Question

A 1.5-kg stone disk is attached to the shaft of an electric motor that exerts a constant torque of 2.5 N ⋅ m on the disk. If the disk has a radius of 0.10 m and starts at rest, determine how many revolutions it makes during the first 3.0 seconds the motor is on.

Answer 1

Step 1: FInd angular acceleration of disk:

We know that torque applied on disk is given by

Torque = I*

= angular acceleration = /I

I = moment of inertia of disk = (1/2)*m*R^2

R = Radius of disk = 0.10 m

m = mass of disk = 1.5 kg

So,

= /(m*R^2/2) = 2*/(m*R^2)

= 2*2.5/(1.5*0.10^2) = 333.33 rad/sec^2

Step 2: Find angular displacement of disk

Using 2nd rotational kinematic equation:

= wi*t + (1/2)**t^2

wi = Initial angular speed = 0 rad/sec

t = time interval = 3.0 sec

So,

= 0*3.0 + (1/2)*333.33*3.0^2

= 1499.985 rad = 1500 rad

Step 3: Number of revolutions will be:

1 rev = 2*pi rad, So

Number of revolutions = /(2*pi) = 1500/(2*pi) = 238.73 rev

Let me know if you've any query.