The Question is from neet-physics rotational-motion rotational-kinematics, and it was created by the author P.NARASIMHA

Question

A solid sphere of radius R and mass M rolls down an inclined plane without slipping, making an angle $	heta$ with the horizontal. A solid cylinder of the same mass and radius rolls down the same incline. If the sphere takes time $t_s$ and the cylinder takes time $t_c$ to reach the bottom, then the ratio $t_c/t_s$ is:

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Rotational Motion and Energy Conservation
	**Related Topics**: Moment of Inertia, Rolling without slipping, Conservation of Energy
	**Prerequisites**: Kinematics, Energy principles
## SOLUTION PATHWAY
**Given Information**:
	* Solid sphere and solid cylinder with same mass (M) and radius (R)
	* Rolling down an inclined plane without slipping at angle $	heta$
	* Time taken by sphere: $t_s$
	* Time taken by cylinder: $t_c$
**Method & Steps**:
	1. **Relate time taken to acceleration**: For constant acceleration, the distance travelled (length of the incline, say L) is related to time by $L = \frac{1}{2}at^2$. Since both objects travel the same distance, we have $t \propto \frac{1}{\sqrt{a}}$.
		Reasoning:  The time taken is inversely proportional to the square root of the acceleration.
		Formula/Rule: $s = ut + \frac{1}{2}at^2$, where initial velocity $u=0$.
	2. **Determine the acceleration**: For rolling without slipping, the acceleration down the incline is given by $a = \frac{g\sin	heta}{1 + \frac{I}{MR^2}}$, where I is the moment of inertia.
		Working: For a sphere, $I_s = \frac{2}{5}MR^2$, and for a cylinder, $I_c = \frac{1}{2}MR^2$.
		Key Point: The moment of inertia influences the acceleration. The larger the moment of inertia, the smaller the acceleration.
	3. **Calculate the ratio $t_c/t_s$**: Since $t \propto \frac{1}{\sqrt{a}}$, we have $\frac{t_c}{t_s} = \sqrt{\frac{a_s}{a_c}} = \sqrt{\frac{1 + \frac{I_c}{MR^2}}{1 + \frac{I_s}{MR^2}}} = \sqrt{\frac{1 + \frac{1}{2}}{1 + \frac{2}{5}}} = \sqrt{\frac{\frac{3}{2}}{\frac{7}{5}}} = \sqrt{\frac{15}{14}}$.
		Result: $\frac{t_c}{t_s} = \sqrt{\frac{15}{14}}$
		Verification: The cylinder has a larger moment of inertia and hence a smaller acceleration, thus taking longer to roll down.
 ## OPTION ANALYSIS
[A] :x: $\sqrt{\frac{14}{15}}$: This is the inverse of the correct ratio.
[B] ✓: $\sqrt{\frac{15}{14}}$: This is the correct ratio derived from the calculations.
[C] :x: $\frac{14}{15}$: This is the square of the incorrect ratio.
[D] :x: $\frac{15}{14}$: This is the square of the correct ratio.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly using the formula for acceleration without considering the moment of inertia: [Remember to account for the rotational inertia].
		2. Confusing the relationship between time and acceleration: [Time is inversely proportional to the square root of acceleration, not directly proportional].
	**Quick Tips**:
		1. Remember the moments of inertia for common shapes like spheres and cylinders.
		2. Visualize the rolling motion: The object with higher moment of inertia will resist rolling more and thus take longer.

P.NARASIMHA · Answer

√(14/15)

P.NARASIMHA · Answer

The correct Option is "√(15/14)"

P.NARASIMHA · Answer

14/15

P.NARASIMHA · Answer

15/14

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