How Is Maximum Kinetic Energy Calculated for an Object on a Circular Path?

[duplaimp]

Hi,
I am trying to solve a problem but I can't figure how to continue..
There's an object moving in a circular path with radius 0.5m in an horizontal surface. The rope will break if the tension exceeds 16N. What is the maximum kinect energy?

So r = 0.5m

T = 16N

K=$\frac{1}{2}$m$v^{2}$

How can I find m and v? I tried with T = mg <=> 16 = m(9.8) <=> m = 1.63kg but I don't know if it is correct. But besides that I can't figure how to find v

 

[Doc Al]

Hint: Make use of the fact that the motion is circular.

 

[duplaimp]

I also tried that

v = $\sqrt{a*r}$

But how to find a? Will it be g?

 

[MrWarlock616]

yes..

 

[duplaimp]

It is solved :) Thanks!

 

[Doc Al]

I also tried that

v = $\sqrt{a*r}$

But how to find a?

Just rearrange that to write a in terms of v and r. Then apply Newton's 2nd law.

Will it be g?

No.

 

[duplaimp]

Just rearrange that to write a in terms of v and r. Then apply Newton's 2nd law.

No.

But with that I don't get the right answer. Just to confirm: is the mass correct?

 

[Doc Al]

But with that I don't get the right answer.

Show what you did with it.

Just to confirm: is the mass correct?

You mean the value of the mass you calculated in post #1? No, that's not correct. For some reason, you set the tension equal to the weight of the mass. Why would you do that?

Hint: You don't need the mass to answer the question. It asks for kinetic energy, not mass.

 

[duplaimp]

Show what you did with it.

a = $\frac{v^{2}}{r}$ and then F = m*a <=> F = m*$\frac{v^{2}}{r}$
<=> 16 = 1.63 * $\frac{v^{2}}{0.5}$

But since 1.63 isn't right it will give wrong results

You mean the value of the mass you calculated in post #1? No, that's not correct. For some reason, you set the tension equal to the weight of the mass. Why would you do that?

Hint: You don't need the mass to answer the question. It asks for kinetic energy, not mass.

I did that because I couldn't figure any other way to find the mass.. trial and error

But K=$\frac{1}{2}$m$v^{2}$ so I need the mass

 

[Doc Al]

a = $\frac{v^{2}}{r}$ and then F = m*a <=> F = m*$\frac{v^{2}}{r}$

Perfect so far.

<=> 16 = 1.63 * $\frac{v^{2}}{0.5}$

But since 1.63 isn't right it will give wrong results

Yes, your value for mass is wrong. (Hint: There's not enough information to determine the mass. But you don't need it!)

I did that because I couldn't figure any other way to find the mass.. trial and error

But K=$\frac{1}{2}$m$v^{2}$ so I need the mass

You do not need the mass, you need the kinetic energy. So you just need to solve for 1/2mv². Go back and stare at the first equation in this post.

 

[duplaimp]

Perfect so far.

Yes, your value for mass is wrong. (Hint: There's not enough information to determine the mass. But you don't need it!)

You do not need the mass, you need the kinetic energy. So you just need to solve for 1/2mv². Go back and stare at the first equation in this post.

Ok, finally I figured that :P

So, i did this:
F = ma <=> a = $\frac{F}{m}$

a = $\frac{v^{2}}{r}$ <=> $v^{2}=a*r$ <=> $v^{2}=\frac{F}{m}*r$

K = $\frac{1}{2}*m*\frac{F*r}{m} = \frac{1}{2}*F*r$

Thanks!

 

[Doc Al]

Ok, finally I figured that :P

So, i did this:
F = ma <=> a = $\frac{F}{m}$

a = $\frac{v^{2}}{r}$ <=> $v^{2}=a*r$ <=> $v^{2}=\frac{F}{m}*r$

K = $\frac{1}{2}*m*\frac{F*r}{m} = \frac{1}{2}*F*r$

Thanks!

Cool.

A short cut would be to recognize mv² when you see it:

$$F = \frac{m v^2}{r}$$
Thus:
$$m v^2 = F r$$
$$\frac{1}{2}m v^2 = \frac{1}{2}F r$$