A bowling ball [I= 2mr/5] rolls down an inclined plane of length L and angle 0. The friction between the plane and the bowling ball is sufficient for the ball to roll without slipping. Find the bowling ball's (linear) speed at the bottom.
A bowling ball [I= 2mr/5] rolls down an inclined plane of length L and angle 0. The friction between the plane and the bowling ball is sufficient for the ball to roll without slipping. Find the bowling ball's (linear) speed at the bottom.
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![**Problem:**
A bowling ball \([I_{\text{cm}} = 2mr^2/5]\) rolls down an inclined plane of length \(L\) and angle \(\theta\). The friction between the plane and the bowling ball is sufficient for the ball to roll without slipping. Find the bowling ball's (linear) speed at the bottom.
**Solution Description:**
To solve this problem, we need to apply principles of energy conservation and rotational motion. Since the ball rolls without slipping, both translational and rotational motions are involved.
1. **Energy Conservation:**
- Initial potential energy at the top of the incline is converted to kinetic energy at the bottom.
- The potential energy at the top is given by \(mgh\), where \(h = L \sin \theta\).
2. **Kinetic Energy:**
- At the bottom, the total kinetic energy is the sum of translational and rotational kinetic energy.
- Translational kinetic energy: \(\frac{1}{2} mv^2\).
- Rotational kinetic energy: \(\frac{1}{2} I_{\text{cm}} \omega^2\), where \(I_{\text{cm}} = \frac{2}{5} mr^2\) for a solid sphere, and \(v = r\omega\).
3. **Equations:**
- Setting initial potential energy equal to the sum of translational and rotational kinetic energy:
\[
mgh = \frac{1}{2} mv^2 + \frac{1}{2} I_{\text{cm}} \omega^2
\]
- Substituting \(\omega = v/r\) and \(I_{\text{cm}} = \frac{2}{5} mr^2\), and simplifying, you can find \(v\).
4. **Conclusion:**
- Solve for \(v\) to find the linear speed at the bottom of the incline.
This problem illustrates the application of physics concepts to solve a practical problem involving both linear and rotational motion.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F50a0a1ab-f671-42c0-bd3d-0403645711b0%2F41bf0392-f8a6-47ee-9ecb-52a1ba2ded61%2Frftmohq_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
A bowling ball \([I_{\text{cm}} = 2mr^2/5]\) rolls down an inclined plane of length \(L\) and angle \(\theta\). The friction between the plane and the bowling ball is sufficient for the ball to roll without slipping. Find the bowling ball's (linear) speed at the bottom.
**Solution Description:**
To solve this problem, we need to apply principles of energy conservation and rotational motion. Since the ball rolls without slipping, both translational and rotational motions are involved.
1. **Energy Conservation:**
- Initial potential energy at the top of the incline is converted to kinetic energy at the bottom.
- The potential energy at the top is given by \(mgh\), where \(h = L \sin \theta\).
2. **Kinetic Energy:**
- At the bottom, the total kinetic energy is the sum of translational and rotational kinetic energy.
- Translational kinetic energy: \(\frac{1}{2} mv^2\).
- Rotational kinetic energy: \(\frac{1}{2} I_{\text{cm}} \omega^2\), where \(I_{\text{cm}} = \frac{2}{5} mr^2\) for a solid sphere, and \(v = r\omega\).
3. **Equations:**
- Setting initial potential energy equal to the sum of translational and rotational kinetic energy:
\[
mgh = \frac{1}{2} mv^2 + \frac{1}{2} I_{\text{cm}} \omega^2
\]
- Substituting \(\omega = v/r\) and \(I_{\text{cm}} = \frac{2}{5} mr^2\), and simplifying, you can find \(v\).
4. **Conclusion:**
- Solve for \(v\) to find the linear speed at the bottom of the incline.
This problem illustrates the application of physics concepts to solve a practical problem involving both linear and rotational motion.
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