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

Question

A solid sphere of mass $M$ and radius $R$ rolls down an inclined plane without slipping. 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, what is the ratio $\frac{t_c}{t_s}$?

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Rotational Motion and Energy Conservation
	**Related Topics**: Rolling without slipping, Moment of Inertia, Kinematics
	**Prerequisites**: Understanding of kinetic energy (translational and rotational), moment of inertia for standard shapes, conservation of energy, kinematics equations.
## SOLUTION PATHWAY
**Given Information**:
	* Mass of sphere = Mass of cylinder = M
	* Radius of sphere = Radius of cylinder = R
	* Inclined plane is same for both
	* Rolling without slipping
**Method & Steps**:
	1.	Determine the acceleration of the sphere and cylinder.
		Reasoning: Using energy conservation, we can relate the potential energy at the top to the kinetic energy (translational and rotational) at the bottom. Then, we can use kinematics to relate the acceleration to the time taken.
		Formula/Rule: Conservation of Energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$, where $I$ is the moment of inertia and $\omega = v/R$ for rolling without slipping.
	2.	Apply energy conservation for the sphere.
		Working: $mgh = \frac{1}{2}mv_s^2 + \frac{1}{2}I_s\omega_s^2 = \frac{1}{2}mv_s^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v_s}{R})^2 = \frac{1}{2}mv_s^2 + \frac{1}{5}mv_s^2 = \frac{7}{10}mv_s^2$. So $v_s^2 = \frac{10}{7}gh$. Using kinematics, $v_s^2 = 2a_s d$. Thus, $a_s = \frac{5}{7}g\sin	heta$, where $h=d\sin	heta$
		Key Point: Moment of inertia for sphere $I_s = \frac{2}{5}MR^2$ and rolling condition $v = R\omega$.
	3.	Apply energy conservation for the cylinder.
		Working: $mgh = \frac{1}{2}mv_c^2 + \frac{1}{2}I_c\omega_c^2 = \frac{1}{2}mv_c^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v_c}{R})^2 = \frac{1}{2}mv_c^2 + \frac{1}{4}mv_c^2 = \frac{3}{4}mv_c^2$. So $v_c^2 = \frac{4}{3}gh$. Using kinematics, $v_c^2 = 2a_c d$. Thus, $a_c = \frac{2}{3}g\sin	heta$.
		Key Point: Moment of inertia for cylinder $I_c = \frac{1}{2}MR^2$.
	4.	Relate time taken to acceleration and distance.
		Working: Using kinematics, $d = \frac{1}{2}at^2$. So, $t = \sqrt{\frac{2d}{a}}$. Therefore $t_s = \sqrt{\frac{2d}{a_s}} = \sqrt{\frac{2d}{\frac{5}{7}g\sin	heta}}$ and $t_c = \sqrt{\frac{2d}{a_c}} = \sqrt{\frac{2d}{\frac{2}{3}g\sin	heta}}$.
	5.	Calculate the ratio $\frac{t_c}{t_s}$.
		Working: $\frac{t_c}{t_s} = \frac{\sqrt{\frac{2d}{\frac{2}{3}g\sin	heta}}}{\sqrt{\frac{2d}{\frac{5}{7}g\sin	heta}}} = \sqrt{\frac{\frac{5}{7}}{\frac{2}{3}}} = \sqrt{\frac{5}{7} 	imes \frac{3}{2}} = \sqrt{\frac{15}{14}}$
		Result: $\frac{t_c}{t_s} = \sqrt{\frac{15}{14}}$
		Verification: The ratio is unitless and the calculations are consistent.
 ## OPTION ANALYSIS
[A] :x: This option has an inverse ratio.
[B] ✓: This is the correct ratio as derived.
[C] :x: This ratio is incorrect.
[D] :x: This ratio is incorrect.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. [Error 1]: Incorrectly using moment of inertia for different objects. [How to avoid]: Always use the correct moment of inertia for the specified object (sphere, cylinder, etc.).
		2. [Error 2]: Forgetting the rolling condition $v = R\omega$. [How to avoid]: Always remember to use the correct relationship between linear and angular velocity for rolling without slipping.
	**Quick Tips**:
		1. [Helpful shortcut]: Remember the general form of acceleration for rolling objects is $a = \frac{g\sin	heta}{1 + \frac{I}{mr^2}}$.
		2. [Memory aid]: The object with the smaller moment of inertia will reach the bottom faster for the same mass and radius.

P.NARASIMHA · Answer

$\sqrt{\frac{14}{15}}$

P.NARASIMHA · Answer

The correct Option is "$\sqrt{\frac{15}{14}}$"

P.NARASIMHA · Answer

$\frac{7}{5}$

P.NARASIMHA · Answer

$\frac{5}{7}$

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