Question

A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 44 N through a distance of 0.9 m. Now what is the angular speed?

Answer 1

22.65 rad/s

Step-by-step explanation:

m = 7 kg, r = 0.21 m, F = 44 N, d = 0.9 m

Moment of inertia, I = 1/2 mr^2 = 0.5 x 7 x 0.21 x 0.21 = 0.15435 kg m^2

Work done, W = F x d = 44 x 0.9 = 39.6 J

according to the work energy theorem

Change in the kinetic energy of rotation = Work done

1/ 2 x I x ω^2 = W

0.5 x 0.15435 x ω^2 = 39.6

ω^2 = 513.12

ω = 22.65 rad/s

Answer 2

`\omega = 22.67 rad/s`

Step-by-step explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

`W = (1)/(2)I\omega^2`

now we know that work done is product of force and displacement

so here we have

`W = F.d`

`W = (44 N)(0.9 m) = 39.6 J`

now for moment of inertia of the disc we will have

`I = (1)/(2)mR^2`

`I = (1)/(2)(7 kg)(0.21^2)`

`I = 0.154 kg m^2`

now from above equation we will have

`39.6 = (1)/(2)(0.154)\omega^2`

`\omega = 22.67 rad/s`