Weight after sliding off a ramp

[Arquon]

Homework Statement

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

Homework Equations

E_k = mv² / 2
E_p = mgh
Work = F * h

The Attempt at a Solution

[/B]
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

 

[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 ?

Weight = Work / height = 2Rmg / 2R = mg

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,
( [edit] 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.

 

[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.

 

[BvU]

No given equations or acceleration

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)

 

[Arquon]

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

 

[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 ?

 

[Arquon]

a = v² / R ?

 

[BvU]

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

 

[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.

 

[Arquon]

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

 

[BvU]

Velocity at the top is zero

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

 

[Arquon]

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

 

[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' ?

 

[Arquon]

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

 

[BvU]

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

 

[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

 

[BvU]

At the top it's not 6