What is the force on the ball's lowest point in a vertical revolving motion?

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In summary, the ball has a centripetal force of -1.4 Newton at the top of the circle and 1.4 Newton at the bottom.
  • #1
fara0815
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Hello there!

My study mate and I have been trying to figure this one out but have not had success. This is the problem we would appreciate to get some help with:

A ball with mass of 20 grams is attached to a robe and is revolving vertikal in a circle. There is no energy beeing added to it. At the ball highest point, a force of 0.2 Newton is pulling on the robe. What is the force on the ball"s lowest point?

The statet result says 1.4 Newton.

The most logical way of solving this problem is that at the highest point, the sum of all forces is the the centripedal force minus the gravitational force:

[tex] 0.2 N = - mg + m\omega^2r [/tex]

At its lowest, the sum of the forces that act on the ball should be
[tex] F = + mg + m\omega^2r [/tex]

Is this assumption right? With those to equations we get 0.4 Newton, which sounds actually good to me. Maybe my professor is wrong ?!

Any help will be appreciated!
 
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  • #2
Your method looks OK to me - but using that same method I don't get 0.4N

Use the first eqn to find a value for mw^2 rAlthough neither do I get 1.4N - any suggestions anyone?
 
  • #3
fara0815 said:
The most logical way of solving this problem is that at the highest point, the sum of all forces is the the centripedal force minus the gravitational force:

[tex] 0.2 N = - mg + m\omega^2r [/tex]

At its lowest, the sum of the forces that act on the ball should be
[tex] F = + mg + m\omega^2r [/tex]
Two comments:
(1) You assume the same angular speed at top and bottom. Not so: At the bottom, the ball will be going faster. (Remember conservation of energy.)

(2) "centripetal force" is not a force! It's a term used to describe the net force acting towards the center in circular motion. So, the net force on the ball at the top is:
[tex]F_{net} = -T -mg[/tex]

This equals the "centripetal force", via Newton's 2nd law applied to circular motion:
[tex]F_{net} = - m v^2/r[/tex]

So:
[tex]-T -mg = - m v^2/r[/tex]

Where I take down to be negative. At the bottom, the tension and acceleration both point up. (And the speed and tension are different, of course.)
 
  • #4
Hmm, When we do this at A-level we always have the stone traveling at constant speed.

is it posible to solve this otherwise, with no further info?
 
  • #5
Sure you can solve it: Use conservation of energy.
 
  • #6
Yes, but when I tried that, I was left with r in the equation, which we don't have.

What have I missed?

...I'll keep trying...
 
Last edited:
  • #7
Cancel that, think I've got rid of the r and yes, it gives the answer required.

fara, from the top of the circle, when you find a value for mv^2/r, you have to be able to convert this to kinetic energy. The at the bottom, after you've added the KE gained, you have to be able to convert it back to a new value of mv^2/r

Unless there's a quicker way, Dr Al?
 
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  • #8
thank you very much guys for your help but unfortunately, I am totally clueless about converting the [tex] \frac{mv^2}{r} [/tex] into kinetic energy. It just does not appear to me ! :(
 
  • #9
(mv^2)/r and 0.5 mv^2

So to convert mv^2/r to KE you multiply by r/2

Use the data you've been given for the top pf the circle to find mv^2/r
Convert it to ke.

Add on the ke you gain (from GPE) from top to bottom of circle.

Now you have the total ke at the bottom. This time, to turn it back to mv^2/r you multiply by 2/r - handy because otherwise you´d be left with an r in the expression.

Now use your original T = mg + mv^2/r at the bottom to find the tension.

I got 1.38 doing it this way - which is close enough to what you want.
 
  • #10
yeah! I have got 1.377 N but I do not understand why you can multiply it by [tex]\frac{r}{2}[/tex] to have it converted into KE.
Here is what I figured out:

Situation 1, top :
[tex]-T_1-mg=-ma_1[/tex]
Situation 2, bottom:
[tex]T_2-mg=ma_2[/tex]

Thus
[tex]a_1=g+\frac{T_1}{m}=w^2_1r[/tex]
[tex]a_2=\frac{T_2}{m}-g=w^2_2r[/tex]

So with that I get the velocities:
[tex]V^2_1=(g+\frac{T_1}{m})r[/tex]
[tex]V^2_2=(\frac{T_2}{m}-g)r[/tex]

Conservation of Energy says:

KE+KP=KE
[tex]\frac{1}{2}mV^2_1+mg2r=\frac{1}{2}mV^2_2[/tex]

After solving it to [tex]T_2[/tex] I got:
[tex]2gm+T_1+4gm=T_2[/tex]
and that is 1.37 N!

We really appreciate your help guys!
 
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  • #11
hurrah!

as for the conversion, well they both contain mv^2 , one divided by 2 and one divided by r. So you use the r and the 2 to convert one to the other.
 
  • #12
well, sometimes it is so obvious that one cannot see it ;)
 

Related to What is the force on the ball's lowest point in a vertical revolving motion?

1. What is a stone revolving vertical?

A stone revolving vertical is a type of structure that consists of a large stone or boulder that is mounted on a vertical axis and can rotate or revolve around its axis.

2. How does a stone revolving vertical work?

A stone revolving vertical typically works through the use of a bearing system that allows the stone to rotate smoothly and with minimal friction. It may also be powered by a motor or manually rotated by a person.

3. What are the benefits of a stone revolving vertical?

A stone revolving vertical can serve as a unique and eye-catching piece of artwork or a functional structure for grinding grains or other materials. It can also provide a calming and meditative effect when watching the stone rotate.

4. Where can a stone revolving vertical be found?

Stone revolving verticals can be found in various locations, including parks, gardens, museums, and historical sites. They may also be found in private residences as decorative features.

5. How is a stone revolving vertical different from other types of rotating structures?

A stone revolving vertical is different from other rotating structures, such as windmills or watermills, because it typically relies on a single large stone for its rotation rather than multiple blades or paddles. It also serves a different purpose, such as for decorative or meditative purposes rather than energy production.

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