Rotational Motion: Stop a 10 kg Disk in 20 sec

[pradeepk]

Homework Statement

A solid disk of mass 10 kg and radius 1 m is spinning around its central
axis at a rate of ω = 20 rad/s. A force of magnitude 5 N is applied to the disk. Recall that the
moment of inertia of a solid disk is I=(1/2)mr^2

.
(a) Draw the disk and indicate the direction of rotation. Then draw the direction that the force
should be applied to make the disk stop as quickly as possible.
(b) What is the minimum time needed for the disk to stop?

Homework Equations

$$\tau$$= I$$\alpha$$

$$\tau$$=Fr

The Attempt at a Solution

For part a, I said that the disk was moving counterclockwise, and the fastest way to stop it would be to apply a force that is perpendicular to the radius, because that would produce the largest negative torque. Is that correct?

For part b, I solved for I and got 5 kgm² then I did this:
$$\tau$$=I $$\alpha$$
Fr=5$$\alpha$$
(5N)(1m)=5$$\alpha$$
$$\alpha$$=1rad/²

I then solved for time with the equation:
$$\omega$$f=$$\omega$$i +$$\alpha$$t
0=20rad/s - 1rad/s²(t)
t=20 seconds I don't think this is correct but I'm not sure what else to do. Thank you

 

[gneill]

Looks okay to me!

 

[tiny-tim]

hi pradeepk!

all looks fine

 

[pradeepk]

Thanks Guys!

I just thought that 20 seconds was a very long time to make it stop, but I guess 5 Newtons is a very small force

 

[rwisz]

for your help!Your approach for part a is correct. To stop the disk as quickly as possible, the force should be applied perpendicular to the radius of the disk, creating the largest negative torque.

For part b, your initial calculations for the moment of inertia and torque are correct. However, when solving for time, you should use the equation:

\omega_f = \omega_i + \alpha t

Where \omega_f is the final angular velocity (which is 0 in this case), \omega_i is the initial angular velocity (20 rad/s), and \alpha is the angular acceleration (which you have correctly calculated as 1 rad/s^2). Solving for t gives:

t = (\omega_f - \omega_i)/\alpha = (0 - 20 rad/s)/1 rad/s^2 = 20 seconds

So, the minimum time needed for the disk to stop is indeed 20 seconds. Your final answer is correct. Good job!