Question 1

A potter's wheel with a moment of inertia of 8.0 kg-m² has 5.0 N-m applied to it. It starts from rest. 10. s later it has gained what kinetic energy? A) 44. J B) 78. J C) 1.6 kJ D) 0.12 kJ E) 53. J

Accepted Answer

The torque applied causes angular acceleration, which can be calculated using the formula: torque = moment of inertia * angular acceleration. The angular acceleration is 5.0 N-m / 8.0 kg-m² = 0.625 rad/s². The angular velocity after 10 seconds is 0.625 rad/s² * 10 s = 6.25 rad/s. The kinetic energy is (1/2) * moment of inertia * (angular velocity)² = 0.5 * 8.0 kg-m² * (6.25 rad/s)² = 78.125 J, approximately 78 J.

Question 2

A wheel of 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? A) 57.6 J B) 64.0 J C) 91.9 J D) 122.0 J E) 78.8 J

Accepted Answer

The rotational kinetic energy (KE_rot) of the wheel can be calculated using the formula KE_rot = 0.5 * I * ω^2, where I is the moment of inertia and ω is the angular velocity. First, we need to find the angular acceleration (α) using the torque (τ) and moment of inertia (I) with the formula τ = I * α, giving us α = τ / I = 3.00 N-m / 5.00 kg-m^2 = 0.60 rad/s^2. Then, we find the angular velocity (ω) after 8.00 s using ω = α * t = 0.60 rad/s^2 * 8.00 s = 4.80 rad/s. Finally, we calculate the rotational kinetic energy: KE_rot = 0.5 * 5.00 kg-m^2 * (4.80 rad/s)^2 = 57.6 J.

Question 3

The moment of inertia of a solid cylinder about its axis is given by 0.5MR². If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is A) 1:2 B) 2:1 C) 1:1 D) 1:3 E) 3:1

Accepted Answer

The answer of The moment of inertia of a solid...

Question 4

A hoop of radius 0.50 m and a mass of 0.20 kg is released from rest and allowed to roll down an inclined plane. How fast is it moving after dropping a vertical distance of 3.0 m?
A) 7.7 m/s
B) 3.8 m/s
C) 2.2 m/s
D) 6.2 m/s
E) 5.4 m/s

Accepted Answer

The answer of A hoop of radius 0.50 m and...

Question 5

An 8.0 kg-m² wheel with 155. Joules of rotational kinetic energy requires what torque to bring it to rest in 4.0 s? A) -1.2 N-m B) -1.8 N-m C) -1.0 N-m D) -1.6 N-m E) +0.88 N-m

Accepted Answer

The answer of An 8.0 kg-m² wheel with 155. Joules...

Question 6

A proton of mass M rotates with an angular speed of 2 × 10⁶ rad/s in a circle of radius 0.80 m in a cyclotron. What is the orbital angular momentum of the proton? (M = 1.67 × 10^-27 kg) A) 2.1 × 10^-2¹ N-m-s B) 5.3 × 10^-2¹ N-m-s C) 3.9 × 10^-2¹ N-m-s D) 2.9 × 10^-2¹ N-m-s E) 1.1 × 10^-2⁰ N-m-s

Accepted Answer

The orbital angular momentum of the proton is given by L = mvr, where m is the mass of the proton, v is its speed, and r is the radius of the circle. Substituting the given values, we get L = (1.67 × 10^-27 kg)(2 × 10^6 rad/s)(0.80 m) = 2.1 × 10^-21 N-m-s.

Question 7

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) Figure 8-9 The string has been slowly winding around a vertical rod, and a few seconds later the length of the string has shortened to 0.250 m. 
 
 What is the instantaneous speed of the mass at the moment the string reaches a length of 0.250 m?
A) 3.9 m/s
B) 9.3 m/s
C) 15 m/s
D) 225 m/s
E) 75 m/s

Accepted Answer

The answer of Initially, a 2.00-kg mass is whirling at...

Question 8

An object's angular momentum changes by 10 kg-m²/s in 2.0 s. What magnitude average torque acted on this object? A) 10 N-m B) 5 N-m C) 2.5 N-m D) 40 N-m E) 20 N-m

Accepted Answer

The magnitude of the average torque can be found using the formula $\tau = \frac{\Delta L}{\Delta t}$, where $\Delta L$ is the change in angular momentum and $\Delta t$ is the change in time. Given $\Delta L = 10 \, \text{kg-m}^2/\text{s}$ and $\Delta t = 2.0 \, \text{s}$, the average torque is $\tau = \frac{10}{2} = 5 \, \text{N-m}$.

Quiz 8: Rotational Motion and Equilibrium

A potter's wheel with a moment of inertia of 8.0 kg-m2 has 5.0 N-m applied to it. It starts from rest. 10. s later it has gained what kinetic energy?

A wheel of 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?

The moment of inertia of a solid cylinder about its axis is given by 0.5MR2. If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is

A hoop of radius 0.50 m and a mass of 0.20 kg is released from rest and allowed to roll down an inclined plane. How fast is it moving after dropping a vertical distance of 3.0 m?

An 8.0 kg-m2 wheel with 155. Joules of rotational kinetic energy requires what torque to bring it to rest in 4.0 s?

A proton of mass M rotates with an angular speed of 2 × 106 rad/s in a circle of radius 0.80 m in a cyclotron. What is the orbital angular momentum of the proton? (M = 1.67 × 10-27 kg)

An object's angular momentum changes by 10 kg-m2/s in 2.0 s. What magnitude average torque acted on this object?

A potter's wheel with a moment of inertia of 8.0 kg-m² has 5.0 N-m applied to it. It starts from rest. 10. s later it has gained what kinetic energy?

A wheel of 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?

The moment of inertia of a solid cylinder about its axis is given by 0.5MR². If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is

An 8.0 kg-m² wheel with 155. Joules of rotational kinetic energy requires what torque to bring it to rest in 4.0 s?

A proton of mass M rotates with an angular speed of 2 × 10⁶ rad/s in a circle of radius 0.80 m in a cyclotron. What is the orbital angular momentum of the proton? (M = 1.67 × 10^-27 kg)

An object's angular momentum changes by 10 kg-m²/s in 2.0 s. What magnitude average torque acted on this object?