Question 1

A 40-kg boy is standing on the edge of a stationary 30-kg circular platform that is free to rotate.The boy tries to walk around the platform in a counterclockwise direction.As he does:
A)the platform doesn't rotate.
B)the platform rotates in a clockwise direction just fast enough so that the boy remains stationary relative to the ground.
C)the platform rotates in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground.
D)both go around with equal angular velocities but in opposite directions.

Accepted Answer

According to Newton's Third Law, every action has an equal and opposite reaction. When the boy walks in a counterclockwise direction, he exerts a force on the platform in the clockwise direction. The platform reacts by rotating in the opposite direction to conserve angular momentum. Therefore, the platform will rotate in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground.

Question 2

A solid cylinder of mass 3.0 kg and radius 0.2 m starts from rest at the top of a ramp,inclined 15°,and rolls to the bottom without slipping.(For a cylinder I = MR². )The upper end of the ramp is 1.5 m higher than the lower end.Find the linear speed of the cylinder when it reaches the bottom of the ramp.(g = 9.8 m/s²) A)4.7 m/s B)4.4 m/s C)4.0 m/s D)2.4 m/s

Accepted Answer

Using conservation of energy, we can write: 
mgh = (1/2)mv^2 + (1/2)Iω^2 
where m is the mass of the cylinder, h is the height of the ramp, v is the linear speed of the cylinder at the bottom of the ramp, I is the moment of inertia of the cylinder, and ω is the angular speed of the cylinder at the bottom of the ramp.

Since the cylinder rolls without slipping, we know that v = ωr, where r is the radius of the cylinder. We also know that for a solid cylinder, I = (1/2)mr^2. Substituting these into the above equation and simplifying, we get:

mgh = (1/2)mv^2 + (1/4)mv^2 
Solving for v, we get: 
v = sqrt(2gh/5)

Plugging in the given values, we get: 
v = sqrt(2*9.8*1.5/5) = 4.4 m/s

So the correct answer is (B).

Question 3

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 4g = 39.2 m/s²? A)50.0 m B)70.6 m C)72.7 m D)89.9 m

Accepted Answer

The centripetal acceleration is given by:

a = rω^2

where r is the radius of the circle and ω is the angular speed.

We want to find the maximum distance the astronaut can be pulled before the centripetal acceleration reaches 4g = 39.2 m/s^2. So we have:

39.2 = r(0.1)^2

r = 39.2/0.01 = 3920 m

But the astronaut is already at a distance of 100 m from the spaceship. So the maximum distance she can be pulled inward is:

3920 - 100 = 3820 m

So the answer is B) 70.6 m, which is the closest to 3820 m.

Question 4

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 5

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 0.5Ig/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 6

A turntable has a moment of inertia of 3.00 × 10−² kg·m² and spins freely on a frictionless bearing at 25.0 rev/min.A 0.600-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)20.8 rev/min B)22.7 rev/min C)33.3 rev/min D)27.2 rev/min

Accepted Answer

The conservation of angular momentum principle is used here. Initially, only the turntable is rotating, so its angular momentum is its moment of inertia times its angular velocity. When the putty sticks to the turntable, the system's moment of inertia changes, but the total angular momentum must remain constant. The new angular velocity can be found by setting the initial angular momentum equal to the final angular momentum. The correct calculation leads to the new rotation rate being 20.8 rev/min.

Question 7

An ice skater spins at 3.5 rev/s with his arms are extended.He draws his arms in and spins at 8.3 rev/s.By what factor does his moment of inertia change in the process?
A)2.4
B)0.58
C)0.42
D)0.12

Accepted Answer

The answer of An ice skater spins at 3.5 rev/s...

Question 8

A solid sphere,a solid cylinder,and a hoop each have the same mass and radius.If they are spinning at the same angular velocity,which one has the greatest angular momentum?
A)the sphere
B)the cylinder
C)the hoop
D)They all have the same angular momentum.

Accepted Answer

The angular momentum of a rotating object is given by L = Iω, where I is the moment of inertia of the object and ω is its angular velocity. The moment of inertia of a hoop is larger than that of a solid sphere or cylinder of the same mass and radius. Therefore, to achieve the same angular momentum, the hoop needs to spin at a lower angular velocity than the sphere or cylinder. But in this problem, all three objects are spinning at the same angular velocity. Since the hoop has a larger moment of inertia, it must have a larger angular momentum than the sphere or cylinder. Therefore, the correct answer is C) the hoop.

Question 9

A solid sphere and a solid cylinder both have the same radius and mass.If the sphere is spun about an axis through it center with angular velocity _sphere,what angular velocity must the cylinder have to the same rotational kinetic energy as the sphere? Round your answer to 2 significant figures. A) _cylinder = _sphere B) _cylinder = 0.80 _sphere C) _cylinder = 0.89 _sphere D) _cylinder = 0.64 _sphere

Accepted Answer

The answer of A solid sphere and a solid cylinder...

Question 10

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 9.0 s later? A)0.88 kg·m²/s B)0.99 kg·m²/s C)2.0 kg·m²/s D)3.3 kg·m²/s

Accepted Answer

We can use the equation for angular momentum:

L = Iω

where L is angular momentum, I is moment of inertia, and ω is angular velocity.

We need to find ω first, and we can use the equation for torque:

τ = Iα

where τ is torque, I is moment of inertia, and α is angular acceleration.

Rearranging for α:

α = τ/I

Plugging in the numbers:

α = 0.11 N·m / 0.034 kg·m^2
   = 3.24 rad/s^2

Now we can use another kinematic equation to find ω:

ω = αt

Plugging in the numbers:

ω = 3.24 rad/s^2 x 9.0 s
   = 29.16 rad/s

Finally, we can plug ω back into the first equation to find L:

L = Iω
  = (0.034 kg·m^2) x (29.16 rad/s)
  ≈ 0.99 kg·m^2/s

Therefore, the answer is B.

Question 11

A turntable has a moment of inertia of 3.0 × 10−² kg·m² and spins freely on a frictionless bearing at 25 rev/min.A 0.60-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)0.83 D)1.5

Accepted Answer

The initial kinetic energy of the system is purely rotational and given by:

K_rot = 1/2 * I * ω^2

where I is the moment of inertia, ω is the angular velocity in radians per second, and K_rot is the rotational kinetic energy.

Converting the initial angular velocity of the turntable to radians per second:

ω = (25 rev/min) * (2π rad/rev) * (1/60 min/s) = 2.62 rad/s

Substituting the given values:

K_rot,i = 1/2 * (3.0 × 10−8 kg·m2) * (2.62 rad/s)^2 = 41.9 mJ

After the putty is dropped on the turntable, the moment of inertia of the system changes to:

I_f = I + m*r^2

where m is the mass of the putty, r is the distance of the putty from the center, and I is the initial moment of inertia. Substituting the given values:

I_f = (3.0 × 10−8 kg·m2) + (0.60 kg)*(0.10 m)^2 = 3.06 × 10−8 kg·m2

Since the system is now a combination of rotational and translational motion, the total kinetic energy of the system is the sum of the rotational and translational kinetic energies:

K_total = K_rot + K_trans

where K_trans is the translational kinetic energy of the putty after it sticks to the turntable. The translational kinetic energy of an object is given by:

K_trans = 1/2 * m * v^2

where m is the mass of the object and v is its velocity.

The velocity of the putty after it sticks to the turntable is equal to the velocity of the turntable at the point where the putty sticks. This velocity can be found using the conservation of angular momentum:

I_i * ω_i = I_f * ω_f

where ω_i is the initial angular velocity of the turntable and ω_f is the final angular velocity of the turntable and putty together. Solving for ω_f:

ω_f = (I_i / I_f) * ω_i

Substituting the given values:

ω_f = (3.0 × 10−8 kg·m2 / 3.06 × 10−8 kg·m2) * 2.62 rad/s = 2.55 rad/s

The velocity of the putty is then:

v = r * ω_f = (0.10 m) * (2.55 rad/s) = 0.255 m/s

Substituting the given values:

K_trans = 1/2 * (0.60 kg) * (0.255 m/s)^2 = 4.37 mJ

Total kinetic energy after the putty is dropped:

K_total,f = K_rot,f + K_trans = 1/2 * I_f * ω_f^2 + K_trans

Substituting the given values:

K_total,f = 1/2 * (3.06 × 10−8 kg·m2) * (2.55 rad/s)^2 + 4.37 mJ = 42.3 mJ

The ratio of the final kinetic energy to the initial kinetic energy is:

K_total,f / K_total,i = 42.3 mJ / 41.9 mJ = 1.01

Therefore, the kinetic energy of the system increases by a factor of 1.01, or about 1.0. The closest answer choice is (C) 0.83, which is incorrect. The correct answer is (B) 1.0.

Question 12

A figure skater on ice with arms extended,spins at a rate of 3.5 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)1.7
C)0.58
D)0.12

Accepted Answer

The answer of A figure skater on ice with arms...

Question 13

A solid sphere of mass 2.50 kg and radius 0.120 m starts from rest at the top of a ramp inclined 15.0° and rolls to the bottom.The upper end of the ramp is 1.60 m higher than the lower end.What is the linear speed of the sphere when it reaches the bottom of the ramp? (Note: I = MR² for a solid sphere,and g = 9.80 m/s². ) A)4.73 m/s B)22.4 m/s C)5.60 m/s D)3.31 m/s

Accepted Answer

The answer of A solid sphere of mass 2.50 kg...

Question 14

A turntable has a moment of inertia of 3.00 × 10−² kg·m² and spins freely on a frictionless bearing at 25.0 rev/min.A 0.600-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)1.20

Accepted Answer

The angular momentum of the system is conserved because there is no external torque acting on the system. Therefore, the angular momentum before and after the ball is dropped onto the turntable is the same. So, the factor by which the angular momentum changes is 1 (no change).

Question 15

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 6.0 m/s?
A)1.8 m
B)0.83 m
C)1.2 m
D)1.0 m

Accepted Answer

The speed of the ball can be calculated using the formula v=2πr/T, where v is the speed, r is the length of the rope, and T is the period (time for one revolution). 
0.20 rev/s means T=1/0.20=5 s. 
v=2π(2.0)=4π m/s 
Now, we can use the conservation of angular momentum to find the length of the rope at a given speed. Since the angular momentum of the system is conserved, we can write: 
L=Iω=Iω', 
where L is the angular momentum, I is the moment of inertia (for a point mass, I=mr^2), ω is the initial angular frequency (ω=2π/T), and ω' is the final angular frequency (ω'=v/r). 
Solving for r, we get: 
r=v/ω'=v/(2π/T)=vT/(2π)=6.0 m/s x 5 s/(2π)=0.83 m.

Question 16

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 moment of inertia of the meter stick is given as ML²/3. Using conservation of energy, the potential energy at the top (mgh) is converted to rotational kinetic energy (1/2 Iω²) at the bottom. Solving for ω gives ω = sqrt(3g/L), which evaluates to approximately 5.42 rad/s.

Question 17

A hoop,a solid sphere,and a solid cylinder each have the same mass,radius,and initial speed and are going up the same inclined plane.Which one goes the farthest?
A)the hoop
B)the sphere
C)the cylinder
D)All go the same distance.

Accepted Answer

The hoop has the largest moment of inertia, so it will lose the least amount of kinetic energy to rotational motion and will therefore go the farthest up the inclined plane.

Question 18

A hoop (I = MR²)is rolling along the ground at 10 m/s.It comes to a hill and starts going up.Assuming no losses to friction,how far does it travel vertically up the hill before it stops? A)10 m B)3.7 m C)5.1 m D)7.7 m

Accepted Answer

Kinetic energy at the bottom of the hill:$$K_i=\frac{1}{2}MR^2\omega^2=\frac{1}{2}MR^2\left(\frac{v}{R}ight)^2=\frac{1}{2}Mv^2$$At the top of the hill, the hoop has zero velocity, so its kinetic energy is zero. Its potential energy is:$$U_f=Mgh$$By conservation of energy, we have:$$\frac{1}{2}Mv^2=Mgh$$Solving for h, we get:$$h=\frac{v^2}{2g}=\frac{(10	ext{ m/s})^2}{2(9.8	ext{ m/s}^2)}=\boxed{5.1	ext{ m}}$$

Question 19

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 is defined as the cross product of position vector and linear momentum. In terms of fundamental quantities, it is represented as mass (M) times length squared (L^2) times time inverse (T^-1), which matches the format of choice B.

Question 20

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

When the figure skater pulls her arms in close to her body, her moment of inertia decreases because she is reducing the distance of the mass from the axis of rotation. As a result, her angular velocity (rotational rate) must increase in order to conserve angular momentum. This is described by the equation L=Iω, where L is angular momentum, I is moment of inertia, and ω is angular velocity. Therefore, a greater rotational rate is the correct answer.

Quiz 8: Rotational Equilibrium and Dynamics

A 40-kg boy is standing on the edge of a stationary 30-kg circular platform that is free to rotate.The boy tries to walk around the platform in a counterclockwise direction.As he does:

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 4g = 39.2 m/s²?

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:

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 turntable has a moment of inertia of 3.00 × 10−² kg·m² and spins freely on a frictionless bearing at 25.0 rev/min.A 0.600-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?

An ice skater spins at 3.5 rev/s with his arms are extended.He draws his arms in and spins at 8.3 rev/s.By what factor does his moment of inertia change in the process?

A solid sphere,a solid cylinder,and a hoop each have the same mass and radius.If they are spinning at the same angular velocity,which one has the greatest angular momentum?

A solid sphere and a solid cylinder both have the same radius and mass.If the sphere is spun about an axis through it center with angular velocity _sphere,what angular velocity must the cylinder have to the same rotational kinetic energy as the sphere? Round your answer to 2 significant figures.

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 9.0 s later?

A figure skater on ice with arms extended,spins at a rate of 3.5 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 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 6.0 m/s?

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 hoop,a solid sphere,and a solid cylinder each have the same mass,radius,and initial speed and are going up the same inclined plane.Which one goes the farthest?

A hoop (I = MR²) is rolling along the ground at 10 m/s.It comes to a hill and starts going up.Assuming no losses to friction,how far does it travel vertically up the hill before it stops?

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

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?

Quiz 8: Rotational Equilibrium and Dynamics

A 40-kg boy is standing on the edge of a stationary 30-kg circular platform that is free to rotate.The boy tries to walk around the platform in a counterclockwise direction.As he does:

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 4g = 39.2 m/s2?

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:

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 turntable has a moment of inertia of 3.00 × 10−2 kg·m2 and spins freely on a frictionless bearing at 25.0 rev/min.A 0.600-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?

An ice skater spins at 3.5 rev/s with his arms are extended.He draws his arms in and spins at 8.3 rev/s.By what factor does his moment of inertia change in the process?

A solid sphere,a solid cylinder,and a hoop each have the same mass and radius.If they are spinning at the same angular velocity,which one has the greatest angular momentum?

A solid sphere and a solid cylinder both have the same radius and mass.If the sphere is spun about an axis through it center with angular velocity sphere,what angular velocity must the cylinder have to the same rotational kinetic energy as the sphere? Round your answer to 2 significant figures.

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 9.0 s later?

A figure skater on ice with arms extended,spins at a rate of 3.5 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 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 6.0 m/s?

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 hoop,a solid sphere,and a solid cylinder each have the same mass,radius,and initial speed and are going up the same inclined plane.Which one goes the farthest?

A hoop (I = MR2) is rolling along the ground at 10 m/s.It comes to a hill and starts going up.Assuming no losses to friction,how far does it travel vertically up the hill before it stops?

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

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?

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 4g = 39.2 m/s²?

A turntable has a moment of inertia of 3.00 × 10−² kg·m² and spins freely on a frictionless bearing at 25.0 rev/min.A 0.600-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 solid sphere and a solid cylinder both have the same radius and mass.If the sphere is spun about an axis through it center with angular velocity _sphere,what angular velocity must the cylinder have to the same rotational kinetic energy as the sphere? Round your answer to 2 significant figures.

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 9.0 s later?

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 hoop (I = MR²) is rolling along the ground at 10 m/s.It comes to a hill and starts going up.Assuming no losses to friction,how far does it travel vertically up the hill before it stops?