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.
The solid sphere will acquire a greater angular speed after a given time.
When torques of equal magnitude are applied to both bodies, the angular acceleration produced depends on the moment of inertia. From the relation τ = Iα, we have α = τ/I. Since torque is the same, the body with smaller moment of inertia will have greater angular acceleration.
The moment of inertia of a hollow cylinder about its axis is I = MR², while the moment of inertia of a solid sphere about an axis through its centre is I = (2/5)MR². Since (2/5)MR² < MR², the solid sphere has a smaller moment of inertia. Therefore, the solid sphere will have greater angular acceleration and will acquire greater angular speed in a given time.
Explanation
This question tests understanding of rotational dynamics, specifically the relationship τ = Iα. The hollow cylinder has all its mass distributed at distance R from the axis, giving I = MR². The solid sphere has its mass distributed throughout its volume, with most mass closer to the axis, resulting in a smaller moment of inertia I = (2/5)MR². With equal torques, the body with smaller moment of inertia undergoes greater angular acceleration, leading to higher angular speed in the same time interval.
Solution Steps
Step 1: Apply the relation τ = Iα, where τ is torque, I is moment of inertia, and α is angular acceleration.
Step 2: For equal torques, α ∝ 1/I, so smaller I gives larger α.
Step 3: Compare moments of inertia: I_cylinder = MR² and I_sphere = (2/5)MR².
Step 4: Since I_sphere < I_cylinder, the sphere has greater angular acceleration and thus greater angular speed after a given time.