Q. 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.

Answer :

Given that Torque acting on the both the bodies are same.

We know Torque on a body,

τ = Iα

Where

I is moment if inertia of the body

α is angular acceleration of the body

The body with greater angular acceleration will acquire greater angular speed in a given time.

Both hollow cylinder and solid sphere have the same mass and radius (let them be m and R respectively).

Moment of inertia of hollow cylinder, I_{cyl} = mR^{2}

Moment of inertia of solid sphere, I_{sph} = 2mR^{2}/5

Let, α_{cyl} and α_{sph} be the angular acceleration of hollow cylinder and solid sphere respectively on action of torque.

Since the torque acting on the bodies are same,

I_{cyl} α_{cyl} = I_{Sph} α_{Sph}

But I_{sph} = 2I_{cyl}/5

∴ α_{cyl} = 2α_{sph}/5

That is Solid sphere has greater angular acceleration than hollow cylinder when same torque acts on both bodies. Therefore, Solid sphere will attain greater angular speed than hollow cylinder in a given time.

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A solid sphere is set into motion on a rough horizontal surface with a linear speed u in the forward direction and an angular speed u/R in the anticlockwise direction as shown in figure (10-E 16). Find the linear speed of the sphere (a) when it stops rotating and (b) when slipping finally ceases and pure rolling starts.

HC Verma - Concepts of Physics Part 1