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Q7.11:
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.

Solution :

Step 1: Establishing Physical Parameters

Given that both objects have the same mass ($M$) and radius ($R$), we define their respective Moments of Inertia ($I$) about their given axes of rotation [Per the definitions of rotational inertia for rigid bodies]:

  • For a hollow cylinder rotating about its axis of symmetry: $I_{cyl} = MR^2$
  • For a solid sphere rotating about its central axis: $I_{sph} = \frac{2}{5}MR^2$

Step 2: Relational Analysis of Dynamics

The rotational analogue of Newton's Second Law states that the applied torque ($\tau$) is related to the angular acceleration ($\alpha$) by the equation:

$\tau = I \alpha$

Rearranging for angular acceleration:

$\alpha = \frac{\tau}{I}$

Given that the applied torque ($\tau$) is equal for both objects, the angular acceleration is inversely proportional to the moment of inertia: $\alpha \propto \frac{1}{I}$.

Step 3: Calculating Angular Speed

Assuming the objects start from rest ($\omega_0 = 0$), the angular speed ($\omega$) after a given time ($t$) is given by [Per the Kinematic Equation for Rotational Motion]:

$\omega = \alpha t = \left( \frac{\tau}{I} \right) t$

Since $\tau$ and $t$ are constant for both cases, the final angular speed depends entirely on the magnitude of the moment of inertia ($I$). Specifically, a smaller $I$ results in a larger $\omega$ for a fixed time interval.

Step 4: Comparative Evaluation

Object Moment of Inertia ($I$) Relative Magnitude
Hollow Cylinder $MR^2$ $1.0 \cdot MR^2$
Solid Sphere $0.4 \cdot MR^2$ $0.4 \cdot MR^2$

Since $I_{sph} < I_{cyl}$ (as $0.4 < 1.0$), it follows mathematically that:

$\frac{\tau}{I_{sph}} > \frac{\tau}{I_{cyl}}$

Therefore, $\alpha_{sph} > \alpha_{cyl}$.

Hollow Cylinder Solid Sphere

Final Solution: The solid sphere will acquire a greater angular speed. Because the solid sphere has a smaller moment of inertia ($0.4 MR^2$) compared to the hollow cylinder ($MR^2$), the same applied torque produces a higher angular acceleration, resulting in a higher angular speed after the same duration.


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