Question

Suppose you exert a force of 337 N tangential to a grindstone (a solid disk) with a radius of 0.398 m and a mass of 98.0 kg What is the resulting angular acceleration of the grindstone assuming negligible opposing friction?

Answer 1

Answer:

17.284 rad/s^2

Explanation:

F = 337 N

R = 0.398 m

M = 98 kg

let the angular acceleration is α

torque = I x α = F x R

Where I is the moment of inertia

I =  1/2 MR^2 = 0.5 x 98 x 0.398 x 0.398 = 7.76 kg m^2

So,

7.76 x α = 337 x 0.398

α = 17.284 rad/s^2