Question

A light cord is wrapped about a disk with radius of 30 cm and a mass of 1.4 kg. the disk is free to rotate about an anxle. The cord is pulled with a tension T= 1.5 newtons for 10 sec.

a. What was the torque applied to the disk?

b. What was the angular acceleration of the disk?

c. Calculate the angular velocity of the disk and the total angular distance turned after 10 secs

d. Show that the final rotational kinetic energy of the disks equals the work done in pulling on the cord.

Answer 1

a)

torque=r*F*sin(theta)

=0.3*1.5*sin(90)

=0.45 Nm

b)

alpa=torque/I

=0.45/0.5*mr^2

=0.45/0.5*1.4*0.3^2

=7.143 rad/sec^2

c)

w=alppa*t

=7.143*10

=71.428 rad/sec

and

theta=1/2*alpa*t^2

=0.5*7.143*10^2

=357.142 rad

d)

rotational K.E=1/2*I*W^2

=(1/2)*(1/2)*mr^2*w^2

=(1/4)*1.4*(0.3^2)*(71.428^2)

=160.7117 J .......................(1)

and

work done=torque*theta

=(0.45*357.142)

=160.7139 J ........................(2)

from above two equation rotational K.E is equals to work done