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Question

A ring of mass $m$ and radius $r$ rolls down an inclined plane of angle $	heta$ without slipping. A block of the same mass slides down the same incline. The ratio of the velocities of the ring and the block when they reach the bottom is:

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [B]
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of Energy
	**Related Topics**: Rotational Kinetic Energy, Translational Kinetic Energy, Gravitational Potential Energy
	**Prerequisites**: Basic understanding of energy conservation, moments of inertia, and rolling motion.
## SOLUTION PATHWAY
**Given Information**:
	* Mass of ring = mass of block = $m$
	* Radius of ring = $r$
	* Inclination angle = $	heta$
	* Ring rolls without slipping, block slides without friction
**Method & Steps**:
	1.  **Energy Conservation for the Ring**:
		Reasoning: The potential energy at the top is converted to both translational and rotational kinetic energy at the bottom.
		Formula/Rule:  $PE = KE_{trans} + KE_{rot}$
	2.  **Apply Energy Conservation for the Ring**:
		Working: $mgh = \frac{1}{2}mv_r^2 + \frac{1}{2}I\omega^2$. For a ring, $I = mr^2$ and $v_r = r\omega$. Substituting, we get $mgh = \frac{1}{2}mv_r^2 + \frac{1}{2}(mr^2)(\frac{v_r}{r})^2 = \frac{1}{2}mv_r^2 + \frac{1}{2}mv_r^2 = mv_r^2$. Therefore, $v_r = \sqrt{gh}$.
		Key Point:  The moment of inertia of the ring and the relationship between linear and angular velocity are crucial.
	3. **Energy Conservation for the Block**:
		Reasoning: The potential energy at the top is converted to translational kinetic energy at the bottom.
		Formula/Rule: $PE = KE_{trans}$
	4. **Apply Energy Conservation for the Block**:
		Working: $mgh = \frac{1}{2}mv_b^2$. Therefore, $v_b = \sqrt{2gh}$.
		Key Point: The block has only translational kinetic energy.
	5. **Calculate the Ratio of Velocities**:
		Working: $\frac{v_r}{v_b} = \frac{\sqrt{gh}}{\sqrt{2gh}} = \frac{1}{\sqrt{2}}$
		Result:  The ratio of velocities of the ring and the block is $\frac{1}{\sqrt{2}}$
		Verification: The answer matches the derived ratio.
 ## OPTION ANALYSIS
[A] :x: Incorrect. The ratio is not 1.
[B] ✓: Correct. The ratio of the velocities of the ring and the block is $\frac{1}{\sqrt{2}}$.
[C] :x: Incorrect. The ratio is not $\sqrt{2}$.
[D] :x: Incorrect. The ratio is not 2.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. **Forgetting Rotational Kinetic Energy**: Forgetting to account for the rotational kinetic energy of the ring leads to an incorrect result.
		2. **Incorrect Moment of Inertia**: Using the wrong moment of inertia for the ring will lead to an error. Make sure to use $I = mr^2$ for a ring about its central axis.
	**Quick Tips**:
		1.  **Rolling without slipping**: Remember the relation $v = r\omega$ for rolling without slipping.
		2. **Energy Conservation**: Use energy conservation to relate potential energy at the top to kinetic energy (translational and/or rotational) at the bottom.

SUBHANI · Answer

1

SUBHANI · Answer

The correct Option is "$\frac{1}{\sqrt{2}}$"

SUBHANI · Answer

$\sqrt{2}$

SUBHANI · Answer

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