# Kinetic and Potential energy problem

1. May 17, 2014

### BrainMan

1. The problem statement, all variables and given/known data
A simple pendulum is constructed by supporting a .375 kg steel sphere by a thin string 1.2 in length and of negligible mass. The pendulum is released form rest at a position where the string makes an angle of 35° with the vertical. What is the speed of the sphere as it reaches the lowest point of its arc.

2. Relevant equations
KE= 1/2mv^2
PE=mgy

3. The attempt at a solution
I attempted to find the potential energy when it is at the top of its path and then ,since it will have the same amount of energy at the bottom, find the velocity using the formula for KE (air friction I believe is not taken into account). I unfortunately don't know how to find the height of the sphere based on the angle from the vertical.

2. May 17, 2014

### dauto

Draw a picture and use Geometry/Trigonometry

3. May 17, 2014

### BrainMan

Do I have to use circle geometry? If so what is the relationship between the y distance, the radius, and the vertical angle?

4. May 17, 2014

### SammyS

Staff Emeritus
Use trigonometry to come up with that.

5. May 17, 2014

### Regtic

if angle B was 35° then $a$ would be (c)(cosB)

Hope that helps

6. May 17, 2014

### BrainMan

What I tried to do now was to make a right triangle and do 1.2 cos theta to get the upper distance and then subtract that number from 1.2 to get x. I still got the wrong answer.

7. May 17, 2014

### BrainMan

The correct answer is .323 m/s

8. May 17, 2014

### haruspex

9. May 17, 2014

### BrainMan

1.2 cos 37 = .958
1.2 - .958 = .242
mgy = .889
1/2mv^2= .889
v^2 = 4.74
v = 2.18

10. May 17, 2014

### haruspex

It isn't 37.

11. May 17, 2014

### BrainMan

I still am not getting the correct answer even though I am using 35°. It still comes out larger than the answer.

12. May 17, 2014

### haruspex

Strictly speaking, you are using the wrong length for the pendulum. You need to add the radius of the sphere. However, that involves looking up the density of steel to figure out the radius.
Similarly, not all the energy goes into linear motion. The sphere rotates as it swings. That also appears to involve finding the radius, but I think you'll find it cancels out in this case.
Try taking the rotation into account but continuing to use 1.2m for the length.

13. May 17, 2014

### BrainMan

What do you mean by taking the rotation into account?

14. May 17, 2014

### haruspex

I mean as in $\frac 12 I \omega^2$.
But let me correct an error in my previous post - I don't think the radius will cancel out. That happens in the case of a rolling sphere, but not here. So I would suggest: Let the radius of the sphere be r and obtain a purely symbolic expression for the answer. Don't plug any numbers in yet. Post what you get, maybe an outline of the working too. Let's get that right before putting in the numbers.

15. May 17, 2014

### BrainMan

Can you clarify what that formula means? I am not aware of what the symbols represent.

16. May 17, 2014

### BrainMan

Is that the formula for the KE?

17. May 17, 2014

### haruspex

Yes, it's the formula for rotational KE. I is the moment of inertia about the point of rotation, ω is the angular speed.
I recommend treating the sphere (at its lowest point) as having horizontal linear motion speed v at the sphere's centre, and rotating about that centre.
Do you know the formula for moment of inertia of a uniform solid sphere about its centre?
Can you write out the relationship between v, ω, r and l?

18. May 18, 2014

### BrainMan

No I do not

19. May 18, 2014

### haruspex

20. May 18, 2014

### BrainMan

I unfortunately am having trouble understanding what your approach to the problem is.