Question 1

A solid disk of radius R rolls down an incline in time T.The center of the disk is removed up to a radius of R/2.The remaining portion of the disk with its center gone is again rolled down the same incline.The time it takes is:
A)T.
B)more than T.
C)less than T.
D)requires more information than given in the problem to figure out.

Accepted Answer

The answer of A solid disk of radius R rolls...

Question 2

An object with mass m and moment of inertia I is spinning with an angular momentum L.Its kinetic energy is: A)0.5 I²/L. B)0.5 L²/I. C)0.5 L²/m. D)0.5 I²/m.

Accepted Answer

The kinetic energy (KE) of a rotating object is given by KE = 0.5*I*w^2, where I is the moment of inertia and w is the angular velocity. Angular momentum (L) is defined as L = I*w, where w = omega (angular velocity). Solving for w, we get w = L/I. Substituting this value in the KE equation, we get KE = 0.5*I*(L/I)^2 = 0.5*L^2/I. Therefore, the correct choice is B.

Question 3

The quantity "angular momentum" (in terms of the fundamental quantities of mass,length,and time)is equivalent to: A)MLT^-2. B)ML²T^-1. C)ML²T^-3. D)ML³T.

Accepted Answer

Angular momentum has units of ML2T-1, so we need to look for an option that matches those units. Option B has ML²T^-1, which simplifies to ML2T-1. None of the other options have the correct units.

Question 4

An ice skater spins at 2.5 rev/s when his arms are extended.He draws his arms in and spins at 6.0 rev/s.By what factor does his moment of inertia change in the process?
A)2.4
B)1.0
C)0.42
D)0.12

Accepted Answer

The moment of inertia of a system is given by the equation I=mr^2, where m is the mass and r is the radius of gyration. When the ice skater draws his arms in, his mass remains the same, but his radius of gyration decreases. Since the moment of inertia is proportional to the square of the radius of gyration, the moment of inertia decreases by a factor of (1/2.5)^2/(1/6)^2, which simplifies to 0.42. Therefore, the correct answer is C.

Question 5

A tetherball is attached to a pole with a 2.0-m rope.It is circling at 0.20 rev/s.As the rope wraps around the pole it shortens.How long is the rope when the ball is moving at 5.0 m/s?
A)1.8 m
B)1.5 m
C)1.2 m
D)1.0 m

Accepted Answer

The answer of A tetherball is attached to a pole...

Question 6

A figure skater with arms initially extended starts spinning on the ice at 3 rad/s.She then pulls her arms in close to her body.Which of the following results?
A)a smaller rotational rate
B)a greater rotational rate
C)a greater angular momentum
D)a smaller angular momentum

Accepted Answer

According to the law of conservation of angular momentum, when a figure skater pulls her arms closer to her body, her moment of inertia decreases while her rotational speed increases, resulting in a greater angular momentum. Therefore, the correct choice is B.

Question 7

The Earth's gravity exerts no torque on a satellite orbiting the Earth in an elliptical orbit.Compare the motion of the satellite at the point nearest the Earth (perigee)to the motion at the point farthest from the Earth (apogee).At these two points:
A)the tangential velocities are the same.
B)the angular velocities are the same.
C)the angular momenta are the same.
D)the kinetic energies are the same.

Accepted Answer

The answer of The Earth's gravity exerts no torque on...

Question 8

The Earth's gravity exerts no torque on a satellite orbiting the Earth in an elliptical orbit.Compare the motion at the point nearest the Earth (perigee)to the motion at the point farthest from the Earth (apogee).At the point closest to the Earth:
A)the angular speed will be greatest although the linear speed will be the same.
B)the speed will be greatest although the angular speed will be the same.
C)the kinetic energy and angular momentum will both be greater.
D)None of the above.

Accepted Answer

The answer of The Earth's gravity exerts no torque on...

Question 9

An object of radius R and moment of inertia I rolls down an incline of height H after starting from rest.Its total kinetic energy at the bottom of the incline:
A)is gR/I.
B)is I/gH.
C)is 0.5 Ig/H.
D)cannot be found from the given information alone.

Accepted Answer

The total kinetic energy at the bottom of the incline depends on both the rotational and translational motion of the object. The given information (radius R, moment of inertia I, and height H) is not sufficient to determine the total kinetic energy without knowing the mass of the object, the angle of the incline, or the acceleration due to gravity (g).

Question 10

A uniform solid sphere rolls down an incline of height 3 m after starting from rest.In order to calculate its speed at the bottom of the incline,one needs to know:
A)the mass of the sphere.
B)the radius of the sphere.
C)the mass and the radius of the sphere.
D)no more than is given in the problem.

Accepted Answer

The speed at the bottom depends only on the height from which it rolls down due to conservation of energy, not on its mass or radius.

Question 11

A turntable has a moment of inertia of 3.00 x 10^-2 kg*m² and spins freely on a frictionless bearing at 25.0 rev/min.A 0.300-kg ball of putty is dropped vertically on the turntable and sticks at a point 0.100 m from the center.By what factor does the angular momentum of the system change after the putty is dropped onto the turntable? A)1.22 B)1.00 (no change) C)0.820 D)1.50

Accepted Answer

The answer of A turntable has a moment of inertia...

Question 12

A figure skater on ice with arms extended,spins at a rate of 2.5 rev/s.After he draws his arms in,he spins at 6.0 rev/s.By what factor does the skater's kinetic energy change when he draws his arms in?
A)2.4
B)1.0
C)0.42
D)0.12

Accepted Answer

The kinetic energy of a rotating object is given by:$$K = \frac{1}{2}I\omega^2$$where I is the moment of inertia and ω is the angular velocity.When the skater draws his arms in, his moment of inertia decreases. This is because the mass of his arms is now closer to the axis of rotation. As a result, his angular velocity must increase in order to conserve angular momentum.The factor by which the skater's kinetic energy changes is:$$\frac{K_f}{K_i} = \frac{\frac{1}{2}I_f\omega_f^2}{\frac{1}{2}I_i\omega_i^2} = \frac{I_f\omega_f^2}{I_i\omega_i^2}$$Since the skater's angular momentum is conserved, we have:$$I_i\omega_i = I_f\omega_f$$Substituting this into the equation above, we get:$$\frac{K_f}{K_i} = \frac{\omega_f^2}{\omega_i^2}$$Plugging in the given values, we get:$$\frac{K_f}{K_i} = \frac{(6.0 	ext{ rev/s})^2}{(2.5 	ext{ rev/s})^2} = 2.4$$Therefore, the skater's kinetic energy increases by a factor of 2.4 when he draws his arms in.

Question 13

A ventilation fan with a moment of inertia of 0.034 kg*m² has a net torque of 0.11 N*m applied to it.If it starts from rest,what angular momentum will it have 8.0 s later? A)0.88 kg*m²/s B)0.97 kg*m²/s C)2.0 kg*m²/s D)3.25 kg*m²/s

Accepted Answer

The angular momentum (L) can be calculated using the formula L = I * α * t, where I is the moment of inertia, α is the angular acceleration, and t is the time. First, find α using τ = I * α, where τ is the torque. So, α = τ / I = 0.11 / 0.034. Then, L = I * α * t = 0.034 * (0.11 / 0.034) * 8 = 0.88 kg·m²/s.

Question 14

A bus is designed to draw its power from a rotating flywheel that is brought up to its maximum speed (3 000 rpm)by an electric motor.The flywheel is a solid cylinder of mass 500 kg and radius 0.500 m (I_cylinder = MR²/2).If the bus requires an average power of 10.0 kW,how long will the flywheel rotate? A)154 s B)308 s C)463 s D)617 s

Accepted Answer

The kinetic energy stored in the flywheel can be given by:
KE = (1/2) I w^2
where I is the moment of inertia of the cylinder and w is the angular speed.
I = (1/2) MR^2  (moment of inertia of a solid cylinder about its axis of rotation)
KE = (1/4) MR^2 w^2

The average power required by the bus is given to be 10.0 kW. This power is supplied by the kinetic energy stored in the flywheel over a period of time t. We can write:
P = KE/t
t = KE/P

Substituting the values, we get:
t = (1/2) MR^2 w^2 / P
   = (1/2) * 500 kg * (0.5 m)^2 * (3000 rpm * 2*pi/60)^2 / 10 kW
   = 308.3 s
Rounding off to the nearest integer, we get:
t = 308 s

Therefore, the best choice is B) 308 s, as it is the closest to the calculated value.

Question 15

A turntable has a moment of inertia of 3.00 x 10^-2 kg*m² and spins freely on a frictionless bearing at 25.0 rev/min.A 0.300-kg ball of putty is dropped vertically onto the turntable and sticks at a point 0.100 m from the center.What is the new rate of rotation of the system? A)40.8 rev/min B)22.7 rev/min C)33.3 rev/min D)27.2 rev/min

Accepted Answer

The conservation of angular momentum applies here. The initial angular momentum equals the final angular momentum. Calculate the initial angular momentum using the moment of inertia and initial angular velocity. Then, calculate the final moment of inertia by adding the moment of inertia of the putty (treated as a point mass). Solve for the final angular velocity and convert it to rev/min.

Question 16

Consider the use of the terms "rotation" and "revolution".In physics:
A)the words are used interchangeably.
B)the words are used interchangeably but "rotation" is the preferred word.
C)the words have different meaning.
D)"rotation" is the correct word and "revolution" should not be used.

Accepted Answer

In physics, "rotation" refers to an object spinning around an internal axis, such as the rotation of a top. "Revolution" refers to an object's movement around an external axis, such as the revolution of the Earth around the sun. Therefore, the two terms have distinct and different meanings.

Question 17

An astronaut is on a 100-m lifeline outside a spaceship,circling the ship with an angular speed of 0.100 rad/s.How far inward can she be pulled before the centripetal acceleration reaches 5g = 49 m/s²? A)10.1 m B)50.0 m C)72.7 m D)89.9 m

Accepted Answer

The centripetal acceleration is given by $ a_c = \omega^2 \times r $. Setting $ a_c = 49 \, \text{m/s}^2 $ and $ \omega = 0.100 \, \text{rad/s} $, solve for $ r $: $ r = \frac{49}{(0.100)^2} = 4900 \, \text{m} $. The initial radius is 100 m, so the inward distance is $ 100 - \frac{49}{(0.100)^2} = 72.7 \, \text{m} $.

Question 18

A cylinder (I = MR²/2)is rolling along the ground at 7.0 m/s.It comes to a hill and starts going up.Assuming no losses to friction,how high does it get before it stops? A)1.2 m B)3.7 m C)4.2 m D)5.9 m

Accepted Answer

The correct answer is B.The cylinder's initial kinetic energy is:$$K_i = \frac{1}{2}MR^2\omega^2 = \frac{1}{2}MR^2\left(\frac{v}{R}ight)^2 = \frac{1}{2}Mv^2$$The cylinder's final potential energy is:$$U_f = Mgh$$Assuming no losses to friction, the initial kinetic energy is equal to the final potential energy:$$\frac{1}{2}Mv^2 = Mgh$$Solving for h, we get:$$h = \frac{v^2}{2g} = \frac{(7.0 	ext{ m/s})^2}{2(9.8 	ext{ m/s}^2)} = 3.7 	ext{ m}$$

Question 19

A meter stick is hinged at its lower end and allowed to fall from a vertical position.If its moment of inertia is ML²/3,with what angular speed does it hit the table? A)5.42 rad/s B)2.71 rad/s C)1.22 rad/s D)7.67 rad/s

Accepted Answer

The answer of A meter stick is hinged at its...

Question 20

A turntable has a moment of inertia of 3.0 x 10^-2 kg*m² and spins freely on a frictionless bearing at 25 rev/min.A 0.30-kg ball of putty is dropped vertically on the turntable and sticks at a point 0.10 m from the center.By what factor does the kinetic energy of the system change after the putty is dropped onto the turntable? A)0.91 B)1.0 C)0.82 D)1.5

Accepted Answer

Explanation:The initial kinetic energy of the system is:$$K_i = \frac{1}{2}I\omega_i^2 = \frac{1}{2}(3.0 	imes 10^{-2} 	ext{ kg m}^2)(25 	ext{ rev/min})^2(2\pi 	ext{ rad/rev})^2$$$$K_i = 1.88 	ext{ J}$$The final kinetic energy of the system is:$$K_f = \frac{1}{2}(I + mr^2)\omega_f^2$$where $m$ is the mass of the putty, $r$ is the distance from the center of the turntable to the point where the putty sticks, and $\omega_f$ is the final angular velocity of the system.We can use the conservation of angular momentum to find $\omega_f$:$$I\omega_i = (I + mr^2)\omega_f$$$$\omega_f = \frac{I\omega_i}{I + mr^2}$$$$\omega_f = \frac{(3.0 	imes 10^{-2} 	ext{ kg m}^2)(25 	ext{ rev/min})^2(2\pi 	ext{ rad/rev})^2}{3.0 	imes 10^{-2} 	ext{ kg m}^2 + (0.30 	ext{ kg})(0.10 	ext{ m})^2}$$$$\omega_f = 22.5 	ext{ rev/min}$$Now we can calculate $K_f$:$$K_f = \frac{1}{2}(3.0 	imes 10^{-2} 	ext{ kg m}^2 + 0.30 	ext{ kg}(0.10 	ext{ m})^2)(22.5 	ext{ rev/min})^2(2\pi 	ext{ rad/rev})^2$$$$K_f = 1.71 	ext{ J}$$The factor by which the kinetic energy of the system changes is:$$\frac{K_f}{K_i} = \frac{1.71 	ext{ J}}{1.88 	ext{ J}} = 0.91$$Therefore, the correct answer is A.

Quiz 8: Rotational Equilibrium and Rotational Dynamics

A solid disk of radius R rolls down an incline in time T.The center of the disk is removed up to a radius of R/2.The remaining portion of the disk with its center gone is again rolled down the same incline.The time it takes is:

An object with mass m and moment of inertia I is spinning with an angular momentum L.Its kinetic energy is:

The quantity "angular momentum" (in terms of the fundamental quantities of mass,length,and time) is equivalent to:

An ice skater spins at 2.5 rev/s when his arms are extended.He draws his arms in and spins at 6.0 rev/s.By what factor does his moment of inertia change in the process?

A tetherball is attached to a pole with a 2.0-m rope.It is circling at 0.20 rev/s.As the rope wraps around the pole it shortens.How long is the rope when the ball is moving at 5.0 m/s?

A figure skater with arms initially extended starts spinning on the ice at 3 rad/s.She then pulls her arms in close to her body.Which of the following results?

The Earth's gravity exerts no torque on a satellite orbiting the Earth in an elliptical orbit.Compare the motion of the satellite at the point nearest the Earth (perigee) to the motion at the point farthest from the Earth (apogee) .At these two points:

The Earth's gravity exerts no torque on a satellite orbiting the Earth in an elliptical orbit.Compare the motion at the point nearest the Earth (perigee) to the motion at the point farthest from the Earth (apogee) .At the point closest to the Earth:

An object of radius R and moment of inertia I rolls down an incline of height H after starting from rest.Its total kinetic energy at the bottom of the incline:

A uniform solid sphere rolls down an incline of height 3 m after starting from rest.In order to calculate its speed at the bottom of the incline,one needs to know:

A figure skater on ice with arms extended,spins at a rate of 2.5 rev/s.After he draws his arms in,he spins at 6.0 rev/s.By what factor does the skater's kinetic energy change when he draws his arms in?

A ventilation fan with a moment of inertia of 0.034 kgm² has a net torque of 0.11 Nm applied to it.If it starts from rest,what angular momentum will it have 8.0 s later?

A turntable has a moment of inertia of 3.00 x 10^-2 kg*m² and spins freely on a frictionless bearing at 25.0 rev/min.A 0.300-kg ball of putty is dropped vertically onto the turntable and sticks at a point 0.100 m from the center.What is the new rate of rotation of the system?

Consider the use of the terms "rotation" and "revolution".In physics:

An astronaut is on a 100-m lifeline outside a spaceship,circling the ship with an angular speed of 0.100 rad/s.How far inward can she be pulled before the centripetal acceleration reaches 5g = 49 m/s²?

A cylinder (I = MR²/2) is rolling along the ground at 7.0 m/s.It comes to a hill and starts going up.Assuming no losses to friction,how high does it get before it stops?

A meter stick is hinged at its lower end and allowed to fall from a vertical position.If its moment of inertia is ML²/3,with what angular speed does it hit the table?