# Homework Help: Help on TOTAL kinetic energy

1. Apr 9, 2004

### psruler

Can someone help me with this problem:

A cylindrical wooden log rolls (without slipping) down a smal ramp. Tha log has radius of 15 cm, a length of 2.0m and a mass of 85kg.
What was the log's TOTAL kinetic energy at the bottom of the ramp? (Recall that Icylinder = (1/2)MR^2) .

The total angular velocity i got was 200 rad/s.

I don't know if the angular velocity will help solve the problem.

THANKS!!

2. Apr 9, 2004

Not enough information. How high is the log on the ramp?

3. Apr 9, 2004

### psruler

Well, that is the next question I have to solve. That's all the information. Is there a way of solving this problem if the height is not given?

4. Apr 9, 2004

No. The energy is determined by the height, since the only energy the log has at first is gravitational potential energy, which depends on height.

5. Apr 9, 2004

### psruler

The previous question states that the log rolls at a speed of 3.0m/s when it reaches the bottom of the ramp.

6. Apr 9, 2004

### Staff: Mentor

Well now you have enough information!

The total KE is the sum of (1) translational KE of the center of mass (1/2MV2), and (2) rotational KE about center of mass(1/2Iω2). Since the log rolls without slipping, V =ωR.

7. Apr 9, 2004

### psruler

Thanks Doc Al!

8. Apr 9, 2004

### psruler

Then can I find the height using this equation:

h = v^2/2g ?

9. Apr 9, 2004

### Staff: Mentor

Find the height using energy conservation. The initial gravitational PE at the top gets transformed into the total KE at the bottom.

10. Apr 9, 2004

### Chen

No because the log also has rotational KE (1/2Iω2), not just translational KE.

11. Apr 10, 2004

### psruler

The answer i got is: 0.62m. Can you verify if that is correct?

12. Apr 10, 2004

### Chen

I get a total kinetic energy of 573.75J at the bottom and a height of 6.75/g. So if you take g to be 10.9m/s2 it's right.

13. Apr 10, 2004

### psruler

how did you get 6.75 as the height?

14. Apr 10, 2004

### Staff: Mentor

I get a (slightly) different answer. Maybe we should average them all together?

Show us the equations you used, what numbers you assumed, and how you calculated the answer. (Show it algebraically first, then with numbers plugged in.)

15. Apr 10, 2004

### Chen

I got:

$$mgh = 573.75J$$

So:

$$h = \frac{573.75J}{mg} = \frac{6.75}{g}$$

My equation:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega ^2 = \frac{1}{2}mv^2 + \frac{1}{4}mr^2\omega ^2 = \frac{3}{4}mv^2$$

$$h = \frac{3v^2}{4g}$$

Last edited: Apr 10, 2004
16. Apr 10, 2004

### Staff: Mentor

17. Apr 10, 2004

### Chen

I have a question now. Let's say it is a sphere now, instead of a cylinder. The equation would be:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega ^2 = \frac{1}{2}mv^2 + \frac{1}{5}mr^2\omega ^2 = \frac{7}{10}mv^2$$

$$v^2 = \frac{10gh}{7}$$

But now let's take that sphere and reduce its radius until it tends to zero (so we find ourselves with a "point" particle). Obviously the mass and the radius of the object do not affect the speed it would reach at the bottom of the hill, as that only depends on the height of it. So on one hand, we know that for point particles there is no such thing as rotational kinetic energy, so the speed squared would be 2gh. However, if we do count the rotational KE we find the speed squrared to be of 10gh/7. So which answer is the correct one? And where's the limit between a point particle and a sphere? i.e when in the process of reducing the object's radius will the speed change from one expression to the other?

Last edited: Apr 10, 2004
18. Apr 10, 2004

### Staff: Mentor

An excellent question!

This is an excellent question, Chen! (I think you have the cases mixed, though.) I think when you reach that limit, the normal analysis of static friction causing rotation will break down. Surface effects will dominate.