Question

Suppose you exert a force of 197 N tangential to a 0.290-m-radius 75.0-kg grindstone (a solid disk, so I = 1 2 M R 2 ).

What torque is exerted?

N*m

What is the angular acceleration assuming negligible opposing friction?

rad/s2

What is the angular acceleration if there is also an opposing frictional force of 20 N exerted 1.90 cm from the axis?

rad/s2

Answer 1

here,

tangential force , Ft = 197 N

radius of grindstone , r1 = 0.29 m

mass , m = 75 kg

the torque exerted , T = Ft * r1

T = 197 * 0.29 N.m = 57.13 N.m

the angular acceleration , alpha = T /moment of inertia

alpha = 57.13 /(0.5 * m * r^2)

alpha = 57.13 /(0.5 * 75 * 0.29^2)

alpha = 18.1 rad/s^2

if there is a force at Ff = 20 N is at r2 = 1.9 cm = 0.019 m

equating the torques

Ft * r1 - Ff * r2 = (0.5 * m * r1^2) * alpha'

197 * 0.29 - 20 * 0.019 = (0.5 * 75 * 0.29^2) * alpha'

alpha' = 18 rad/s^2

the angular acceleration is 18 rad/s^2