# Homework Help: Energy problem

1. Mar 20, 2015

### goonking

1. The problem statement, all variables and given/known data

2. Relevant equations
1/2 mv^2 = mgh

3. The attempt at a solution
first, a side question, if p = mv, can't we just place p into 1/2 mv^2 so it becomes 1/2 p^2?

anyway, on to the question, at the point where the cube falls off, the Fnormal should just about equal 0, correct?
is this the way to approach the problem? Or we can find the velocity where the cube loses contact with the sphere and plug that v into 1/2 mv^2 = mgh, solving for h.

Last edited: Mar 20, 2015
2. Mar 20, 2015

### Pierce610

3. No you can't because there is only v square while m isn't.
Personally I would prefer to consider the balance of two force, gravity and centripetal, looking at which speed the mass can still move of a rotational motion.

3. Mar 20, 2015

### goonking

so at the point the cube loses contact with the half-sphere, acceleration becomes 9.8 m/s^2.

a = v^2/r

9.8 x 10.3m = v^2
v= 10.04 m/s

1/2 m v^2 = mgh
mass cancels out

1/2 (10.04)^2 = g h

h = 5.15 meters.

is that correct?

4. Mar 20, 2015

### Satvik Pandey

When the ball will freely fall then it's total acceleration will be $g$. You just need here radial acceleration.

If the block looses contact with the sphere then what can you say about the magnitude of Normal reaction force acting on the block.
If the block has velocity at $v$ at the instant shown then what should be the centripetal force acting on it? This centripetal force it provided by which force? Equate them and get the answer.

5. Mar 20, 2015

### goonking

Centripetal force is provided by Fnormal? which is 0 when it loses contact with the half-sphere.

correct?

6. Mar 20, 2015

### Satvik Pandey

Yes Normal force will be zero when it looses contact. But centripetal force is provided by the component of mg. Can you find it?

7. Mar 20, 2015

### goonking

Centri. Force = m a

a = v^2/r = mg ?

8. Mar 20, 2015

### Satvik Pandey

I said a $component$ of mg. Can you find it in terms of mg and $\theta$?

9. Mar 20, 2015

### goonking

it should be the hypotenuse in the picture, which is mg / cos theta

how do we solve for theta?

10. Mar 20, 2015

### Satvik Pandey

It will be $mgcos\theta$

First equate this to centripetal force. You can find relation between $v$ and $\theta$ by conservation of energy.

11. Mar 20, 2015

### goonking

mgcosθ = m v^2 / r

gcosθ = v^2/r

correct?

12. Mar 20, 2015

### Satvik Pandey

Yes. Now use conservation of energy to find relation between v and theta.

13. Mar 20, 2015

### goonking

conservation of energy = 1/2 mvi^2 + mghi = 1/2 mvf^2 + mghf ?

14. Mar 20, 2015

### Satvik Pandey

What is the initial velocity of the block?
Can you represent hi and hf in terms of R and theta?

15. Mar 20, 2015

### goonking

initial velocity should be 0 since it was at rest.

hi is 10.3m

hf is what we are looking for

and i have no idea how to find theta.

16. Mar 20, 2015

### goonking

.

17. Mar 20, 2015

### Satvik Pandey

hf is $Rcos\theta$. Now proceed.

18. Mar 20, 2015

### goonking

but we have 2 unknowns, hf and theta.

19. Mar 20, 2015

### Satvik Pandey

$h_{f}=Rcos\theta$. Put this in energy equation. You would have two unknowns v and $\theta$. Remember the equation which you made by using centripetal force. Use these two equation and find $\theta$.

20. Mar 20, 2015

### goonking

is v = 14 m/s?

21. Mar 20, 2015

### Satvik Pandey

I think you have done some mistake. The equations are
$2mgR(1-cos\theta)=mv^{2}$

and $Rgcos\theta=v^{2}$

Find $\theta$. You don't have to find v.

22. Mar 20, 2015

### goonking

angle should be 48 degrees

23. Mar 20, 2015

### Satvik Pandey

Yes $cos\theta=\frac{2}{3}$. From this can you find the height above which it looses contact with the ball?

24. Mar 20, 2015

### goonking

6.89 m

thank you!

25. Mar 20, 2015

### Satvik Pandey

You are welcome!

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