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Question

A solid cylinder of mass $M$ and radius $R$ rolls without slipping down a rough inclined plane of angle $	heta$. A block of the same mass $M$ slides down a frictionless inclined plane of the same angle $	heta$.  If both start from rest at the same height, what is the ratio of the linear speeds of the cylinder and the block when they reach the bottom?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [Option A]
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of Energy
	**Related Topics**: Rotational Kinetic Energy, Translational Kinetic Energy, Potential Energy
	**Prerequisites**: Understanding of energy conservation, moments of inertia, and rotational motion.
## SOLUTION PATHWAY
**Given Information**:
	Mass of cylinder and block: $M$
	Radius of cylinder: $R$
	Inclination angle: $	heta$
	Initial height: $h$ (same for both)
	Initial velocities: 0 (both start from rest)
	Cylinder rolls without slipping
	Block slides without friction
**Method & Steps**:
	1. Apply conservation of energy to the cylinder
		Reasoning: The initial potential energy is converted to translational and rotational kinetic energy.
		Formula/Rule: $PE_i = KE_{trans} + KE_{rot}$
	2. Express the energy terms:
		Working:  $Mgh = \frac{1}{2}Mv_c^2 + \frac{1}{2}I\omega^2$. For a solid cylinder, $I=\frac{1}{2}MR^2$. Since the cylinder rolls without slipping, $v_c = R\omega$ or $\omega=\frac{v_c}{R}$. Substituting these gives:
$Mgh = \frac{1}{2}Mv_c^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v_c}{R})^2 = \frac{1}{2}Mv_c^2 + \frac{1}{4}Mv_c^2 = \frac{3}{4}Mv_c^2$
		Key Point: The rotational kinetic energy is included in the cylinder's total kinetic energy.
	3.  Solve for the linear velocity of the cylinder ($v_c$):
		Working:  $v_c = \sqrt{\frac{4gh}{3}}$
	4. Apply conservation of energy to the block
		Reasoning: The initial potential energy is converted to translational kinetic energy.
		Formula/Rule: $PE_i = KE_{trans}$
	5. Express the energy terms:
		Working: $Mgh = \frac{1}{2}Mv_b^2$
	6. Solve for the linear velocity of the block ($v_b$):
		Working: $v_b = \sqrt{2gh}$
	7. Calculate the ratio of the cylinder's speed to the block's speed:
		Working: $\frac{v_c}{v_b} = \frac{\sqrt{\frac{4gh}{3}}}{\sqrt{2gh}} = \sqrt{\frac{4gh}{3} \cdot \frac{1}{2gh}} = \sqrt{\frac{2}{3}}$
		Result: The ratio is $\sqrt{\frac{2}{3}}$
		Verification: The ratio is dimensionless and consistent with what we expect.
## OPTION ANALYSIS
[A] ✓: The ratio of the linear speeds of the cylinder and the block is $\sqrt{\frac{2}{3}}$.
[B] :x: This is the inverse of the correct ratio.
[C] :x: This value is not obtained from the correct calculation.
[D] :x: This value would only be true if the cylinder did not rotate or if there were no rotational energy.
## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting to include rotational kinetic energy for the cylinder.  
			[How to avoid]:  Always consider rotational energy for rolling objects.
		2.[Error 2]: Using the incorrect moment of inertia for the cylinder.
			[How to avoid]: Remember the formula for the moment of inertia of a solid cylinder: $I = \frac{1}{2}MR^2$.
	**Quick Tips**:
		1.[Helpful shortcut]: Remember the ratio of energies for rolling objects: $KE_{rot} : KE_{trans}$ is often a simple fraction.
		2.[Memory aid]: Rolling objects are slower because some energy is used for rotation.

Mr. ANIL KUMAR.G · Answer

The correct Option is "$\sqrt{\frac{2}{3}}$"

Mr. ANIL KUMAR.G · Answer

$\sqrt{\frac{3}{2}}$

Mr. ANIL KUMAR.G · Answer

$\sqrt{2}$

Mr. ANIL KUMAR.G · Answer

$1$

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