# Conservation Laws in Rotational Motion

1. Apr 28, 2008

### Lma12684

1. The problem statement, all variables and given/known data
A uniform solid sphere, with diameter 28 cm and mass 2.5 kg, rolls without slipping on a horizontal surface, at constant speed of 2.0 m/s.
1) What is the rotational kinetic energy?
2) What is its total kinetic energy?
3) What is its angular momentum?

2. Relevant equations

1) KE=1/2 Iw^2
2) KE=1/2 Mv(cm) + 1/2*I(cm)*w^2
3) L=Iw

3. The attempt at a solution

1) KE=1/2(2/5MR^2)(v/r)
=1/2(19.6)(.14)
=1.372

2) KE=1/2 (2.5)(2.0)+ 1/2 (19.6)*(.14^2)
=2.69

3) L=(19.6)(.14)
=2.744

Does this solution look good? Thanks.

2. Apr 28, 2008

### CrazyIvan

$$KE_{linear} = \frac{1}{2} m v^{2}_{cm}$$

3. Apr 28, 2008

### Lma12684

Did I forget the square on #2? Is that what you are saying? Thanks.

4. Apr 28, 2008

### alphysicist

Hi Lma12684,

Two things about 1): You did not square the (v/r) term; also, since you don't have units, I'm assuming that you want them in SI units. However, 19.6 is not correct for the moment of inertia. It's probably better to convert the radius to meters before you enter it in your calculator; or, if you wait until the end remember that there are two factor of centimeters in I, and you have to convert both of them to meters.

5. Apr 28, 2008

### Lma12684

Ok, I recalculated and found:

1) 2 J
2) 4.99 J
3) 3.99 J

6. Apr 28, 2008

### alphysicist

The first one looks right to me, but not the answers for #2 and #3. It's difficult to tell, though, since you haven't posted the numbers you used.

For #2, was the 4.99 J for the total energy, or did you actually just calculate the translational kinetic energy? It looks like you may not have added the rotational KE to the translational KE.

For #3, it won't have units of Joules. Also, did you perhaps (incorrectly) square the omega term?

7. Apr 28, 2008

### Lma12684

Here is what I did:

2) KE=1/2MV(cm) + 1/2 I(cm)w^2
=1/2(2.5)(2.0) + 1/2(.0196)(204.08)
=4.99 J

3) L=Iw
=(.0196)(204.08)
=3.99 J

8. Apr 28, 2008

### alphysicist

In #2, you did not square the 2.0 for the speed in the translational kinetic energy.

In #3, you can see from #2 that 204.08 is w^2, but here you need w.

9. Apr 28, 2008

### Lma12684

Thank You Again!!!!!!!!!!!!!

10. Aug 27, 2009

### mtimmerman123

Since it is a sphere then the moment of inertia is

I = 2/5mr^2

so the equations I used for a,b, and c are:

a) KE = (1/2)(2/5mr^2)(v^2/r^2) = 2.0J
b) KE = 1/2mv^2 + 1/2Iw^2 = 5.0J + 2.0J = 7.0J
c) L = Iw = 4.0 kg*m^2/s

I think this is right, what do you think?

also, w^2 = (v^2/r^2) so the radius cancels out and I didn't have to use r = 0.14 m