Circular and Rotational Motion
Two children sit 2.60 m apart on a very low-mass horizontal seesaw with a movable fulcrum. The child on the left has a mass of 29.0 kg, and the child on the right has a mass of 38.0 kg. At what distance, as measured from the child on the left, must the fulcrum be placed in order for them to balance?
(29.0kg)(9.8)(r) = (38.0kg)(9.8)(2.6 - r) r = 1.47 m
You move a roller coaster with a loop-the-loop from the Earth to your new amusement park on the Moon. (a) How does the minumum speed to complete the loop-the-loop compare between the Earth and the Moon? (b) A roller coaster car starts from rest on a hill that preceeds the loop. On the Earth, if the car is released on the hill from a height h above the bottom of the loop, it will have the minimum speed required to get around the loop. Is the release height required to just get around the same loop on the Moon greater than, less than, or equal to h?
(a) It is less on the Moon (b) Equal
A thin, small hoop is rolling slowly along the ground. An ant is walking on the inside of the hoop and a beetle is walking on top of the rolling hoop, in the directions shown. They walk at such a speed that they maintain their positions at the bottom and top of the hoop, respectively. (a) Which insect has to walk at a higher speed? (b) According to a nearby spider sitting on the ground watching the spectacle, which insect is moving faster with respect to the Earth?
(a) They walk at the same speed (b) They move at the same speed
Use the simulation in the interactive problem in this section to answer the following questions. (a) Does changing the distance from the axis of rotation affect the angular acceleration? (b) Where should you push to create the maximum torque and angular acceleration?
(a) Yes (b) Farthest from the axis of rotation
(a) Can an object have an angular velocity if there is no net torque acting on it? (b) Can an object have a net torque acting on it if it has zero angular velocity?
(a) Yes (b) Yes
Use the simulation in the interactive problem in this section to answer the following questions. (a) Is the centripetal acceleration of the car higher when it is moving faster? (b) If the speed of the car remains constant, do the x and y components of the car's velocity change as the car goes around the track?
(a) Yes (b) Yes
A ladder leans against a wall making a 55.0° angle to the floor. The ladder is 4.50 m long, and weighs 415 N. The wall is frictionless and so is the floor. A horizontal wire is attached to the base of the ladder and attached to the wall. (a) What is the tension in the wire? (b) A person who weighs 655 N stands on a rung of the ladder located 2.00 m from its lower end. What is the new tension in the wire?
(a) mg+FGy=0 -FN+FGx=0 FGx = FN net torque =0 FN(4.5* sin55)- (415)(2.25cos55) = 0 FGx= (a) 145 N (a) 145 N b) Xcg=(2.25*415+2*655)/415+655 = 2.09696 FN(4.5* sin55)- (415+655)(2.09696cos55) = 0 b) FGx= 349 N
An object is moving at a constant speed around a circle. (a) In which of these cases does the magnitude of the centripetal acceleration of the object increase? (Assume all other factors are kept the same.) (b) In which case does the centripetal acceleration increase the most?
(a)The object's speed doubles The radius of the circle is halved (b) The object's speed doubles
Horses that move with the fastest linear speed on a merry-go-round are located
. near the outside.
A 1-kg rock is suspended from the tip of a meter stick at the 0-cm mark so that the meter stick balances like a seesaw when the fulcrum is at the 25-cm mark. From this information, what is the mass of the meter stick?
1 kg
A toy airplane connected by a guideline to the top of a flagpole flies in a circle at a constant speed. If the plane takes 4.5 s to complete one loop, and the radius of the circular path is 11 m, what is the magnitude of the plane's centripetal acceleration?
1- T=(2*pi*r)/v 2- Ac= (v)^2/r 21 m/s2
Consider the radius of the Earth to be 6.38×10^6 m. What is the magnitude of the centripetal acceleration experienced by a person (a) at the equator and (b) at the North Pole due to the Earth's rotation?
1- T=(2*pi*r)/v 2- Ac= (v)^2/r a) 0.0337 m/s2 (b) 0 m/s
A string's tension force supplies the centripetal force needed to keep a yo-yo whirling in a circle. (a) What force supplies the centripetal force keeping a satellite in uniform circular motion around the Earth? (b) What kinds of forces keep a roller coaster held to a looping track?
1. gravity 2. centrifical
On a balanced seesaw, a boy three times as heavy as his partner sits
1/3 the distance from the fulcrum.
What is the centripetal acceleration of a point on the perimeter of a bicycle wheel of diameter 70.0 cm when the bike is moving 8.00 m/s?
183 m/s2
Suppose the circumference of a bicycle wheel is 2 meters. If it rotates at 1 revolution per second when you are riding the bicycle, then your speed will be
2 m/s.
A straight 2.60 m rod pivots around a vertical axis so that it can swing freely in a horizontal plane. Jane pushes perpendicular to the rod at its midpoint with 98.0 N of force, directed horizontally, to create a torque. Matt has attached a rope to the end of the rod, and is pulling on it horizontally to create an opposing torque. The rope creates a 67.0° angle with the rod. With what force should Matt pull so that the net torque on the rod is zero?
2.6/2 * 98 = F*2.6*sin67 => F = 1.3*98/2.6*sin67 = 53.2 N
How many radians does a 0.300-m radius automobile tire rotate after starting from rest and accelerating at a constant angular acceleration of 2.00 rad/s2 over a 5.00-s interval?
25.0 rad
A dancer completes 3.4 revolutions in a pirouette. What is her angular displacement?
2•π•N =2•π•3.4=21.36 rad
What is the average angular velocity of the Earth around the sun? Assume a circular counterclockwise orbit, and 365 days in a year.
As the earth moves around the circular orbit one time, it moves an angle of 2 π radians. To determine the earth's average angular velocity in rad/s, divide this number by the time in seconds. t = 365 days * 24 hr/day * 60 min/hr * 60 s/min = 3.1536 * 10^7 seconds ω = 2 π ÷ 3.1536 * 10^7 The angular velocity is approximately 1.99 * 10^-7 rad/s.
A boy and a girl are riding on a merry-go-round which is turning at a constant rate. The boy is near the outer edge, and the girl is closer to the center. Who has the greater angular displacement?
Both have the same non-zero angular displacement.
A bee loaded with pollen flies in a circular path at a constant speed of 3.20 m/s. If the mass of the bee is 133 mg and the radius of its path is 11.0 m, what is the magnitude of the centripetal force?
Centripetal force = mv^2/r 133mg = 0.000133kg. 1.24e−4 N
An astronaut in training rides in a seat that is moved in uniform circular motion by a radial arm 5.00 meters long. If her speed is 15.0 m/s, what is the centripetal force on her in "G's," where one G equals her weight on the Earth?
Centripetal force is equal to (mass * velocity^2) / (radius). In this case, (centripetal force) = (mass * 15.0^2) / (5.00) = (mass) * (45.0 m/s^2), essentially Newton's second law, F = ma. One G is the force caused by the Earth's acceleration. Again, F=ma, so the force of one G is equal to (mass) * (9.81 m/s^2). Her own mass does not play a part in the number of Gs she experiences, since it is multiplied into both equations (her centripetal force and the equation for G), so it can be canceled out. The problem then is just to find how many Gs fit into her centripetal acceleration, or (45.0 m/s^2) / (9.81 m/s^2) = 4.59 Gs, which is the answer.
An object of fixed mass and rigid shape has one unique value for its moment of inertia. True or false? Explain your answer.
False
A horizontal log just barely spans a river, and its ends rest on opposite banks. The log is uniform and weighs 2400 N. If a person who weighs 840 N stands one fourth of the way across the log from the left end, how much weight does the bank under the right end of the log support?
First off, the weight of the log alone, is evenly distributed on both banks. (2400/2) = 1200N. each side. Then, the weight of the person is split unevenly, but related to the position on the log in terms of distances. The person is 3/4 the length from the right end, and 1/4 the length from the left end. The ratio is 3:1. Note, that means there's 4 parts of the log, 3 on one side of the person, 1 on the other. So what you do is divide the person's weight by 4. You can also see that most of the weight will be on the left end, as the person is closer to it. (840/4) = 210N. Add that to the 1200N. the log weight puts on the right bank, = 1410N = 1.4e3 N
7 The hammer throw is a track-and-field event, popular in Scotland, in which a ball on a rope (the "hammer") is whirled around the thrower in a circle before being released. The goal is the send the ball as a projectile as far down the field as possible. At a track meet, the circle in which a ball is whirled by a hammer thrower is at a 45° angle to the ground. To achieve the longest distance, at what point in its tilted circular orbit should the thrower release the ball?
Halfway as it moves from the lowest to the highest point
A 1.6 kg disk with radius 0.63 m is rotating freely at 55 rad/s around an axis perpendicular to its center. A second disk that is not rotating is dropped onto the first disk so that their centers align, and they stick together. The mass of the second disk is 0.45 kg and its radius is 0.38 m. What is the angular velocity of the two disks combined?
L = Iω, angular momentum is rotational inertia times angular velocity. Now if m is the mass and r is the radius, the rotational inertia of a disk is I = 0.5mr². So the rotational inertia of the 1.6 kg disk is I = 0.5 * 1.6 * 0.63² = 0.31752 kg m². And it's angular momentum is L = I ω = 0.31752 kg m² * 55 rad/s = 17.4636 kg m²/s. Angular momentum will be conserved, so the total angular momentum of the two disks combined will still be 17.4636 kg m²/s. But the rotational inertia will increase, so the angular velocity will decrease. But how will the rotational inertia increase? Well, first let's calculate the rotational inertia of the 0.45 kg disk. It's a disk, so it can still be calculated by the same method: I = 0.5 * 0.45 * 0.38² = 0.03249 kg m². And the total rotational inertia of the two disks is just the sum of the rotational inertias of each disk: I = 0.31752 kg m² + 0.03249 kg m² = 0.35001 kg m². We know the angular momentum of the two disks, we know the rotational inertia of the two discs, so we can find the angular velocity using the equation I gave at the start of my answer. Rearrange to ω = L/I = (17.4636 kg m²/s)/(0.35001 kg m²) = 49.9 rad/s.
Saturn travels at an average speed of 9.66×10^3 m/s around the Sun in a roughly circular orbit. Its distance from the Sun is 1.43×10^12 m. How long (in seconds) is a "year" on Saturn?
Period for uniform circular motion v=(2*pi*r)/T 9.66×10^3 m/s= (2*pi*1.43×10^12 m)/T T= 9.3*10^8s= 29.5 years
Jupiter's distance from the Sun is 7.78×10^11 meters and it takes 3.74×10^8 seconds to complete one revolution of the Sun in its roughly circular orbit. What is Jupiter's speed?
Period for uniform circular motion T=(2*pi*r)/v 1.31e4 m/s
The pulley shown in the illustration has a radius of 2.70 m and a moment of inertia of 39.0 kg∙m2. The hanging mass is 4.20 kg and it exerts a force tangent to the edge of the pulley. What is the angular acceleration of the pulley?
Radius: 2.70 m moment of inertia: 39.0 kg*m^2 Mass: 4.20 kg Tension = T ΣF = ma Torque = I*alpha mg - T = ma g - T/m = a (T)(r) = I(a/r) 1.60 rad/s2
Suppose you have four rods made of four different materials-aluminum, copper, steel and titanium. They all have the same dimensions, and the same force is applied to each. Order the rods from smallest to largest by how much each stretches.
Steel, Copper, Titanium, Aluminum
A ball with mass 0.48 kg moves at a constant speed. A centripetal force of 23 N acts on the ball, causing it to move in a circle with radius 1.7 m. What is the speed of the ball?
The equation for centripetal force is F = mv²/r, where F is the force, m the mass, v is the speed and r the radius of the circle. Rearranging the equation, v² = Fr/m = (23 N)(1.7 m)/(0.48 kg) = 81.458, and take the square root to get a speed of 9.0 m/s.
Bob and Ray push on a door from opposite sides. They both push perpendicular to the door. Bob pushes 0.63 m from the door hinge with a force of 89 N. Ray pushes 0.57 m from the door hinge with a force of 98 N, in a manner that tends to turn the door in a clockwise direction. What is the net torque on the door?
Torque = 89 * 0.63 = 56.07 Torque = 98 * 0.57 = 55.86 56.07 - 55.86 = 0.21 N∙m
A 1.1 kg birdfeeder hangs from a horizontal tree branch. The birdfeeder is attached to the branch at a point that is 1.1 m from the trunk. What is the amount of torque exerted by the birdfeeder on the branch? The origin is at the pivot point, where the branch attaches to the trunk.
Torque = Force x arm length (perpendicular to Force) weight of bird feeder = 1.1(9.81) = 10.8 N arm length = 1.1 m Torque of feeder on branch = (10.8)(1.1) = 11.9 N-m ANS
Your bicycle tires have a radius of 0.33 m. It takes you 850 seconds to ride 14 times counterclockwise around a circular track of radius 73 m at constant speed. (a) What is the angular velocity of the bicycle around the track? (b) What is the magnitude of the angular velocity of a tire around its axis? (That is, don't worry about whether the tire's rotation is clockwise or counterclockwise.)
Use the following equation to determine angular velocity. ω = θ ÷ t To determine the time for the bicycle to move around the track, divide 850 by 14. t = 850 ÷ 14 This is approximately 60.7 seconds. As the bicycle moves around the track, it rotates an angle of 2 π radians. To determine its angular velocity, use the equation above. ω = 2 π ÷ (850 ÷ 14) This is approximately 0.103 rad/s. Let's determine the circumference of the track. C = 2 * π * 73 = 146 * π C = 2 * π * 0.33 = 0.66 * π To determine the number of times the tire rotates as the bicycle moves around the track, divide these two numbers. N = 146 * π ÷ 0.66 * π This is approximately 221. Total angle = 2 π * 146/0.66 = 292/0.66 * π ω = 292/0.66 * π ÷ (850 ÷ 14) This is approximately 22.9 rad/s.
A woman with weight 637 N lies on a bed of nails. The bed has a weight of 735 N and a length of 1.72 m. The bed is held up by two supports, one at the head and one at the foot. Underneath each support is a scale. When the woman lies in the bed, the scale at the foot reads 712 N. How far is the center of gravity of the system from the foot of the bed of nails?
Xcg = (w1 x1 + w2 x2+ w3 x3)/w1 + w2 + w3 0.827 m
A skateboarder stands on a skateboard so that 62% of her weight is located on the front wheels. If the distance between the wheels is 0.85 m, how far is her center of gravity from the front wheels? Ignore the weight of this rather long skateboard.
Xcg = (w1 x1 + w2 x2+ w3 x3)/w1 + w2 + w3 Xcg = (62*0.85)/ 38+62 = 0.527m 0.85 m-0.527m = 0.32 m
A weightlifter has been given a barbell to lift. One end has a mass of 5.5 kg while the other end has a mass of 4.7 kg. The bar is 0.20 m long. (Consider the bar to be massless, and assume that the masses are thin disks, so that their centers of mass are at the ends of the bar.) How far from the heavier end should she hold the bar so that the weight feels balanced?
Xcg = (w1 x1 + w2 x2+ w3 x3)/w1 + w2 + w3 xcg = 0.108 How far from the heavier end should she hold 0.20 - 0.108 = 0.092 m 9.2e−2 m
Two beads are tied to a string at different positions, and you swing the string around your head at a constant rate so that the beads move in uniform circular motion. Bead A is closer to your hand than bead B. Compare (a) the periods of A and B; (b) the speeds of A and B;
a) A and B have the same period b)B has greater speed
The solar system moves around the galactic center in a roughly circular path at a radius of about 2.6×10^20 m. The system's orbital speed around the center is 2.2×10^5 m/s. (a) What is the period of this circular motion in years? (b) What is the sun's centripetal acceleration? (c) What is the magnitude of the centripetal force keeping the sun in this path given that the sun's mass is 1.99×10^30 kg? a) b) c)
a) A roughly circular path, radius = 2.6e20m has a length of 2π*2.6e20 = 16.34e20m The orbital period in years is simply the number of years it takes to traverse the length of the orbit at a speed of 2.2e5 m/s. T = 16.24e20/2.2e5 s = 7.43e15 s. There are 31.56 megaseconds in a year, so T = 7.43e15/31.56e6 = 2.354e8 years b) The centripetal acceleration = v^2/r = 4.84e10m^2/s^2 / 2.6e20 = 1.86e-10m/s^2 c) F = ma = 1.99e30 * 1.86e-10 = 3.70e20N
The platter of a modern hard disk drive spins at 7.20×10^3 rpm (revolutions per minute). (a) How much time, in seconds, does it take for the disk to make a complete revolution? (b) Starting from rest, suppose the disk reaches full speed in 5.00 seconds. What is the average angular acceleration of the disk in radians per second? (c) Assuming constant angular acceleration, how many revolutions has the hard disk turned while spinning up to its final angular velocity?
a) t = 1/7200 min = 60/7200 sec = .00833 sec b) α = dw/dt = (7200*2π/60)/5.00 = 150.8 rad/sec² (not rad/sec...) c) Θ = ½α*t² = ½*150.8*5.00² = 1885 radians = 300 revolutions
What type of acceleration does an object moving with constant speed in a circular path experience?
centripetal acceleration
A runner rounds a circular curve of radius 24.0 m at a constant speed of 5.25 m/s. What is the magnitude of the runner's centripetal acceleration?
centripetal acceleration: Ac= (v)^2/r =(5.25 m/s)^2/24m =1.15m/s^2
Your pet hamster sits on a record player whose angular speed is constant. If he moves to a point twice as far from the center, then his linear speed
doubles.
The blades of a kitchen blender rotate counterclockwise at 2.2×104 rpm (revolutions per minute) at top speed. It takes the blender 2.1 seconds to reach this top speed after being turned on. What is the average angular acceleration of the blades?
max rotational speed of blender = 2.2e4 rpm = (2.2/60)e4 rps = (2.2/60)(2π)e4 rad/s min rotational speed of blender = 0 change in speed of blender = 0.230e4 rad/s time to make this change = 2.1 s avg rate of angular change = 0.230e4/2.1 = 0.110e4 rad/s² ANS
Horses that move with the fastest linear speed on a merry-go-round are located
near the outside.
Does an object moving in uniform circular motion have constant centripetal acceleration?
no
Mars travels at an average speed of 2.41×10^4 m/s around the Sun, and takes 5.94×10^7 s to complete one revolution. How far is Mars from the Sun?
r = v*T / 2pi Plug v = 2.41 x 10^4 m/s, T = 5.94 x 10^7 s and pi = 3.14 We get r = 2.28e11 m
In a carnival ride, passengers are rotated at a constant speed in a seat at the end of a long horizontal arm. The arm is 8.30 m long, and the period of rotation is 4.00 s. (a) What is the magnitude of the centripetal acceleration experienced by a rider? (b) State the acceleration in "gee's," that is, as a multiple of the gravitational acceleration constant g.
radius = 8.3 m circ = 8.3 x 2pi = 52.15 m 52.15/4 = 13.04 m/s tangential velocity = v=(2*pi*r)/T Acc = v^2/r = 13.04^2/8.3 = 20.49 m/s^2 g = 9.8 m/s^2 20.49/9.8 = 2.09 gee's
Rotational Kinematics - Angular Displacement? The radius of the tires on your car is 0.33 m. You drive 1600 m in a straight line. What is the angular displacement of a point on the outer rim of a tire in rad, around the center of the tire, during this trip? Assume the tire rotates in the counterclockwise direction.
the circumference of the wheel is 2pir ie 2*3.14*0.33 m The total distance covered = 1600 m Hence the number of rotations of the wheel = 1600/2*3.14*0.33 = 772.1 For one rotation the angular displacement = 2pi radian. So for 772.1 rotations, the angular displacement = 2*3.14*772.1 = 4848.8 radian (approx)
You want to exert a torque of at least 35.0 N·m on a wrench whose handle is 0.150 m long. If you can provide a force of 355 N to the end of the wrench, what is the minimum angle at which you can apply the force in order to achieve the desired torque?
torque = r * F * sin(angle) 35 = .150 * 355 * sin(angle) 35/(.150*355) = sin(angle) sin-1(35/(.150*355) = angle by the way, that sin-1 means the inverse sine. Angle is 41.1 degree.
When tires are installed or reinstalled on a car, they are usually first balanced on a device that spins them to see if they wobble. A tire with a radius of 0.380 m is rotated on a tire balancing device at exactly 460 revolutions per minute. A small stone is embedded in the tread of the tire. What is the magnitude of the centripetal acceleration experienced by the stone?
v = 2πr/T v = 2πr * f v = 2π * 0.38 * 460 / 60 ≈ 18.3 m/s So.. a = v²/r = 18.3² / 0.38 ≈ 881 m/s²
Two brothers, Jimmy and Robbie, sit 3.00 m apart on a horizontal seesaw with its fulcrum exactly midway between them. Jimmy sits on the left side, and his mass is 42.5 kg. Robbie's mass is 36.5 kg. Their sister Betty sits at the exact point on the seesaw so that the entire system is balanced. If Betty is 29.8 kg, at what location should she sit? Take the fulcrum to be the origin, and right to be positive. Assume that the mass of the seesaw is negligible.
τ = r*F tl= mg*r = 42 kg*9.8*(1.5m) = 624.75 N.m tr= mg*r = 36.5 kg*9.8*(1.5m) = 536.55 N.m 624.75-536.55=88.2 Nm tb=mg*r 88.2/29.8*9.8 = rb rb= 0.302 m
A 3.30 kg birdfeeder hangs from the tip of a 1.60 m pole that sticks up from the ground at a 65.0° angle. What is the magnitude of the torque exerted on the pole by the birdfeeder? Treat the bottom end of the pole as the pivot point.
τ = r*F *cos θ (ojo) use the horizontal component ot the pole's length ( r cos(angle) ) and the weight ( m g ) to find the torque T = (m g) (r cos(angle)) = 3.30 (9.81) (1.60cos65) = 21.9 M*N
A simple pendulum consists of a 0.500 kg "bob" (a small but massive object) at the end of a light rod, which is allowed to swing back and forth from a pivot point. The mass of the rod is negligible compared to the mass of the bob. The bob on the pendulum is positioned so the rod makes a 12.0° angle with the vertical. The rod is 2.70 m long. What is the magnitude of the torque exerted on the rod by the weight of the bob?
τ = r*F *sin θ 2.75 N∙m
The wheel on a car is held in place by four nuts. Each nut should be tightened to 94.0 N∙m of torque to be secure. If you have a wrench with a handle that is 0.250 m long, what minimum force do you need to exert perpendicular to the end of the wrench to tighten a nut correctly?
τ = rF 376 N
