Calculating Work Done by Block on Pulley

[DH214]

The pulley in the illustration is a uniform disk of mass 2.40 kg and radius 0.220 m. The block applies a contant torque to the pulley, which is free to rotate without friction, resulting in an angular acceleration of magnitude 0.180 rad/s2 for the pulley. As the block falls 0.500 m, how much work does it do on the pulley?

 

[djeitnstine]

what illustration :S

 

[thomate1]

I would approach this problem by first understanding the basic principles of work and torque. Work is defined as the force applied over a distance, while torque is the rotational force applied to an object. In this scenario, the block is applying a constant torque to the pulley, resulting in an angular acceleration. Therefore, the work done by the block on the pulley can be calculated by using the formula W = τθ, where τ is the torque and θ is the angular displacement.

To calculate the torque, we can use the formula τ = Iα, where I is the moment of inertia and α is the angular acceleration. In this case, the moment of inertia of the pulley can be calculated as I = ½mr^2, where m is the mass and r is the radius of the pulley. Plugging in the given values, we get I = ½(2.40 kg)(0.220 m)^2 = 0.119 kgm^2.

Next, we can calculate the torque by using the given angular acceleration of 0.180 rad/s^2. Therefore, τ = (0.119 kgm^2)(0.180 rad/s^2) = 0.0214 Nm.

Finally, we can calculate the angular displacement of the pulley by converting the linear displacement of the block (0.500 m) into an angular displacement. Since the circumference of the pulley is 2πr, the angular displacement can be calculated as θ = 0.500 m/0.220 m = 2.27 radians.

Plugging in the values, we get W = (0.0214 Nm)(2.27 radians) = 0.0486 J. Therefore, the work done by the block on the pulley is 0.0486 Joules.