A torque of 1 N m is applied to one end of a solid rubber cylinder of radius 5 cm and length 30 cm. By what angle does the end rotate?
A)0.3 degrees.
B)5 degrees.
C)10 degrees.
D)17.5 degrees.
E)20 degrees.

Question

Quizplus · Accepted Answer

The moment of inertia of a cylinder is given by:

I = (1/2)mr^2

where m is the mass of the cylinder and r is the radius.

The torque applied to the cylinder is given by:

τ = Iα

where α is the angular acceleration.

We are given that the torque is 1 N m, the radius is 5 cm, and the length is 30 cm.

The mass can be found using the density of rubber, which is approximately 1.2 g/cm^3.

m = ρV = ρπr^2h = (1.2)(π)(0.05^2)(0.3) = 0.034 kg

The moment of inertia is:

I = (1/2)mr^2 = (1/2)(0.034)(0.05^2) = 4.25 × 10^-5 kg m^2

Solving for the angular acceleration:

α = τ/I = (1)/(4.25 × 10^-5) = 23529 rad/s^2

The angular displacement can be found using:

θ = (1/2)αt^2

where t is the time the torque is applied. We are not given the time, so we assume that the torque is applied for a very short period of time, such that t ≈ 0.

θ = (1/2)(23529)(0)^2 = 0

Therefore, the end of the cylinder does not rotate. However, if a torque is applied for a longer period of time, the end of the cylinder would rotate.

A Torque of 1 N M Is Applied to One