Question

A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to rotate by pulling on the rope with a constant force of 195 N. If the disk starts from rest, what is its angular speed in rev/s at the end of 2.05 s?

Answer 1

The angular speed is 0.13 rev/s

Step-by-step explanation:

From the formula

`\tau = I\alpha`

Where
`\tau` is the torque

`I` is the moment of inertia

`\alpha` is the angular acceleration

But, the angular acceleration is given by

`\alpha = (\omega)/(t)`

Where
`\omega` is the angular speed

and
`t` is time

Then, we can write that

`\tau = (I\omega)/(t)`

Hence,

`\omega = (\tau t)/(I)`

Now, to determine the angular speed, we first determine the Torque
`\tau` and the moment of inertia
`I`.

Here, The torque is given by,

`\tau = rF`

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N

∴
`\tau = 3.00 * 195`

`\tau = 585` Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

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

Where M is the mass and

R is the radius

∴
`I = (1)/(2) * 325 * (3.00)^(2)`

`I = 1462.5` kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

`\omega = (\tau t)/(I)`

`\omega = (585 * 2.05)/(1462.5)`

`\omega = 0.82` rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,