Question

A 5 kg pully has a string round around it and a force of 1.0 N pulls down on the string. The pully is frictionless, 10 cm in radius, and is a uniform disk. The pully is initially at rest.

A) What is the torque of the pully?

B) What is the angular acceleration of the pully?

C) How many times will the pully rotate in 3.0 seconds?

D) What will the pully's angular velocity be after 3.0 seconds?

Answer 1

A.)

Torque is given by,

= F*R*sinA

here, F = 1.0 N

R = radius = 10 cm = 0.1 m

A = angle between F vector and R vector = 90 deg

So, = 1.0*0.1*sin 90 deg

= 0.1 N*m

B.

also, = I*alpha

here, I = moment of inertia of pulley = 0.5m*R^2

m = mass = 5 kg

alpha = angular acceleration = /I = 0.1/(0.5*5*0.1^2)

alpha = 4 rad/s^2

C.

From second rotational kinematics law,

= wi*t + 0.5*alpha*t^2

here, t = 3.0 sec.

wi = 0 (given)

So, = 0*3.0 + 0.5*4*3.0^2 = 18 rad

= 18/(2*pi) = 2.86 revolutions

D.

From first rotational kinematics law,

wf = wi + alpha*t

wf = final angular speed = 0 + 4*3.0

wf = 12.0 rad/s