Calculating Rotational Parameters of a Rotating Sphere

[mistermangos]

Hello everyone, I have been working on a problem and I am having a hard time figuring out if this is the right answer.

Homework Statement

A solid sphere of mass 10 kg and radius 10 cm is rotating about its axis.

A. Find its moment of inertia

B. What is the kinetic energy of the sphere if it has an angular velocity of 10 revolutions per minute?

C. A tangential force of 100N is applied for 10 seconds. How much will the sphere slow down if it starts with the same angular velocity as part B?

Homework Equations

Moment of inertia of solid sphere = I = $\frac{2}{5}$*MR^2

Rotational Kinetic Energy = $\frac{1}{2}$I$\omega$^2

The Attempt at a Solution

A. I = 2/5 (10)(.1)^2 = 2/50 = .04 kg $\bullet$ m^2

B. $\frac{10}{60}$ rev/s = $\frac{1}{6}$ rev/s = $\frac{\pi}{3}$

Rot. Ke = 1/2 (.04)($\pi$/3)^2 = .0219 J

C. A tangential force is applied, so the equivalent torque is $\tau$ = Fr = (100)(.1) = 10

$\tau$ = I$\alpha$, so $\alpha$ = 10/.04 = -250 rad/sec^2 (negative because it has a clockwise direction I guess, since its asking for the slowdown)

$\omega$(t) = w$_{0}$ + $\alpha$t

$\omega$(10) = $\frac{\pi}{3}$ + (-250)(10)$\omega$(10) = -2498.95 rad/sec, or -397.72 rev/sec

seems a little fast for me... any help is appreciated

 

[TSny]

Hello and welcome to Physics Forums!

All of your work looks correct to me. I agree that the numbers work out to be a little odd in part C. But unless I'm overlooking something, you got the right answer.

Maybe someone else will check it, too.

 

[Simon Bridge]

Welcome to PF;

seems a little fast for me..

What are you comparing it to?Hmmm ... if I've got this right: the ball has a density of 2387.3kg/m^3 ... about as dense as graphite. So it should be pretty easy to set it spinning compared with, say, the same size ball of steel or lead.

How would you apply a 100N force to it? Well:
Wrap string around it and tie another one of those balls to the other end, drop the ball, you get a couple of Newtons less force ... for 10secs, that's like dropping the weight 50m (actually a bit less). A free-falling ball dropped that far would be going at v=at=98m/s when it hit... that would translate into w=v/r=980rad/s as a lower-bound for the additional speed (since some of the gravitational PE goes to rotation).

2500 rad/s is faster ...
So we expect fast. We'd expect about twice as fast because the example involves twice the mass (remember all those falling object fbds you did?)
Can you think of anything else to use as a reality check?

There are some interesting figures in there ... pi/3 is approximately 1, MR=1 FR=10, FT=1000 ... all "nice" numbers.

$\tau = -FR = I\alpha = \frac{2}{5}MR^2(\omega - \omega_0)/T$
$$\Rightarrow \omega = \omega_0 - \frac{5}{2}\frac{FT}{MR}$$
... so $$\omega \approx 1 - \frac{2}{5}\frac{1000}{1} = 1-2500=-2499\text{rad/s}$$

 

[mistermangos]

thank you guys very much for the help!

for the poster above, in my recent memory i don't recall ever seeing an object or a problem where something rotates at a speed of almost 400 revolutions per second... using v = rw, v = 25000, its going about 73 times the speed of sound! (at the surface, at least)

 

[Simon Bridge]

Well 465.1m/s is the surface tangential speed of the Earth.
at v=rw, v=2500rad/s x0.1m = 250m/s ... somewhat lower than the speed of sound.

Notice that the rotational KE is very small to get this?
I suspect that's the lesson.