Question

A cable passes over a pulley. Because the cable grips the pulley and the pulley has nonzero mass, the tension in the cable is not the same on opposite sides of the pulley. The force on one side is 162 N, and the force on the other side is 63 N. Assuming that the pulley is a uniform disk of mass 1.29 kg and radius 1.184 m, find the magnitude of its angular acceleration. [For a uniform disk, I

Answer 1

Answer:

a = 129.663 $rad/s^{2}$

Explanation:

We know that:

T = Ia

Where T is the torque, I is the moment of inertia and a is the angular aceleration:

First, we will find the moment of inertia using the following equation:

I = $\frac{1}{2}MR^2$

Where M is the mass and R is the radius of the disk. Replacing values, we get:

I = $\frac{1}{2}(1.29kg)(1.184m)^2$

I = 0.904 kg*m^2

Second, we will find the torque using the following equation:

T = ($F_1-F_2$)*(R)

Where $F_1$ is the force on one side and $F_2$ is the force on the other side. Replacing values, we get:

T = (162N-63N)(1.184m)

T = 117.216N*m

Finally, we replace T and I on the initial equation as:

T = Ia

117.216N = (0.904)(a)

Solving for a:

a = 129.663 $rad/s^{2}$