# Kinetic Energy and Rolling motion

1. Oct 22, 2007

### Nitrag

A 500g, 8.0-cm-diameter can rolls across the floor at 1.0 m/s. What's the can's kinetic energy.

I solved this using:

Code (Text):
K = $$\frac{1}{2}$$*I*($$\frac{v}{r}$$)$$^{2}$$
and for some reason I get 1/3 the answer. So I multiplied it by 3 at the end. Is there something wrong with the formula I'm using?

If i do the same thing with a similar problem I don't get the right answer. I think the difference lies in the 1.50 m/s.

Similar problem:
Code (Text):

A 600 g, 6.50-cm-diameter solid cylinder rolls across the floor at 1.50 m/s.

2. Oct 22, 2007

### Sourabh N

Can you show the formula u used for I(in terms of r) ?

3. Oct 22, 2007

### Nitrag

Oops sry,

I = (.5*m*r^2)

4. Oct 22, 2007

### Sourabh N

This is the value of I about COM, the cylinder is not rolling about the COM.

5. Oct 22, 2007

### rl.bhat

Since the can is hollow, you must know the inner and outer diameter of the can to find the MI.

6. Oct 22, 2007

### rl.bhat

If the walls of the can are very small, the moment of inertia = MR^2

7. Oct 22, 2007

### Sourabh N

Hey karnataki, specify the axis (coz here I is 3/2MR^2)

8. Oct 23, 2007

### rl.bhat

Moment of inertia of a hollow cylinder about the axis parallel to the axis of cylinder is = M(R1^2 + R2^2)/2
If the can is thin, in that case R1 is nearly equal to R2.
and MI = MR^2. You have taken the axis tangential to the cylinder and parallel to the axis of the cylinder.

9. Oct 23, 2007

### Sourabh N

Thats the axis along which pure rotation is taking place, I suppose.

10. Oct 23, 2007

### rl.bhat

K.E. of rolling can = 1/2*I(com)(v/r)^2 + 1/2*Mv(com)^2
= 1/2*Mr^2(v/r)^2 + 1/2*Mv(com)^2
= 1/2*Mv^2 +1/2*Mv^2
= Mv^2
In such problems radius of the object is not needed.

11. Oct 23, 2007

### Sourabh N

Hey I(com) = 1/2Mr^2.

12. Oct 23, 2007

### Nitrag

Just mv^2 does not work.

My buddy has a different values and he got 0.443 for the question:
Code (Text):
A 500g, 8.0-cm-diameter can rolls across the floor at 1.0 m/s. What's the can's kinetic energy.

13. Oct 24, 2007

### rl.bhat

I (com) = 1/2(MR^2) is true for solid cylinder.

Total KE of rolling body = KE of translation + KE of rotaion
If possible go through Halliday & Resnic Book of Physics

14. Oct 24, 2007

### Sourabh N

Oh ya, I'm sorry for the confusion.
So, K = Mv^2 = 0.5 * (1)^2 = 0.5 J

and for the other ques. K = 0.6 * (1.5)^2 = 1.35 J
Are your numerical values same, Nitrag?

15. Oct 24, 2007

### Nitrag

Yes, I got 1.35 J also but it is not correct.

16. Oct 24, 2007

### rl.bhat

For the other question K = 1/2*1/2*(MR^2)(V/R)^2 + 1/2*MV^2
= 1/4MV^2 + 1/2MV^2 =3/4MV^2 = 1.0125 J

17. Oct 25, 2007

### Nitrag

ah, I see the connection now. thanks for the help!