Calculate Torque to Accelerate .22kg Disk to 1500rpm in 5s

[guru]

A .22kg , .22-m-diameter plastic disk is spun on an axle through its center by an electric motor. What torque must the motor supply to take the disk from 0 to 1500 rpm in 5.00 ?

I used the equations
T(thau)=alpha/I
w_f=w_i + alpha*(t_f - t_i)
I found I to be equal to .021kgm^2
then found w_f for the first .5s to be:
(1500rev/1min) * (1min/60s) *(2pi/1rev) = 157.08rad/s
157.08=0+alpha*(.5s)
alpha = 31.42rad/s^2
T= 31.42/.021 = 1496.2Nm

Am not totally sure of my calculations. I will greatly appreciate any help

 

[NeutronStar]

I haven't done this in a very long time, but isn't $$\tau=I\alpha$$

And $$I=mr^2$$

Also, is the time 5 seconds? or 0.5 seconds?

P.S.

You might want to convert your answer into foot pounds and think about the problem intuitively. Is it really going to take over 1000 ft/lbs of torque to spin a little lightweight plastic disk to 1500 rpm?

 

[guru]

I haven't done this in a very long time, but isn't $$\tau=I\alpha$$

And $$I=mr^2$$

Also, is the time 5 seconds? or 0.5 seconds?

P.S.

You might want to convert your answer into foot pounds and think about the problem intuitively. Is it really going to take over 1000 ft/lbs of torque to spin a little lightweight plastic disk to 1500 rpm?

Yes, you are right $$\tau=I\alpha$$
However, i believe $$I=(1/2)mr^2$$
Also, it is 5seconds not .5s. I made a mistake
Am supposed to give my answer in Nm

 

[NeutronStar]

Yes, you are right $$\tau=I\alpha$$
However, i believe $$I=(1/2)mr^2$$
Also, it is 5seconds not .5s. I made a mistake
Am supposed to give my answer in Nm

You're right about the moment of inertia for a solid disk.

It would be $$I=(1/2)mr^2$$ for a solid disk.

I was thinking of a pendulum because I work with clocks a lot.

I didn't mean for you to turn your answer in in units of ft-lbs. I simply meant to convert it to ft-lbs to think about it intuitively to check whether or not the answer makes sense. Although that's because I identify with ft-lbs easier. Maybe Nm work better for you intuitively.

I just thought that your answer is an aweful lot of torque to spin such a small lightweight disk to 1500 rpm (even in a half a second). But if you have 5 seconds to get it spinning 1500 rpm it shouldn't take much torque at all. I would expect a very low amount of torque to be required. That's just my intuitive guess.

 

[UMDstudent]

Heres the proper way to compute this answer:

Torque = Moment of Inertia * Angular Acceleration.
Moment of Inertia = 1/2 * MR^2. (For a disk)
Angular Acceleration = change in omega / change in time.

*Make sure you understand this. If not, please reply or PM me.

Physics is Phun

**Sorry about posting in an older thread, but as many students find, mastering physics repeats its questions so I am sure we will have students searching through the forums.