Rotational Energy and Momentum

Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m, rotate independently around a common axis through their centers. Two of the flywheels rotate in one direction at 8.94 rad/s, but the other one rotates in the opposite direction at 3.42 rad/s. Calculate the magnitude of the net angular momentum of the system. 940 kg • m2/s 1500 kg • m2/s 975 kg • m2/s 1020 kg • m2/s

1020 kg • m2/s

An ice skater has a moment of inertia of 5.0 kg • m2 when her arms are outstretched, and at this time she is spinning at 3.0 rev/s. If she pulls in her arms and decreases her moment of inertia to 2.0 kg • m2, how fast will she be spinning? 2.0 rev/s 10 rev/s 7.5 rev/s 3.3 rev/s

7.5 rev/s

A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.120 kgkg and length 80.0 cmcm . https://session.masteringphysics.com/problemAsset/1011182/15/1011182B.jpg With a skillful move, the majorette changes the rotation of her baton so that now it is spinning about an axis passing through its end at the same angular velocity 3.00 rad/srad/s as before. (Figure 2) What is the new magnitude of the angular momentum of the rod about a point where the axis of rotation intersects the end of the baton? Express your answer in kilogram meters squared per second.

7.68×10−2kg⋅m2/s

Two children, Ahmed and Jacques, ride on a merry-go-round. Ahmed is at a greater distance from the axis of rotation than Jacques. Which of the following are true statements? (There could be more than one correct choice.) Check all that apply. Jacques and Ahmed have the same angular speed. Jacques has a greater angular speed than Ahmed. Jacques has a smaller angular speed than Ahmed. Ahmed has a greater tangential speed than Jacques. Jacques and Ahmed have the same tangential speed.

Ahmed has a greater tangential speed than Jacques. Jacques and Ahmed have the same angular speed.

Suppose you are standing on the center of a merry-go-round that is at rest. You are holding a spinning bicycle wheel over your head so that its rotation axis is pointing upward. The wheel is rotating counterclockwise when observed from above. Suppose you now grab the edge of the wheel with your hand, stopping it from spinning. What happens to the merry-go-round? It remains at rest. It begins to rotate counterclockwise (as observed from above). It begins to rotate clockwise (as observed from above).

It begins to rotate counterclockwise (as observed from above).

When is the angular momentum of a system constant? Only when no net external force acts on the system. Only when its total kinetic energy is constant. Only when the moment of inertia is constant. Only when the linear momentum and the energy are constant. Only when no net external torque acts on the system.

Only when no net external torque acts on the system.

Learning Goal: To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular acceleration. Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant nonzero angular acceleration. The kinematic equations for such motion can be written as θ=θ0+ω0t+12αt2θ=θ0+ω0t+12αt2 and ω=ω0+αtω=ω0+αt. Here, the meaning of the symbols is as follows: θθtheta is the angular position of the particle at time ttt. θ0θ0theta_0 is the initial angular position of the particle. ωωomega is the angular velocity of the particle at time ttt. ω0ω0omega_0 is the initial angular velocity of the particle. ααalpha is the angular acceleration of the particle. True or false: The quantity represented by ω0ω0omega_0 is a function of time (i.e., is not constant).

false

A net torque applied to an object causes __________. the moment of inertia of the object to change the angular velocity of the object to change the object to rotate at a constant rate a linear acceleration of the object

the angular velocity of the object to change

Which factor does the torque on an object NOT depend on? the magnitude of the applied force the angle at which the force is applied the object's angular velocity

the object's angular velocity

Moment of inertia is the rotational equivalent of mass. the point at which all forces appear to act. the time at which inertia occurs. an alternative term for moment arm.

the rotational equivalent of mass.

The graph in the figure shows the x component F of the net force that acts for 10 s on a 100-kg crate. What is the change in the momentum of the crate during the 10 s that this force acts? https://session.masteringphysics.com/problemAsset/1901693/2/8514107006.jpg -100 kg • m/s 75 kg • m/s -75 kg • m/s 25 kg • m/s -25 kg • m/s

-75 kg • m/s

A 40.0-kg child running at 3.00 m/s suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. The child is traveling tangential to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 600 kg • m2 and very little friction at its rotation axis. What is the angular speed of the merry-go-round just after the child has jumped onto it? 0.788 rad/s 3.14 rev/s 6.28 rev/s 0.261 rad/s 2.00 rad/s

0.261 rad/s

A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg • m2 is rotating freely with an angular speed of 1.0 rad/s. Two people, each of mass 60 kg, are standing right outside the edge of the merry-go-round and suddenly step onto the edge with negligible speed relative to the ground. What is the angular speed of the merry-go-round right after the two people have stepped on? 0.67 rad/s 0.20 rad/s 0.60 rad/s 0.80 rad/s 0.40 rad/s

0.40 rad/s

A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.120 kgkg and length 80.0 cmcm . https://session.masteringphysics.com/problemAsset/1011182/15/1011182A.jpg Initially, the baton is spinning about an axis through its center at angular velocity 3.00 rad/srad/s . (Figure 1)What is the magnitude of its angular momentum about a point where the axis of rotation intersects the center of the baton? Express your answer in kilogram meters squared per second.

1.92×10−2 kg⋅m2/s

A wheel having a moment of inertia of 5.00 kg • m2 starts from rest and accelerates under a constant torque of 3.00 N • m for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s? 57.6 J 64.0 J 122 J 78.8 J

57.6 J

Consider a uniform hoop of radius R and mass M rolling without slipping. Which is larger, its translational kinetic energy or its rotational kinetic energy? Rotational kinetic energy is larger. Translational kinetic energy is larger. Both are equal. You need to know the speed of the hoop to tell.

Both are equal.

Two uniform solid balls, one of radius R and mass M, the other of radius 2R and mass 8M, roll down a high incline. They start together from rest at the top of the incline. Which one will reach the bottom of the incline first? The large sphere arrives first. The small sphere arrives first. Both reach the bottom at the same time.

Both reach the bottom at the same time.

The figure below shows two blocks suspended by a single cord over a pulley. (Figure 1) The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord's weight can be ignored. https://session.masteringphysics.com/problemAsset/1013838/23/1013838.png Which of the following statements correctly describes the system shown in the figure? Check all that apply. The acceleration of the blocks is zero. The net torque on the pulley is zero. The angular acceleration of the pulley is nonzero.

The angular acceleration of the pulley is nonzero.

The figure below shows two blocks suspended by a single cord over a pulley. (Figure 1) The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord's weight can be ignored. https://session.masteringphysics.com/problemAsset/1013838/23/1013838.png What happens when block B moves downward? The left cord pulls on the pulley with greater force than the right cord. The left and right cord pull with equal force on the pulley. The right cord pulls on the pulley with greater force than the left cord.

The right cord pulls on the pulley with greater force than the left cord.

A solid sphere and a solid cylinder, both uniform and of the same mass and radius, roll without slipping at the same forward speed. It is correct to say that the total kinetic energy of the solid sphere is equal to the total kinetic energy of the cylinder. less than the total kinetic energy of the cylinder. more than the total kinetic energy of the cylinder.

less than the total kinetic energy of the cylinder.

An object's angular momentum is proportional to its __________. linear momentum kinetic energy mass moment of inertia

moment of inertia

While spinning down from 500 rpm to rest, a flywheel does 3.9 kJ of work. This flywheel is in the shape of a solid uniform disk of radius 1.2 m. What is the mass of this flywheel? 4.0 kg 3.4 kg 5.2 kg 4.6 kg

4.0 kg

A light-weight potter's wheel, having a moment of inertia of 24 kg ∙ m2 , is spinning freely at 40.0 rpm. The potter drops a small but dense lump of clay onto the wheel, where it sticks a distance 1.2 m from the rotational axis. If the subsequent angular speed of the wheel and clay is 32 rpm, what is the mass of the clay? 4.6 kg 2.8 kg 3.7 kg 4.2 kg

4.2 kg

A little girl is going on the merry-go-round for the first time, and wants her 46 kgkg mother to stand next to her on the ride, 2.7 mm from the merry-go-round's center. If her mother's speed is 4.5 m/sm/s when the ride is in motion, what is her angular momentum around the center of the merry-go-round? Express your answer using two significant figures.

560 kg⋅m2/s

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg • m2. If he pulls in his arms so his moment of inertia decreases to 1.80 kg • m2, what will be his new angular speed? 2.25 rad/s 4.60 rad/s 0.81 rad/s 6.25 rad/s

6.25 rad/s

Learning Goal: To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular acceleration. Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant nonzero angular acceleration. The kinematic equations for such motion can be written as θ=θ0+ω0t+12αt2θ=θ0+ω0t+12αt2 and ω=ω0+αtω=ω0+αt. Here, the meaning of the symbols is as follows: θθtheta is the angular position of the particle at time ttt. θ0θ0theta_0 is the initial angular position of the particle. ωωomega is the angular velocity of the particle at time ttt. ω0ω0omega_0 is the initial angular velocity of the particle. ααalpha is the angular acceleration of the particle. A turntable is rotating at 3313rpm3313rpm. You then flip a switch, and the turntable speeds up, with constant angular acceleration, until it reaches 78rpm78rpm. Suppose you are asked to find the amount of time ttt, in seconds, it takes for the turntable to reach its final rotational speed. Which of the following equations could you use to directly solve for the numerical value of ttt? θ=θ0+ω0t+12αt2 ω=ω0+αt ω2=ω20+2α(θ−θ0) More information is needed before ttt can be found.

More information is needed before ttt can be found.

A disk and a hoop of the same mass and radius are released at the same time at the top of an inclined plane. If both are uniform, which one reaches the bottom of the incline first if there is no slipping? The disk The hoop Both reach the bottom at the same time.

The disk

Consider a solid uniform sphere of radius R and mass M rolling without slipping. Which form of its kinetic energy is larger, translational or rotational? Rotational kinetic energy is larger. Both are equal. Translational kinetic energy is larger. You need to know the speed of the sphere to tell.

Translational kinetic energy is larger.

Learning Goal: To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular acceleration. Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant nonzero angular acceleration. The kinematic equations for such motion can be written as θ=θ0+ω0t+12αt2θ=θ0+ω0t+12αt2 and ω=ω0+αtω=ω0+αt. Here, the meaning of the symbols is as follows: θθtheta is the angular position of the particle at time ttt. θ0θ0theta_0 is the initial angular position of the particle. ωωomega is the angular velocity of the particle at time ttt. ω0ω0omega_0 is the initial angular velocity of the particle. ααalpha is the angular acceleration of the particle. True or false: The quantity represented by θ0θ0theta_0 is a function of time (i.e., is not constant).

false

If an object is rotating clockwise, this corresponds to a _____ angular velocity. negative positive

negative

Suppose a uniform solid sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The linear velocity of the sphere at the bottom of the incline depends on the radius of the sphere. the mass of the sphere. both the mass and the radius of the sphere. neither the mass nor the radius of the sphere.

neither the mass nor the radius of the sphere.

The angular displacement of a rotating object is measured in _____. radians degrees degrees per second

radians

A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneously at the top of an inclined plane and do not slip, which one will reach the bottom first? pipe sphere cylinder The pipe and cylinder arrive together before the sphere. They all reach the bottom at the same time.

sphere

A disk, a hoop, and a solid sphere are released at the same time at the top of an inclined plane. They are all uniform and roll without slipping. In what order do they reach the bottom? hoop, sphere, disk sphere, disk, hoop hoop, disk, sphere disk, hoop, sphere

sphere, disk, hoop

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that the moment of inertia of the system decreases and the angular speed decreases. the moment of inertia of the system decreases and the angular speed increases. the moment of inertia of the system decreases and the angular speed remains the same. the moment of inertia of the system increases and the angular speed increases. the moment of inertia of the system increases and the angular speed decreases.

the moment of inertia of the system decreases and the angular speed increases.

A small uniform disk and a small uniform sphere are released simultaneously at the top of a high inclined plane, and they roll down without slipping. Which one will reach the bottom first? the sphere the one of greatest mass the one of smallest diameter the disk They will reach the bottom at the same time.

the sphere

Learning Goal: To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular acceleration. Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant nonzero angular acceleration. The kinematic equations for such motion can be written as θ=θ0+ω0t+12αt2θ=θ0+ω0t+12αt2 and ω=ω0+αtω=ω0+αt. Here, the meaning of the symbols is as follows: θθtheta is the angular position of the particle at time ttt. θ0θ0theta_0 is the initial angular position of the particle. ωωomega is the angular velocity of the particle at time ttt. ω0ω0omega_0 is the initial angular velocity of the particle. ααalpha is the angular acceleration of the particle. In the equation ω=ω0+αtω=ω0+αt, what does the time variable ttt represent? Choose the answer that is always true. Several of the statements may be true in a particular problem, but only one is always true. the moment in time at which the angular velocity equals ω0 the moment in time at which the angular velocity equals ω the time elapsed from when the angular velocity equals ω0 until the angular velocity equals ω

the time elapsed from when the angular velocity equals ω0ω0 until the angular velocity equals ωω

Learning Goal: To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular acceleration. Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant nonzero angular acceleration. The kinematic equations for such motion can be written as θ=θ0+ω0t+12αt2θ=θ0+ω0t+12αt2 and ω=ω0+αtω=ω0+αt. Here, the meaning of the symbols is as follows: θtheta is the angular position of the particle at time t. θ0 theta is the initial angular position of the particle. ω omega is the angular velocity of the particle at time t. ω0 omega is the initial angular velocity of the particle. α alpha is the angular acceleration of the particle. True or false: The quantity represented by θθtheta is a function of time (i.e., is not constant).

true

Learning Goal: To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular acceleration. Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant nonzero angular acceleration. The kinematic equations for such motion can be written as θ=θ0+ω0t+12αt2θ=θ0+ω0t+12αt2 and ω=ω0+αtω=ω0+αt. Here, the meaning of the symbols is as follows: θθtheta is the angular position of the particle at time ttt. θ0θ0theta_0 is the initial angular position of the particle. ωωomega is the angular velocity of the particle at time ttt. ω0ω0omega_0 is the initial angular velocity of the particle. ααalpha is the angular acceleration of the particle. True or false: The quantity represented by ωωomega is a function of time (i.e., is not constant).

true

A string is wrapped around a uniform solid cylinder of radius rrr, as shown in (Figure 1). The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass mmm. https://session.masteringphysics.com/problemAsset/1011172/22/MAD_ia_6.jpg Find the magnitude ααalpha of the angular acceleration of the cylinder as the block descends. Express your answer in terms of the cylinder's radius rrr and the magnitude of the free-fall acceleration ggg.

α=2/3(g/r)

Learning Goal: To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular acceleration. Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant nonzero angular acceleration. The kinematic equations for such motion can be written as θ=θ0+ω0t+12αt2θ=θ0+ω0t+12αt2 and ω=ω0+αtω=ω0+αt. Here, the meaning of the symbols is as follows: θθtheta is the angular position of the particle at time ttt. θ0θ0theta_0 is the initial angular position of the particle. ωωomega is the angular velocity of the particle at time ttt. ω0ω0omega_0 is the initial angular velocity of the particle. ααalpha is the angular acceleration of the particle. Which of the following equations is not an explicit function of time ttt, that is, does not involve ttt as a variable, and is therefore useful when you do not know or do not need the time? θ=θ0+ω0t+1/2αt2 ω=ω0+αtω=ω0+αt ω2=ω20+2α(θ−θ0)

ω2=ω20+2α(θ−θ0)

Learning Goal: To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular acceleration. Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant nonzero angular acceleration. The kinematic equations for such motion can be written as θ=θ0+ω0t+12αt2θ=θ0+ω0t+12αt2 and ω=ω0+αtω=ω0+αt. Here, the meaning of the symbols is as follows: θθtheta is the angular position of the particle at time ttt. θ0θ0theta_0 is the initial angular position of the particle. ωωomega is the angular velocity of the particle at time ttt. ω0ω0omega_0 is the initial angular velocity of the particle. ααalpha is the angular acceleration of the particle. You are now given an additional piece of information: It takes five complete revolutions for the turntable to speed up from 3313rpm3313rpm to 45rpm45rpm. Which of the following equations could you use to directly solve for the numerical value of the angular acceleration ααalpha? θ=θ0+ω0t+12αt2 ω=ω0+αt ω2=ω02+2α(θ−θ0) More information is needed before ααalpha can be found.

ω2=ω20+2α(θ−θ0)