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

Answer:

The angular speed is 0.13 rev/s

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

$\alpha = \frac{\omega}{t}$

Where $\omega$ is the angular speed

and $t$ is time

Then, we can write that

$\tau = \frac{I\omega}{t}$

Hence,

$\omega = \frac{\tau t}{I}$

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

∴ $\tau = 3.00 \times 195$

$\tau = 585$ Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

$I = \frac{1}{2}MR^{2}$

Where M is the mass and

R is the radius

∴$I = \frac{1}{2} \times 325 \times (3.00)^{2}$

$I = 1462.5$ kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

$\omega = \frac{\tau t}{I}$

$\omega = \frac{585 \times 2.05}{1462.5}$

$\omega = 0.82$ 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,