A solid cylinder is pivoted at its center about a frictionless axle. A force is

Question

A solid cylinder is pivoted at its center about a frictionless axle. A force is applied to the outer radius of 1.75 m at an angle of 30 degree above the tangential and exerts a force of 5 N. A second force is applied by wrapping rope around the inner radius of 0.67 m, which exerts a force of 7.74 N tangent to the cylinder. The moment of inertia of the cylinder is 325 kg. m^2 Find the net torque on the cylinder. A counter-clockwise torque is defined to be positive. Answer in units of N m. Find the magnitude of the angular acceleration of the cylinder. Answer in units of rad/s^2.

Explanation / Answer

The diagram would have been helpful. For instance, do the forces produce torque in the opposite direction? Does each one produce a negative or a positive torque?

outer force: torque = 5N*cos30º*1.75m = 7.58 N·m
inner force: = 7.74N * 0.67m = 5.19 N·m

Therefore, the net torque = 2.39 N·m

b) The angular acceleration is
= / I = / 325 kg·m² = 2.39/325 = 0.0074 rad/s^2

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