Question

A pully disk 4 meters in diameter with a mass of 5 kg has a mass hanging from either side exerting a 20 N and 30 N respectively.

1.What minimum static torque exerted by friction at the axle will keep the system at rest?

The kinetic torque, the torque on the axle when the system is in motion, is half of the static value.

2.Once in motion, what is the angular acceleration of the pulley disk?

Answer 1

In equilibrium net torque = 0

F1 = 20 N

F2 = 30 N

radius R = D/2 = 4/2 = 2 m

mass of disk m = 5kg

torque due to F1 = -F1*R ( clockwise )

torque due to F2 = F2*R ( counter clock wise)

torque due to static friction Ts = ?

if the suyste is in rest

net torque = 0

-F1*R + F2*R + Ts = 0

-20*2 + 30*2 + Ts = 0

Ts = 20 Nm <<<------------answer

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If the disk rotates

net torque = I*alpha

I = moment of inertia = (1/2)*m*R^2

-F1*R + F2*R + Tk = I*alpha

-20*2 + 30*2 + 20/2 = (1/2)*5*2^2*alpha

angular acceleration alpha = 3 rad/s^2 <<<<--------------answer