

A; The bucket is traveling in a circle, so the net force is the same as the centripetal force and is acting towards the center of the circle. By definition, the centripetal force is the net force acting towards the center of the circle. Both the force of gravity and the normal force from the bottom of the bucket are acting downward, creating the centripetal force, so gravity alone is not the centripetal force, eliminating two answer choices. The shells are traveling in a circle so their velocity is changing direction and they have non-zero acceleration, therefore the net force cannot be zero, eliminating the remaining incorrect answer choice.

A race car is traveling around a banked curve on a racetrack, meaning the road is not flat, but at an angle with the horizontal. Why can cars travel faster on a banked racetrack? A. The normal force contributes to the centripetal force since the ground is at an angle, so the cars are less likely to slip. B. The friction force is increased since the ground is at an angle, so the cars are less likely to slip. C. The force of gravity is decreased since the ground is at an angle, so the cars are less likely to slip. D. The normal force is increased since the ground is at an angle, so the cars are less likely to slip.

A; The centripetal force on the cars is the force that keeps the cars traveling in a circle and not flying off the track. On a flat road, the centripetal force is the force of static friction. On a banked road, the centripetal force is comprised of static friction and a component of the normal force. Since it is not friction alone that keeps the cars in a circle, they can travel faster without slipping. The banked road decreases the normal force, and the friction force, eliminating the answer choices stating friction or normal force increase. The force of gravity does not depend on the angle of the ground, eliminating the answer choice stating it decreases.

A child's mobile is made from a thin rigid rod 50 cm long of negligible mass. Three objects hang from the rod when suspended: a star (m = 100 g), a moon (m = 200 g), and a planet (m = 200 g). If the star and planet are at opposite ends of the rod, how far from the star must the moon be attached so that the mobile will be balanced with the support string tied to the middle of the rod? A. 12.5 cm B. 15 cm C. 20 cm D. 37.5 cm

A; We can find the value for the distance from the star to the moon by applying the center of mass equation for one dimension, setting the location of the star (x1) as the origin. One nice feature of this equation is that it does not require converting units into SI units because whatever we put in we will get out (there are no constants in the formula). Note that the problem specifies where the center of mass should be, so we will enter a value into the left side of this equation:

For an object that travels in a circular path at constant speed: A. the velocity is constant since the speed is constant. B. the acceleration is not zero and is always directed toward the center of the path. C. the acceleration is not zero and is always directed tangent to the path. D. the acceleration is zero since the speed is constant.

B

A massless rod is attached to the ceiling by a string. Two weights are hung from the rod: a 0.4-lb weight at its left end and a 1.2-lb weight at its right end. If the length of the rod is L, how far from its left end should the string be attached so that the rod (with attached weights) will be horizontal? A.L/4 B. 3L/4 C. 5L/6 D. 2L/3

B;

A uniform plank of mass 12 kg and length L is positioned horizontally, with its two ends supported by sensitive scales. An object of mass 3 kg is placed a distance L/3 from the left end of the plank. What weight does the right-hand scale read? A. 50 N B. 70 N C. 90 N D. 80 N

B;

A uniformly dense plank of length L supports a 1 kg mass on its left end and a 3 kg mass on its right end. Which of the following could possibly be the system's center of mass as measured from the left end of the plank? A. 3L / 4 B. 2L / 3 C. L / 4 D. L / 3

B;

How far from the heavier end must the fulcrum of a massless 5-m seesaw be if an 800-N man on one side is to balance his 200-N daughter at the other end? A. 4 m B. 1 m C. 0.5 m D. 2 m

B;

Three metal blocks are hanging from a 16-foot rod of negligible mass. Blocks #1 and #2 each weigh 0.4 lb, and the weight of Block #3 is 0.8 lb. Block #1 is at the very left end of the rod, Block #2 is at the center of the rod, and Block #3 is at the very right end of the rod. How far from the left end is the center of gravity? A. 9 ft B. 10 ft C. 6 ft D. 12 ft

B;

Two masses are resting on an 8-meter-long, uniform 10-kg plank. Mass #1 is 15 kg and rests 2 meters to the left of the plank's center, and Mass #2 is 5 kg and rests 3 m to the right of the plank's center. How far from the center of the plank is the center of mass? A. 0.5 m to the right of the plank's center B. 0.5 m to the left of the plank's center C. 1.5 m to the left of the plank's center D. 1.5 m to the right of the plank's center

B;

A basketball (0.6 kg, 12 cm radius) rolls across the court without slipping. If the coefficient of static friction between the ball and the court is 1.0, what is the torque provided by friction that causes the basketball to rotate at its maximum speed? A. 0.6 N·m B. 0.7 N·m C. 6 N·m D. 7 N·m

B; fs(macx) = usFN fs(max) = (1)(.6)(10) fs(max) = 6N T = rF T = (.12)(6) T = .6 Nm

Let F represent the net force on an object traveling in a circular path (of radius r) at a constant speed v. If the radius is reduced to (1/2)r, and the speed is increased to 2v, then the net force on this same object becomes: A. 2F. B. 8F. C. 4F. D.F.

B; Let F represent the net force on an object traveling in a circular path (of radius r) at a constant speed v. If the radius is reduced to (1/2)r, and the speed is increased to 2v, then the net force on this same object becomes 8F.

A sphere of uniform density has a circumference of 12 cm. How far from the surface of the sphere is its center of mass? A. 9.5 mm B. 1.9 cm C. 3.8 cm D. It is not possible to determine from the information given.

B; The center of mass of a sphere is located at its geometric center. The distance from the surface of a sphere to its center is its radius, which we can find by using the circumference formula for a circle with the given circumference: 2pir = 12 pir = 6 r = about 2 cm, or choice B

A housefly of mass m is sitting on the horizontal blade of a ceiling fan and has a tangential speed of v as the fan spins. The coefficient of static friction between the fly and the blade is μs and the acceleration due to gravity is g. As the fly walks away from the center of the fan, which of the following must be true so that the fly does not slip off the fan? A. The force of static friction decreases. B. The force of static friction increases. C. The mass increases. D. The acceleration due to gravity increases.

B; The centripetal force on the fly is the force of static friction so FCentripetal = FStatic Friction and mv2/r = FStatic Friction where r is the distance from the fly to the center of rotation of the fan. As the fly walks away from the center of the fan, r increases. However, v also increases since v is equal to r times the angular speed (which is constant at all points of a rotating object). The centripetal force therefore increases and the force of static friction must increase. (Note that mass in this case remains constant, and acceleration due to gravity is constant, eliminating those two answer choices).

A housefly of mass m is sitting on the horizontal blade of a ceiling fan and has a tangential speed of v as the fan spins. The coefficient of static friction between the fly and the blade is μs and the acceleration due to gravity is g. What is the maximum distance the fly can be from the point of rotation of the fan? A. μsg/v2 B. v2/ μsg C. v/ μsg D. v2/ μsmg

B; The centripetal force on the fly is the force of static friction so FCentripetal = FStatic Friction. Since the question asks for the maximum distance, the centripetal force and force of static friction should be maximized so mv2/r = μsmg and r = v2/μsg (note that v increases proportionately to r, so the v2 term means that centripetal force increases with distance from the center).

A snowboarder can tighten a curve by carving deeper into the snow because: A. it lowers their center of mass. B. it increases the centripetal force, decreasing the radius of the turn. C. it helps to overcome the centripetal force. D. it increases the centripetal force, decreasing their speed.

B; The resistance between the snowboard and the snow is the source of centripetal motion. Centripetal force is not an independent force to be overcome. The carving actually opposes the snowboarder's inertia in a straight path. From the equation Fc = mv2 / r we see that centripetal force is inversely proportional to radius. Increasing the force will decrease the radius as stated in the correct answer. Speed does not have this relation. Lowering the center of mass does not affect the circular motion, but it does lower the torque experienced by the snowboarder, making it easier to remain upright.

A car of mass m drives around a circular banked track of radius R at constant speed v, and maintains a constant height h. What best describes the work done by the normal force on the car during one quarter of a revolution around the track? A. WN < 0 B. WN = 0 C. WN > 0 D. The work done by the normal force can be positive or negative depending upon the direction the car is moving.

B; Work done by a force on an object depends upon the component of the force along the direction of the displacement of the object. Anything moving in a circular path is, at any given instant, displacing in a direction tangent to the path (the instantaneous velocity), while it is accelerating along the radius toward the center of the circle (centripetal acceleration). Thus the force and displacement are perpendicular, and so the work done by the force is zero. It is also useful to consider that, because the car is not changing height or speed, it is neither gaining nor losing mechanical energy, so no work is being done on it by an outside force.

A massless meter stick is fixed at Point C, which is 25 cm from its left-hand end. The rod is free to rotate about Point C. If a downward force of 60 N is applied at Point A, what is the minimum force that must be applied at Point B to keep the rod from rotating? A. 20 N, upward B. 15 N, downward C. 20 N, downward D. 15 N, upward

C

Torque has which of the following properties? It is measured in joules. It is a vector. It can be zero when force is nonzero. A. I only B. I and II C. II and III D. I, II, and III

C

Water moves past a water wheel, causing it to turn. The force of the water is 200 N, and the radius of the wheel is 10 m. Calculate the torque around the center of the wheel. A. 20 N-m B. 20,000 N-m C. 2000 N-m D. 200 N-m

C

Which of the following best explains why people with bicep attachment points farther from their elbows tend to have greater elbow flexion strength, and thus an improved ability to perform a dumbbell curling exercise? A. An attachment point that is farther from the elbow increases the force provided by muscle contraction B. At attachment point that is farther from the elbow decreases the force provided by the muscle contraction C. At attachment point that is farther fro the elbow results in a greater torque produced by the bicep as it contracts. D. An attachment point is closer to the hand results in a lesser torque produced by the bicep as it contracts.

C

A 10 kg, 4 m long plank of wood is going to be used as a teeter-totter for a brother and sister. The brother has a mass of 30 kg and the sister a mass of 20 kg. If the brother and sister sit at opposite ends of the plank, how far from the brother should the fulcrum be in order for the teeter-totter to be balanced? A. 1.33 m B. 1.60 m C. 1.67 m D. 2 m

C;

A bicycle tire has a radius of 33 cm. If the bicycle is traveling at 12 m/s, with what approximate frequency are the tires rotating? A. 0.6 rotations/sec B. 3 rotations/sec C. 6 rotations/sec D. 12 rotations/sec

C;

A car is going around a circular turn of radius 12m. The coefficient of friction between the cars and wheels and the ground is .3. What is the maximum speed at which the car can make the turn? A. 3.6 m/s B. 6 m/s C. 36 m/s D. It depends on the mass of the car

C;

A man with a mass of 100 kg sits on a seesaw 5 m from the center. Two children, each with a mass of 20 kg, are seated on the other side of the seesaw. One child sits 10 m from the center. How far from the center should the other child sit to balance the seesaw? A. 10 m B. 5 m C. 15 m D. 20 m

C;

A pendulum consists of a 0.5 kg mass attached to the end of 1-meter-long rod of negligible mass. When the rod makes an angle of 60° with the vertical, find the magnitude of the torque about the pivot. A. 2.5 N·m B. 10.0 N·m C. 4.3 N·m D. 5.0 N·m

C;

A 10 kg, 4 m long plank of wood is going to be used as a teeter-totter for a brother and sister. The brother has a mass of 30 kg and the sister a mass of 20 kg. If the brother and sister sit at opposite ends of the plank, how far from the brother should the fulcrum be in order for the teeter-totter to be balanced? A. 1.33 m B. 1.60 m C. 1.67 m D. 2 m

C; X(CM) = m1x1 + m2x2 / m1+m2 X(CM) = (30)(0) + 10 kg(2) + 20 kg (4) / 60 X(CM) = 20 + 80 / 60 X(CM) = 100/60 X(CM) = 1.67m

A massless rod is attached to the ceiling by a string. Two weights are hung from the rod: a 0.4-lb weight at its left end and a 1.2-lb weight at its right end. The length of the rod is L and a string is attached (3L/4 from its left end) so that the rod (with attached weights) is horizontal. What is the tension in the string supporting the rod and the attached weights? A. 2.4 lb B. 1.2 lb C. 1.6 lb D. 3.2 lb

C; A massless rod is attached to the ceiling by a string. Two weights are hung from the rod: a 0.4-lb weight at its left end and a 1.2-lb weight at its right end. The length of the rod is L and a string is attached (3L/4 from its left end) so that the rod (with attached weights) is horizontal. The tension in the string supporting the rod and the attached weights is 1.6 lb. With the rod in equilibrium, the net force upward (the tension T in the supporting string) must balance the net force downward (the total weight of the hanging objects), so T = 0.4 + 1.2 = 1.6 lb.

The moment of inertia I is the rotational analog of mass found in the rotational form of Newton's second law of motion, τ = Iα (where τ is the net torque on an object and α is its angular acceleration). The value of I depends upon the mass of an object and the distribution of that mass around the axis of rotation, so that the further the mass is from the rotation axis, the greater the value of I. Suppose you have a quarter (a flat disk) balanced on edge, a plain wedding band (a uniform ring) also on edge, and a marble (a solid sphere) that you are going to set spinning on a frictionless surface. Each object has the same mass and the same radius. If you apply the same torque to each object, which object will experience the least angular acceleration? A. The marble B. The quarter C. The wedding band D. They will all experience the same angular acceleration.

C; A ring is hollow and therefore has a greater proportion of its mass concentrated further from the axis of rotation than either of the other objects, eliminating choices A and B (and by extension, choice D). The difference between the quarter and the marble is harder to perceive without doing the math to compute the moments of inertia, but you would be given these formulae on the MCAT if you needed them.

At the beach, Kate collects sea shells of mass 100 g and places them in a bucket. She swings the bucket in a large circle (radius 0.4 m), vertical with the ground, so that the bucket swings over her head. When the bucket is at the bottom of the circle, hanging by her side, the tangential velocity of the shells is 4 m/s. What is the net force on the sea shells inside the bucket at this point? A. 0 N B. 1 N C. 4 N D. 3 N

C; The bucket is traveling in a circle, so the net force is the same as the centripetal force and is acting towards the center of the circle. The centripetal force is given by Fc = mv2/r = 0.1(4)2 / 0.4 = 4 N.

At the beach, Kate collects sea shells of mass 100 g and places them in a bucket. She swings the bucket in a large circle (radius 0.4 m), vertical with the ground, so that the bucket swings over her head. When the bucket is at the top of the circle, upside down, the tangential velocity of the shells is 4 m/s. What is the net force on the sea shells inside the bucket at this point? A. 5 N B. 3 N C. 4 N D. 1 N

C; The bucket is traveling in a circle, so the net force is the same as the centripetal force and is acting towards the center of the circle. The centripetal force is given by Fc = mv2/r = 0.1(4)2 / 0.4 = 4 N.

For which of the following objects does the center of mass fall at a point where none of the object's matter is located? A. A metal disk B. A solid plastic cube C. A golden ring D. A sheet of paper

C; The center of mass of an extended body (a mass in two or three dimensions) of uniform density is the geometric center of that body. For a cylinder, this means the center of the circular cross section at the midpoint of the height of the cylinder. Because a ring is a hollow cylinder, there is no ring material at this point. Each of the other examples has material at its geometric center.

A car is traveling around a banked curve on a road, meaning the road is not flat, but at an angle with the horizontal. The car hits a patch of ice (static friction force is zero) on the road and slides at constant velocity around the curve. Is it possible for the car to slide around the curve and stay on the road? A. No, the car will slide off the road without any static friction to provide the centripetal force. B. Yes, the car will stay on the road if it is heavy enough to provide the centripetal force. C. Yes, the car will stay on the road if the bank is steep enough for the normal force to provide the centripetal force. D. No, the car will slide off the road if it is heavy enough to provide the centripetal force.

C; The centripetal force on the car is the force that keeps the cars traveling in a circle and not flying off the road. On a flat road, the centripetal force is the force of static friction. On a banked road, the centripetal force is typically comprised of static friction and a component of the normal force. In this case, the car is on ice, so the centripetal force is comprised of a component of the normal force only. If the bank is angled enough, the component of the normal force pointing toward the center of the curve can be enough to provide the centripetal force for the car to stay on the road. Using N as the normal force, θ as the angle of the bank with the horizontal, m and v as the mass and velocity of the car, and r as the radius of curvature, yields Nsinθ = mv2/r as the equation to determine the necessary angle for the bank in order for the car to stay on the road. The weight of the car does not contribute to the centripetal force, eliminating two answer choices.

The driver of a 1000 kg car wants to round a curve on a flat road of radius 100 m at a speed of 72 km/hr (20 m/s) without slowing down. What will happen to the car if the pavement is dry and the coefficient of static friction is 0.60? A. There is not enough information to determine what will happen to the car. B. The car will make the turn only if it speeds up. C. The car will make the turn if it stays the same speed. D. The car will make the turn only if it slows down.

C; The net force acting on the car is also known as Fc (centripetal force with acceleration directed towards the center of the circle) provided by friction. We must compare the net force required for the car to make the turn compared to the maximum frictional force that the pavement can provide. Fnet = Fc = mv2 / r = (1000)(20)2 / 100 = 4000 N. Maximum static friction = μsFN = μsmg = (0.6)(1000)(10) = 6000 N. The maximum force of static friction is strictly greater than the force needed for the centripetal force. As such, the car will make the turn.

A rectangular piece of wood with height L and width W is free to rotate about its center. An upward force F acts at the upper right corner as shown. What torque is exerted by this force around the center? A. FL / 2 B. F (sqrt ((L/2^2)+(W/2^2)) C. FW / 2 D. FW / sqrt ((L/2^2)+(W/2^2))

C; The torque exerted by this force around the center is FW / 2. The easiest way of finding the torque is to use the equation τ = Fl, where l is the lever arm (i.e., the distance from the pivot to the line of action). From the picture we can see that l is equal to W / 2. Therefore, τ = (F)(W / 2).

A spring diving board is a long plank composed of flexible material that is fixed at one end and has a second pivot point one third of its length from the fixed end. The second pivot is a fulcrum at which the board will bend when a diver exerts a force by standing or jumping on the board. In which of the following cases is the torque provided by the diver about the fulcrum a maximum? A. When the diver begins running from the fixed end of the board. B. When the diver briefly jumps off from the midpoint of the board. C. When the diver lands on the end of the board opposite the fixed end. D. When the diver leaves the board and goes into his dive.

C; Torque depends upon the force, the distance from the fulcrum to the point where the force is applied, and the sine of the angle between those vectors. In these cases, the force and radial distance vectors are perpendicular each time, so τ = rF. We can rule out choices B and D because, if the diver is not in contact with the board, she is not applying any force to it and therefore produces no torque. Choice A is wrong because the fixed end will cancel out the weight leaving no net force to provide a torque. Even if this were not the case, the fixed end is closer to the fulcrum than is the opposite end (1/3 versus 2/3 of the length of the board), so when the diver lands on the opposite end, exerting at least her weight at that point (actually more due to impulse), that creates the greatest torque.

A bar extends perpendicularly from a vertical wall. The length of the bar is 2 m, and its mass is 10 kg. The free end of the rod is attached to a point on the wall by a light cable, which makes an angle of 30° with the bar. What is the magnitude of the vertical force exerted by the wall on the rod? A. 30 N B. 150 N C. 100 N D. 50 N

D

A bar extends perpendicularly from a vertical wall. The length of the bar is 2 m, and its mass is 10 kg. The free end of the rod is attached to a point on the wall by a light cable, which makes an angle of 30° with the bar. Find the tension in the cable. A. 200 N B. 50 N C. 20 N D. 100 N

D;

A uniform plank of mass 12 kg and length L is positioned horizontally, with its two ends supported by sensitive scales. An object of mass 3 kg is placed a distance L/3 from the left end of the plank. What weight does the left-hand scale read? A. 70 N B. 150 N C. 60 N D. 80 N

D;

An object is moving in a circle at constant speed. Its acceleration vector must be directed: A. tangent to the circle and in the direction of motion. B. tangent to the circle and opposite the direction of motion. C. radially and away from the center of the circle. D. radially and toward the center of the circle.

D;

Assume that a massless bar 5 meters in length is suspended from a rope and that the rope is attached to the bar at a distance x from the bar's left end. If a 20-kg mass hangs from the right side of a bar and a 5-kg mass hangs from the left side of the bar, what value of x will bring about equilibrium? A. 4.5 m B. 3.0 m C. 3.5 m D. 4.0 m

D;

The earth's orbit around the sun is very nearly a perfect circle. The distance from the earth to the sun is 1.5 × 1011 meters, and one earth year is about π × 107 seconds. What is the magnitude of the earth's acceleration toward the sun? A. 3.5 × 10^22 m/s2 B. 1.8 × 10^5 m/s2 C. 6 × 10^4 m/s2 D. 6 × 10^-3 m/s2

D;

Two people are pushing on a 1m door, one trying to open it and the other trying to close it. The person attempting to close the door is pushing on the end with a force of 50N. If the person attempting to open the door is pushing in the middle and the door remains motionless, with what force is the person attempting to open the door pushing? Assume forces are normal to the door. A. 12.5 N B. 25 N C. 50 N D. 100 N

D; T1 = T2 r1F1 = r2F2 F2 = r1F1 / r2 F2 = (1m)(50N) / (.5m) F2 = 100 N

A ball of radius 20 cm rolls down from the top of a hill. By the time it reaches the bottom of the hill, it has completed 100 full rotations. Approximately how long is the hill? A. 12 m B. 30 m C. 60 m D. 120 m

D; d = 2(pi)(r) x rotations d = (2)(3)(.2) x 100 d = 6/5 x 100 d = 120m

A hockey puck is tied to a string and whirled in a circular path on a horizontal table, with the other end of the string threaded through a hole in the center of the table. If the puck has mass m and speed v, and the tension in the string is T, which of the following expressions gives the radius of the circular path? A. (mv2/T)1/2 B. (mv/T)1/2 C. mv/T D. mv2/T

D; A hockey puck is tied to a string and whirled in a circular path on a horizontal table, with the other end of the string threaded through a hole in the center of the table. If the puck has mass m and speed v, and the tension in the string is T, the expression that gives the radius of the circular path is mv2/T. The tension in the string provides the centripetal force. Thus, T = mv2/r. Solving this equation for ryields r = mv2/T.

A carousel ride starts at rest. It takes 1 minute to reach its full rotation speed, then spins at a constant speed for 3 minutes. Finally, the carousel takes 1 minute to come to a stop again. Which best describes the forces acting on a person riding the carousel? I. During the first minute, there is a non-zero net force causing the person on the carousel to speed up. II. During the middle three minutes, there is zero net force since the speed is constant. III. During the last minute, there is a non-zero net force causing the person on the carousel to slow down A. II only B. I, II, and III C. I only D. I and III

D; During the first minute, there is a change in speed, so there is a non-zero acceleration, so there is a non-zero net force. Item I is correct, eliminating the choice "II only." During the last minute, there is a change in speed, so there is a non-zero acceleration, so there is a non-zero net force. Item III is correct. During the middle three minutes, the speed is constant, but the direction of the person on the carousel is constantly changing in order to continue traveling in a circle. Since a change in direction is a change in velocity, and a change in velocity is by definition a non-zero acceleration, then there is also a non-zero net force acting to keep the person traveling in a circle. Item II is false. The correct answer is I and III.

A wooden plank in the shape of a rectangle with side lengths a and b lies on a frictionless surface. Two forces, each of magnitude F, are applied to opposite corners as shown. What is the magnitude and direction of the net torque applied to the plank about its center? A. aF / 2 counterclockwise B. bF / 2 counterclockwise C. aF clockwise D. bF clockwise

D; From the picture, it is clear that each force provides a clockwise force about its center. This eliminates choices A and B. For two dimensional bodies like the rectangular plank, the preferred formula for torque is the moment arm equation, τ = lF, where l, the moment arm (or lever arm), is the component of r that is perpendicular to the force. Another way of thinking about the moment arm is that it is the distance from the pivot to the line containing the force. In this case, l = b / 2 for both forces. Since each torque is in the same direction, the net torque is twice the torque provided by either force.

A uniform meter stick weighing 20 N has a 50-N weight on its left end and a 30-N weight on its right end. The bar is hung from a rope. What is the tension in the rope and how far from the left end of the bar should the rope be attached so that the stick remains level? A. 80 N placed 37.5 cm from the left end of the bar B. 100 N placed 37.5 cm from the left end of the bar C. 80 N placed 40.0 cm from the left end of the bar D. 100 N placed 40.0 cm from the left end of the bar

D; Since the total downward force on the bar is (20 N) + (50 N) + (30 N) = 100 N, the total upward force on the bar—which is provided by the tension in the supporting rope—must also be 100 N to keep the bar in static equilibrium. This eliminates choices "80 N placed 37.5 cm from the left end of the bar" and "80 N placed 40.0 cm from the left end of the bar". Now, let x be the distance (in cm) from the left end of the bar to the suspension point. Using this point as the pivot, we balance the torques. The counterclockwise (CCW) torque due to the 50-N weight at the left end is 50x, and the total clockwise (CW) torque due to the weight of the bar and the 30-N weight at the right end is 20(50-x) + 30(100-x). Setting 50x = 20(50-x) + 30(100-x), we solve for x and find that x = 40 cm.

An umbrella (length L, mass m) leans up against a wall as shown. The bottom of the umbrella makes an angle θ with the ground. Given that there is nonzero static friction between the umbrella and both the floor and the wall, all of the following equations must be true EXCEPT: A. L/2 mg sin theta = LFN(wall) sin theta B. FN(ground) + Fs(wall) = 0 C. Fs(ground) - FN(wall) = 0 D. -(L/2) mgsin (90-theta) + LFs(wall) (sin 90- theta) +LN(wall) sin theta = 0

In static equilibrium (when an object is not moving at all), the forces in the horizontal and vertical directions and the torques must each sum to zero. Note that the force equations will yield choice B for the vertical forces and choice C for the horizontal, eliminating those options. Either A or D must be the equation for the torque, and the easiest way to choose between them is to use process of elimination. Neither includes the forces acting at the ground, so take that point as the pivot. From there, the radial distance to the weight mg of the umbrella is L / 2, so A must be the correct answer choice (the false equation) and D must be wrong (A also fails to include a term for the friction force from the wall).

How would the net force on an object undergoing uniform circular motion have to change if the object's speed doubled?

Increase by a factor of four

If an object undergoing uniform circular motion is being acted upon by a constant force toward the center, why doesn't the object fall into the center?

It is falling toward the center, but because of the speed, the object remains in a circular orbit around the center.

A square plate (of side length s) rests on a flat table, and we exert a force F at the corner, parallel to one of the sides. What is the torque of this force (use the center of the plate as a pivot point

T = 1/2(s)(F)

Lever arm torque equation

T = lF l = lever arm F = force

Torque equation

T=Fr sin theta T = torque (units Nm) F = N r = radius

A 50-g stone is tied to the end of a string and whirled in a horizontal circle of radius 2 m at 20 m/s. Ignoring the force of gravity, determine the tension in the string. A. 500 N B. 100 N C. 10 N D. 5 N

Tension = centripetal force T = mv^2/r T = (.05kg)(20^2)/(2) T = (.05)(400)/2 T = 10N

What is the lever arm?

The shortest distance from the pivot point to the line of action of the force (perpendicular to the line of action)

What is static equilibrium?

a motionless state in which there is no net force or net torque acting

Centripetal acceleration equation

a=v^2/r

Equilibrium means zero ________

acceleration

The acceleration of an object undergoing uniform circular motion always points toward the ________

center

A negative torque is a _______ torque

clockwise

In uniform circular motion: speed is _________ velocity is _________ acceleration is ________

constant changing changing (because v is changing)

A positive porque is a _______ torque

counterclockwise

In uniform circular motion, v is always ______ to the circle

tangent

What is a fulcrum?

the pivot point of a lever

Is torque a vector or scalar?

vector

What is rotational equilibrium?

when the the vector sum of all the torques acting on an object is zero T = 0

center of mass equation

x= (m₁x₁+m₂x₂+m₃x₃...)/(m₁+m₂+m₃...) y= (m₁y₁+m₂y₂+m₃y₃...)/(m₁+m₂+m₃...) z= (m₁z₁+m₂z₂+m₃y₃...)/(m₁+m₂+m₃...)