Angular Acceleration Pulley: m,g,r,I

[jason177]

Homework Statement

A pulley has a radius of 2.70 m and a moment of inertia of 39.0 kg·m2. There is a hanging mass with a mass of 4.20 kg and it exerts a force tangent to the edge of the pulley. What is the angular acceleration of the pulley in rads/sec^2?

r = 2.7m
I=39kg·m2
m=4.2kg

Homework Equations

a = t/I
F = mg
t = Fr

The Attempt at a Solution

I tried finding the torque on the pulley caused by the mass by multiplying the mass * gravity * the radius and then dividing by the moment of inertia to get the angular acceleration but I keep getting 2.8496 which is apparently wrong.

 

[BigJDubs]

F = mgt = Frt = (4.2kg)(9.8m/s^2)(2.7m)t = 114.36N·ma = t/Ia = (114.36N·m)/(39kg·m2)a = 2.8496 rad/s^2

 

[kerimek]

I would like to clarify that the formula for angular acceleration is actually α = τ/I, where τ is the torque applied to the object and I is the moment of inertia. In this case, the force exerted by the hanging mass is equivalent to the torque (τ = Fr), so the correct formula to use is α = Fr/I. Plugging in the given values, we get α = (4.2 kg * 9.8 m/s^2 * 2.7 m) / 39 kg·m^2 = 0.829 rads/sec^2. Therefore, the angular acceleration of the pulley is 0.829 rads/sec^2.