Torque required to spin a disk

[cyclemun]

Homework Statement

A 250 g , 20.0-cm-diameter plastic disk is spun on an axle through its center by an electric motor.

Homework Equations

What torque must the motor supply to take the disk from 0 to 2000 rpm in 4.50 s?

The Attempt at a Solution

I know that Torque = Inertia times angular acceleration or T = Ia.
I tried using the formula for inertia of a solid disk, I = 1/2Mr^2. So with that I get 1/2(.25)(.1^2) = .00125. Then i found angular acceleration: 2000 rpm = 209.44 rad/s divided by 4.5 s so 209.44 rad/s / 4.5s = 46.54 s^2. I multiply these two number together and get .00125*46.54 = .06 Nm. However, this is wrong. What am I doing wrong?

 

[03myersd]

Your working seems correct to me...

 

[thomate1]

There are a few potential issues with your calculation. First, you are using the incorrect formula for inertia. The correct formula for a solid disk is I = 1/2MR^2, where M is the mass of the disk and R is the radius. In this case, M = 0.25 kg and R = 0.10 m, so the inertia would be 0.000125 kgm^2.

Secondly, you are using the incorrect units for angular velocity. Remember that rpm stands for revolutions per minute, so you need to convert this to radians per second by multiplying by 2π/60. This gives an angular velocity of approximately 209.44 rad/s.

Finally, when calculating torque, you also need to consider the direction of rotation. In this case, the disk is rotating clockwise, so the torque would be in the counterclockwise direction. Thus, the final calculation would be T = Iα = (0.000125 kgm^2)(46.54 s^-2) = 0.00582 Nm counterclockwise.

Overall, it is important to carefully check your equations and units to ensure that your calculations are accurate.