A 10 kg solid cylinder with a 50.0 cm radius has a moment of inertia of 1/2 MR².If a torque of 2.0 N.m is applied to the object,the angular acceleration is A) 1.0 rad/s². B) 1.6 rad/s². C) 1.8 rad/s². D) 2.1 rad/s². E) 2.3 rad/s².

Question

Quizplus · Accepted Answer

The moment of inertia of the cylinder can be calculated using the formula:

I = (1/2)MR^2

where M is the mass of the cylinder, and R is its radius.

Substituting the given values, we get:

I = (1/2)(10 kg)(0.5 m)^2 = 1.25 kg m^2

The torque applied to the cylinder is given as:

τ = 2.0 Nm

The angular acceleration of the cylinder can be calculated using the formula:

τ = Iα

where α is the angular acceleration.

Rearranging the equation, we get:

α = τ/I

Substituting the given values, we get:

α = (2.0 Nm)/(1.25 kg m^2) = 1.6 rad/s^2

Therefore, the angular acceleration of the cylinder is 1.6 rad/s^2, which corresponds to choice (B).

A 10 Kg Solid Cylinder with a 50