# Find maximum kinect energy

by duplaimp
Tags: energy, kinect, maximum
 P: 33 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
 Mentor P: 41,543 Hint: Make use of the fact that the motion is circular.
 P: 33 I also tried that v = $\sqrt{a*r}$ But how to find a? Will it be g?
 P: 154 Find maximum kinect energy yes..
 P: 33 It is solved :) Thanks!
Mentor
P: 41,543
 Quote by duplaimp 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.
P: 33
 Quote by Doc Al 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?
Mentor
P: 41,543
 Quote by duplaimp 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.
P: 33
 Quote by Doc Al 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

 Quote by Doc Al 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
Mentor
P: 41,543
 Quote by duplaimp 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.
P: 33
 Quote by Doc Al 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!
Mentor
P: 41,543
 Quote by duplaimp 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:
$$m v^2 = F r$$
$$\frac{1}{2}m v^2 = \frac{1}{2}F r$$

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