Question
rotational motion problems
Answers
GPT-4o mini
Certainly! Rotational motion problems involve the concepts of angular displacement, angular velocity, angular acceleration, torque, moment of inertia, and other rotational dynamics. Below are some problems and their solutions:
### Problem 1: Angular Displacement
A solid disk with a radius of 0.5 meters rolls without slipping along a straight path. If it completes 10 full rotations, what is the distance traveled by the disk?
**Solution:**
The distance traveled \( d \) can be found using the formula:
\[
d = \text{number of rotations} \times \text{circumference of the disk}
\]
The circumference \( C \) of a circle is given by:
\[
C = 2\pi r
\]
For a radius \( r = 0.5 \) m:
\[
C = 2\pi(0.5) = \pi \text{ m}
\]
Thus:
\[
d = 10 \times \pi \approx 31.42 \text{ m}
\]
---
### Problem 2: Angular Velocity
A wheel spins with an angular velocity of 12 rad/s. If it accelerates at a rate of 3 rad/s² for 4 seconds, what will be its final angular velocity?
**Solution:**
The final angular velocity \( \omega_f \) can be found using the equation:
\[
\omega_f = \omega_i + \alpha t
\]
Where:
- \( \omega_i = 12 \text{ rad/s} \) (initial angular velocity)
- \( \alpha = 3 \text{ rad/s}^2 \) (angular acceleration)
- \( t = 4 \text{ s} \)
Substituting the values:
\[
\omega_f = 12 + (3 \times 4) = 12 + 12 = 24 \text{ rad/s}
\]
---
### Problem 3: Torque and Moment of Inertia
A torque of 20 N·m is applied to a wheel with a moment of inertia of 5 kg·m². What is the angular acceleration of the wheel?
**Solution:**
The angular acceleration \( \alpha \) can be calculated using Newton's second law for rotation:
\[
\tau = I \alpha
\]
Where:
- \( \tau = 20 \text{ N·m} \) (torque)
- \( I = 5 \text{ kg·m}^2 \) (moment of inertia)
Rearranging to solve for \( \alpha \):
\[
\alpha = \frac{\tau}{I} = \frac{20}{5} = 4 \text{ rad/s}^2
\]
---
### Problem 4: Kinetic Energy of a Rotating Object
A cylinder of radius 0.1 m and mass 2 kg rolls without slipping. Calculate its rotational kinetic energy when it is spinning at an angular velocity of 10 rad/s.
**Solution:**
Rotational kinetic energy \( KE_{\text{rot}} \) is given by the formula:
\[
KE_{\text{rot}} = \frac{1}{2} I \omega^2
\]
First, we need to find the moment of inertia \( I \) for a solid cylinder:
\[
I = \frac{1}{2} m r^2
\]
Substituting the values:
\[
I = \frac{1}{2} \cdot 2 \cdot (0.1)^2 = \frac{1}{2} \cdot 2 \cdot 0.01 = 0.01 \text{ kg·m}^2
\]
Now substituting \( I \) and \( \omega \):
\[
KE_{\text{rot}} = \frac{1}{2} \cdot 0.01 \cdot (10)^2 = 0.005 \cdot 100 = 0.5 \text{ J}
\]
---
### Problem 5: Rolling Motion
A sphere is rolling down a ramp without slipping. If the sphere has a radius of 0.2 m and mass of 3 kg, calculate its translational speed after descending a height of 1 m.
**Solution:**
Using conservation of energy:
\[
PE = KE_{\text{trans}} + KE_{\text{rot}}
\]
Potential energy at height \( h \):
\[
PE = mgh = 3 \cdot 9.81 \cdot 1 = 29.43 \text{ J}
\]
For a solid sphere, moment of inertia \( I \) is:
\[
I = \frac{2}{5} m r^2 = \frac{2}{5} \cdot 3 \cdot (0.2)^2 = \frac{2}{5} \cdot 3 \cdot 0.04 = 0.048 \text{ kg·m}^2
\]
Using \( v \) for translational speed:
\[
KE_{\text{trans}} = \frac{1}{2} mv^2
\]
And again for \( KE_{\text{rot}} \):
\[
KE_{\text{rot}} = \frac{1}{2} I \omega^2
\]
Since \( v = r\omega \), we have:
\[
KE_{\text{rot}} = \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} \cdot 0.048 \cdot \left(\frac{v}{0.2}\right)^2
\]
Substituting \( \omega = \frac{v}{r} \) into the energy equation:
\[
29.43 = \frac{1}{2} (3) v^2 + \frac{1}{2} (0.048) \left(\frac{v}{0.2}\right)^2
\]
Calculating:
\[
29.43 = \frac{1}{2} (3) v^2 + \frac{1}{2} (0.048)(25v^2) = 1.5 v^2 + 0.6 v^2 = 2.1 v^2
\]
\[
v^2 = \frac{29.43}{2.1} \approx 14.019
\]
\[
v \approx \sqrt{14.019} \approx 3.74 \text{ m/s}
\]
---
Feel free to ask for more problems, clarifications, or additional concepts related to rotational motion!
### Problem 1: Angular Displacement
A solid disk with a radius of 0.5 meters rolls without slipping along a straight path. If it completes 10 full rotations, what is the distance traveled by the disk?
**Solution:**
The distance traveled \( d \) can be found using the formula:
\[
d = \text{number of rotations} \times \text{circumference of the disk}
\]
The circumference \( C \) of a circle is given by:
\[
C = 2\pi r
\]
For a radius \( r = 0.5 \) m:
\[
C = 2\pi(0.5) = \pi \text{ m}
\]
Thus:
\[
d = 10 \times \pi \approx 31.42 \text{ m}
\]
---
### Problem 2: Angular Velocity
A wheel spins with an angular velocity of 12 rad/s. If it accelerates at a rate of 3 rad/s² for 4 seconds, what will be its final angular velocity?
**Solution:**
The final angular velocity \( \omega_f \) can be found using the equation:
\[
\omega_f = \omega_i + \alpha t
\]
Where:
- \( \omega_i = 12 \text{ rad/s} \) (initial angular velocity)
- \( \alpha = 3 \text{ rad/s}^2 \) (angular acceleration)
- \( t = 4 \text{ s} \)
Substituting the values:
\[
\omega_f = 12 + (3 \times 4) = 12 + 12 = 24 \text{ rad/s}
\]
---
### Problem 3: Torque and Moment of Inertia
A torque of 20 N·m is applied to a wheel with a moment of inertia of 5 kg·m². What is the angular acceleration of the wheel?
**Solution:**
The angular acceleration \( \alpha \) can be calculated using Newton's second law for rotation:
\[
\tau = I \alpha
\]
Where:
- \( \tau = 20 \text{ N·m} \) (torque)
- \( I = 5 \text{ kg·m}^2 \) (moment of inertia)
Rearranging to solve for \( \alpha \):
\[
\alpha = \frac{\tau}{I} = \frac{20}{5} = 4 \text{ rad/s}^2
\]
---
### Problem 4: Kinetic Energy of a Rotating Object
A cylinder of radius 0.1 m and mass 2 kg rolls without slipping. Calculate its rotational kinetic energy when it is spinning at an angular velocity of 10 rad/s.
**Solution:**
Rotational kinetic energy \( KE_{\text{rot}} \) is given by the formula:
\[
KE_{\text{rot}} = \frac{1}{2} I \omega^2
\]
First, we need to find the moment of inertia \( I \) for a solid cylinder:
\[
I = \frac{1}{2} m r^2
\]
Substituting the values:
\[
I = \frac{1}{2} \cdot 2 \cdot (0.1)^2 = \frac{1}{2} \cdot 2 \cdot 0.01 = 0.01 \text{ kg·m}^2
\]
Now substituting \( I \) and \( \omega \):
\[
KE_{\text{rot}} = \frac{1}{2} \cdot 0.01 \cdot (10)^2 = 0.005 \cdot 100 = 0.5 \text{ J}
\]
---
### Problem 5: Rolling Motion
A sphere is rolling down a ramp without slipping. If the sphere has a radius of 0.2 m and mass of 3 kg, calculate its translational speed after descending a height of 1 m.
**Solution:**
Using conservation of energy:
\[
PE = KE_{\text{trans}} + KE_{\text{rot}}
\]
Potential energy at height \( h \):
\[
PE = mgh = 3 \cdot 9.81 \cdot 1 = 29.43 \text{ J}
\]
For a solid sphere, moment of inertia \( I \) is:
\[
I = \frac{2}{5} m r^2 = \frac{2}{5} \cdot 3 \cdot (0.2)^2 = \frac{2}{5} \cdot 3 \cdot 0.04 = 0.048 \text{ kg·m}^2
\]
Using \( v \) for translational speed:
\[
KE_{\text{trans}} = \frac{1}{2} mv^2
\]
And again for \( KE_{\text{rot}} \):
\[
KE_{\text{rot}} = \frac{1}{2} I \omega^2
\]
Since \( v = r\omega \), we have:
\[
KE_{\text{rot}} = \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} \cdot 0.048 \cdot \left(\frac{v}{0.2}\right)^2
\]
Substituting \( \omega = \frac{v}{r} \) into the energy equation:
\[
29.43 = \frac{1}{2} (3) v^2 + \frac{1}{2} (0.048) \left(\frac{v}{0.2}\right)^2
\]
Calculating:
\[
29.43 = \frac{1}{2} (3) v^2 + \frac{1}{2} (0.048)(25v^2) = 1.5 v^2 + 0.6 v^2 = 2.1 v^2
\]
\[
v^2 = \frac{29.43}{2.1} \approx 14.019
\]
\[
v \approx \sqrt{14.019} \approx 3.74 \text{ m/s}
\]
---
Feel free to ask for more problems, clarifications, or additional concepts related to rotational motion!