Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two wi

Question

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Accepted Answer

> [!TLDR]
> For the same torque and time, the object with the smaller moment of inertia gains a larger angular acceleration and therefore a larger angular speed. The solid sphere has a smaller moment of inertia than the hollow cylinder, so the sphere acquires the greater angular speed.

The key here is to connect torque, moment of inertia, and angular acceleration — the rotational analogue of Newton’s second law. When equal torques act on two objects, the one with the smaller moment of inertia experiences a larger angular acceleration. Over the same time interval, that larger acceleration leads to a larger final angular speed (assuming both start from rest).

Let’s work through it step by step.

1. **Recall the rotational form of Newton’s second law.**  
   For rotation about a fixed axis,  
   $$ 	au = I \alpha $$  
   where $	au$ is the net torque, $I$ is the moment of inertia about the axis, and $\alpha$ is the angular acceleration.  
   This is the direct analogue of $F = ma$ in linear motion.

2. **Identify the moments of inertia for the two objects.**  
   Both have the same mass $M$ and radius $R$, but different mass distributions.
   - For a **hollow cylinder** rotating about its central axis (the standard axis of symmetry), all the mass is at the same distance $R$ from the axis. Its moment of inertia is  
     $$ I_{	ext{cylinder}} = M R^2 $$
   - For a **solid sphere** rotating about an axis through its centre, the mass is distributed from the centre out to $R$. Its moment of inertia is  
     $$ I_{	ext{sphere}} = \frac{2}{5} M R^2 $$

> [!FORMULA]
   > $$ I_{	ext{cylinder}} = MR^2, \quad I_{	ext{sphere}} = \frac{2}{5}MR^2 $$

3. **Compare the two moments of inertia.**  
   Since $\frac{2}{5} = 0.4$, we have  
   $$ I_{	ext{sphere}} = 0.4\, M R^2 \quad 	ext{and} \quad I_{	ext{cylinder}} = 1.0\, M R^2 $$  
   So $I_{	ext{sphere}} < I_{	ext{cylinder}}$. The sphere is “easier” to spin — it has less rotational inertia.

4. **Find the angular acc

Concept understanding — Rotational Inertia Comparison

Rotational Inertia Comparison: From Intuition to Precision

The Core Idea: Mass × Distance²

Why Comparison Matters

The Intuition Behind the Square

A Quick Comparison Table

Concept: Rotational Analogue of Newton's Second Law ( $\tau=I\alpha$ )

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