Weight after sliding off a ramp

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1. Jan 25, 2017

Arquon

1. The problem statement, all variables and given/known data
A ball with mass of m starts sliding off a ramp with height h = 3R and when it slides off the ball does a "loop" once. ( Just see the picture , it's hard to explain ). I have to find the weight of the ball at the bottom and at the top of the circle.

h = 3R

2. Relevant equations
Ek = mv2 / 2
Ep = mgh
Work = F * h
3. The attempt at a solution

I had found the weight of the ball at the top of the circle, although not sure if correct:

I found the potential energy (or work) at the top of the circle, which is 2Rmg
Weight = Work / height = 2Rmg / 2R = mg

And then I have to find the weight at the bottom of the circle, before doing the first loop, but I am not exactly sure how. A good hint would be awesome

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2. Jan 25, 2017

BvU

Hi,

Your relevant equations don't look as if they have anything with weight in them ? Also, $h$ appears but is never used after that.

Please inform us why the weight of the ball should be different at the top and at the bottom ? What is the definition of weight you intend to use ?
what is the reference point for your 'Work' ?

To me, weight is $mg$ just like what you end up with (by coincidence ?) However, if we agree that weight is mass times vertical acceleration,
(  Not good. If it lies still on the table, the weight isn't zero -- you' ll have to help me there )
then we might get
answers for top and bottom while in the loop that differ from $mg$.

What kind of trajectory is the loop ? Any equations for the acceleration ?

You also have to make an approximation: the size of the ball isn't given, so the best you can do is assume it can be ignored.

PS IMHO this is a rather badly thought out exercise; we'll see how far we can get.

3. Jan 25, 2017

Arquon

My guess is that Weight is different at the top and at the bottom because of different value of acceleration. I have correct answers, but I don't really know how to get the bottom weight. Bottom is 6mg, top is mg.

For Work, I did : Work = 2Rmg ( the work of force of gravity).

The loop has a form of a simple circle, as seen in the picture. No given equations or acceleration.

4. Jan 25, 2017

BvU

You will need something. If it isn't given, you're supposed to know !
Let's try to simplify a little first: what is the (expression for the) horizontal velocity at the end of the ramp when it enters the loop ? (hint: you can use your first two relevant equations for that)

5. Jan 25, 2017

Arquon

v = sqrt ( 2g * 3R) ? Got from kinetic and potential energy formula.

6. Jan 25, 2017

BvU

Excellent. Now, at that moment it is describing a loop with radius R. What is the expression for the acceleration needed to describe a circular loop ?

7. Jan 25, 2017

Arquon

a = v2 / R ?

8. Jan 25, 2017

BvU

The smile indicates you start to see the light !
Now what about the top of the loop ?

9. Jan 25, 2017

Arquon

Ok, I seem to have got correct answer for the bottom, 7mg ! ( I accidently typed 6mg before :P) Trying to get for the top now.

10. Jan 25, 2017

Arquon

Velocity at the top is zero, because potential energy is at maximum ? In that case, I get the answer mg, right?

11. Jan 25, 2017

BvU

Is that so ? What about $E_{\rm kin} = \Delta E_{\rm pot}$ ?

12. Jan 25, 2017

Arquon

I'm a little bit confused. Velocity is still sqrt( 6Rg ) at the top then ? Or rather, sqrt ( 4Rg ), because of 2R height ?

13. Jan 25, 2017

BvU

One way is to ask, the other is to calculate. You found ${1\over 2} mv^2 = mg\; 3R$ for a height difference of 3R. What is the height difference when it's at the top of the loop ? So what's $v^2$ there ? So what's $a$ there ? So what do you get for the 'weight' ?

14. Jan 25, 2017

Arquon

The height difference is 1R, v is sqrt ( 2Rg ), and weight is 3mg, is that right ?

15. Jan 25, 2017

BvU

No; work a bit more meticulous. At the bottom you added 6 and 1. Which way were they pointing ?

16. Jan 25, 2017

Arquon

1 was pointing downwards, while 6 upwards. So do you mean that when the ball is at the top, 6 is pointing downwards as well ? Sorry, I'm a little lost right now

17. Jan 25, 2017

BvU

At the top it's not 6