PY205 Ch5 Questions
A car rounds a curve at a steady 50 km/h. If it rounds the same curve at a steady 70km/h, will its acceleration be any different? Explain.
Yes, the centripetal acceleration will be greater when the speed is greater since centripetal acceleration is proportional to the square of the speed. An object in uniform circular motion has an acceleration, since the direction of the velocity vector is changing even though the speed is constant.
Can a coefficient of friction exceed 1.0?
Yes. Refer to Table 5-1. The coefficient of static friction between rubber and many solid surfaces is typically between 1 and 4. The coefficient of static friction can also be greater than one if either of the surfaces is sticky.
A car maintains a constant speed v as it traverses the hill and valley shown in Fig. 5-31. Both the hill and valley have a radius of curvature R. At which point, A, B, or C, is the normal force acting on the car (a) the largest, (b) the smallest? Explain, (c) Where would the driver feel heaviest and (d) lightest? Explain, (e) How fast can the car go without losing contact with the road at A?
(a) The normal force on the car is largest at point C. In this case, the centripetal force keeping the car in a circular path of radius R is directed upward, so the normal force must be greater than the weight to provide this net upward force. (b) The normal force is smallest at point A, the crest of the hill. At this point the centripetal force must be downward (towards the center of the circle) so the normal force must be less than the weight. (Notice that the normal force is equal to the weight at point B.) (c) The driver will feel heaviest where the normal force is greatest, or at point C. (d) The driver will feel lightest at point A, where the normal force is the least. (e) At point A, the centripetal force is weight minus normal force, or mg - N = mv2/r. The point at which the car just loses contact with the road corresponds to a normal force of zero. Setting N=0givesmg=mv2/r or v= sqrt(gr).
When attempting to stop a car quickly on dry pavement, which of the following methods will stop the car in the least time? (a)Slam on the brakes as hard as possible, locking the wheels and skidding to a stop. (b) Press the brakes as hard as possible without locking the wheels and rolling to a stop. Explain.
(b). If the car comes to a stop without skidding, the force that stops the car is the force of kinetic friction between the brake mechanism and the wheels. This force is designed to be large. If you slam on the brakes and skid to a stop, the force that stops the car will be the force of kinetic friction between the tires and the road. Even with a dry road, this force is likely to be less that the force of kinetic friction between the brake mechanism and the wheels. The car will come to a stop more quickly if the tires continue to roll, rather than skid. In addition, once the wheels lock, you have no steering control over the car.
child on a sled comes flying over the crest of a small hill, as shown in Fig. 5-28. His sled does not leave the ground, but he feels the normal force between his chest and the sled decrease as he goes over the hill. Explain this decrease using Newton's second law.
As the child and sled come over the crest of the hill, they are moving in an arc. There must be a centripetal force, pointing inward toward the center of the arc. The combination of gravity (down) and the normal force (up) provides this centripetal force, which must be greater than or equal to zero. (At the top of the arc, Fy = mg - N = mv2/r ≥ 0.) The normal force must therefore be less than the child's weight.
A bucket of water can be whirled in a vertical circle without the water spilling out, even at the top of the circle when the bucket is upside down. Explain.
At the top of bucket's arc, the gravitational force and normal forces from the bucket provide the centripetal force needed to keep the water moving in a circle. (If we ignore the normal forces, mg = mv2/r, so the bucket must be moving with speed v ≥ gr or the water will spill out of the bucket.) At the top of the arc, the water has a horizontal velocity. As the bucket passes the top of the arc, the velocity of the water develops a vertical component. But the bucket is traveling with the water, with the same velocity, and contains the water as it falls through the rest of its path.
Why is the stopping distance of a truck much shorter than for a train going the same speed?
Because the train has a larger mass. If the stopping forces on the truck and train are equal, the (negative) acceleration of the train will be much smaller than that of the truck, since acceleration is inversely proportional to mass.
Astronauts who spend long periods in outer space could be adversely affected by weightlessness. One way to simulate gravity is to shape the spaceship like a cylindrical shell that rotates, with the astronauts walking on the inside surface (Fig. 5-30). Explain how this simulates gravity. Consider (a) how objects fall, (b) the force we feel on our feet, and (c) any other aspects of gravity you can think of.
For objects (including astronauts) on the inner surface of the cylinder, the normal force provides a centripetal force which points inward toward the center of the cylinder. This normal force simulates the normal force we feel when on the surface of Earth. (a) Falling objects are not in contact with the floor, so when released they will continue to move with constant velocity until the floor reaches them. From the frame of reference of the astronaut inside the cylinder, it will appear that the object falls in a curve, rather than straight down. (b) The magnitude of the normal force on the astronaut's feet will depend on the radius and speed of the cylinder. If these are such that v2/r = g (so that mv2/r = mg for all objects), then the normal force will feel just like it does on the surface of Earth. (c) Because of the large size of Earth compared to humans, we cannot tell any difference between the gravitational force at our heads and at our feet. In a rotating space colony, the difference in the simulated gravity at different distances from the axis of rotation would be significant.
For a drag force of the form F = -bv, what are the units of b?
If we solve for b, we have b = -F/v. The units for b are N·s/m = kg·m·s/(m·s2) = kg/s.
Why do bicycle riders lean in when rounding a curve at high speed?
Leaning in when rounding a curve on a bicycle puts the bicycle tire at an angle with respect to the ground. This increases the component of the (static) frictional force on the tire due to the road. This force component points inward toward the center of the curve, thereby increasing the centripetal force on the bicycle and making it easier to turn.
Sometimes it is said that water is removed from clothes in a spin dryer by centrifugal force throwing the water outward. Is this correct? Discuss.
No. The barrel of the dryer provides a centripetal force on the clothes to keep them moving in a circular path. A water droplet on the solid surface of the drum will also experience this centripetal force and move in a circle. However, as soon as the water droplet is at the location of a hole in the drum there will be no centripetal force on it and it will therefore continue moving in a path in the direction of its tangential velocity, which will take it out of the drum. There is no centrifugal force throwing the water outward; there is rather a lack of centripetal force to keep the water moving in a circular path.
Will the acceleration of a car be the same when a car travels around a sharp curve at a constant 60km/h as when it travels around a gentle curve at the same speed? Explain.
No. The centripetal acceleration depends on 1/r, so a sharp curve, with a smaller radius, will generate a larger centripetal acceleration than a gentle curve, with a larger radius. (Note that the centripetal force in this case is provided by the static frictional force between the car and the road.)
It is not easy to walk on an icy sidewalk without slipping. Even your gait looks different than on dry pavement. Describe what you need to do differently on the icy surface and why.
On an icy surface, you need to put your foot straight down onto the sidewalk, with no component of velocity parallel to the surface. If you can do that, the interaction between you and the ice is through the static frictional force. If your foot has a component of velocity parallel to the surface of the ice, any resistance to motion will be caused by the kinetic frictional force, which is much smaller. You will be much more likely to slip.
A girl is whirling a ball on a string around her head in a horizontal plane. She wants to let go at precisely the right time so that the ball will hit a target on the other side of the yard. When should she let go of the string?
She should let go of the string at the moment that the tangential velocity vector is directed exactly at the target.
A heavy crate rests on the bed of a flatbed truck. When the truck accelerates, the crate remains where it is on the truck, so it, too, accelerates. What force causes the crate to accelerate?
Static friction between the crate and the truck bed causes the crate to accelerate.
The game of tetherball is played with a ball tied to a pole with a string. When the ball is struck, it whirls around the pole as shown in Fig. 5-29. In what direction is the acceleration of the ball, and what causes the acceleration?
The acceleration of the ball is inward, directly toward the pole, and is provided by the horizontal component of the tension in the string.
Suppose two forces act on an object, one force proportional to v and the other proportional to v2. Which force domi nates at high speed?
The force proportional to v2 will dominate at high speed.
You are trying to push your stalled car. Although you apply a horizontal force of 400 N to the car, it doesn't budge, and neither do you. Which force(s) must also have a magnitude of 400 N: (a) the force exerted by the car on you; (b) the friction force exerted by the car on the road; (c) the normal force exerted by the road on you; (d) the friction force exerted by the road on you?
The forces in (a), (b), and (d) are all equal to 400 N in magnitude. (a) You exert a force of 400 N on the car; by Newton's third law the force exerted by the car on you also has a magnitude of 400 N. (b) Since the car doesn't move, the friction force exerted by the road on the car must equal 400 N, too. Then, by Newton's third law, the friction force exerted by the car on the road is also 400 N. (c) The normal force exerted by the road on you will be equal in magnitude to your weight (assuming you are standing vertically and have no vertical acceleration). This force is not required to be 400 N. (d) The car is exerting a 400 N horizontal force on you, and since you are not accelerating, the ground must be exerting an equal and opposite horizontal force. Therefore, the magnitude of the friction force exerted by the road on you is 400 N.
A block is given a push so that it slides up a ramp. After the block reaches its highest point, it slides back down, but the magnitude of its acceleration is less on the descent than on the ascent. Why?
The kinetic friction force is parallel to the ramp and the block's weight has a component parallel to the ramp. The parallel component of the block's weight is directed down the ramp whether the block is sliding up or down. However, the frictional force is always in the direction opposite the block's motion, so it will be down the ramp while the block is sliding up, but up the ramp while the block is sliding down. When the block is sliding up the ramp, the two forces acting on it parallel to the ramp are both acting in the same direction, and the magnitude of the net force is the sum of their magnitudes. But when the block is sliding down the ramp, the friction and the parallel component of the weight act in opposite directions, resulting in a smaller magnitude net force. A smaller net force yields a smaller (magnitude) acceleration.
Describe all the forces acting on a child riding a horse on a merry-go-round. Which of these forces provides the centripetal acceleration of the child?
The three main forces on the child are the downward force of gravity (weight), the normal force up on the child from the horse, and the static frictional force on the child from the surface of the horse. The frictional force provides the centripetal acceleration. If there are other forces, such as contact forces between the child's hands or legs and the horse, which have a radial component, they will contribute to the centripetal acceleration.
Cross-country skiers prefer their skis to have a large coeffi cient of static friction but a small coefficient of kinetic fric tion. Explain why. [Hint: Think of uphill and downhill.]
When a skier is in motion, a small coefficient of kinetic friction lets the skis move easily on the snow with minimum effort. A large coefficient of static friction lets the skier rest on a slope without slipping and keeps the skier from sliding backward when going uphill.
Why do airplanes bank when they turn? How would you compute the banking angle given the airspeed and radius of the turn? [Hint: Assume an aerodynamic "lift" force acts perpendicular to the wings.]
When an airplane is in level flight, the downward force of gravity is counteracted by the upward lift force, analogous to the upward normal force on a car driving on a level road. The lift on an airplane is perpendicular to the plane of the airplane's wings, so when the airplane banks, the lift vector has both vertical and horizontal components (similar to the vertical and horizontal components of the normal force on a car on a banked turn). The vertical component of the lift balances the weight and the horizontal component of the lift provides the centripetal force.
Technical reports often specify only the rpm for centrifuge experiments. Why is this inadequate?
When describing a centrifuge experiment, the force acting on the object in the centrifuge should be specified. Stating the rpm will let you calculate the speed of the object in the centrifuge. However, to find the force on an object, you will also need the distance from the axis of rotation.
When you must brake your car very quickly, why is it safer if the wheels don't lock? When driving on slick roads, why is it advisable to apply the brakes slowly?
When the wheels of a car are rolling without slipping, the force between each tire and the road is static friction, whereas when the wheels lock, the force is kinetic friction. The coefficient of static friction is greater than the coefficient of kinetic friction for a set of surfaces, so the force of friction between the tires and the road will be greater if the tires are rolling. Once the wheels lock, you also have no steering control over the car. It is better to apply the brakes slowly and use the friction between the brake mechanism and the wheel to stop the car while maintaining control. If the road is slick, the coefficients of friction between the road and the tires are reduced, and it is even more important to apply the brakes slowly to stay in control.