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Find maximum kinect energy

by duplaimp
Tags: energy, kinect, maximum
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duplaimp
#1
Jan1-13, 08:42 AM
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=[itex]\frac{1}{2}[/itex]m[itex]v^{2}[/itex]

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
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Doc Al
#2
Jan1-13, 08:52 AM
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Hint: Make use of the fact that the motion is circular.
duplaimp
#3
Jan1-13, 08:54 AM
P: 33
I also tried that

v = [itex]\sqrt{a*r}[/itex]

But how to find a? Will it be g?

MrWarlock616
#4
Jan1-13, 08:55 AM
P: 154
Find maximum kinect energy

yes..
duplaimp
#5
Jan1-13, 09:01 AM
P: 33
It is solved :) Thanks!
Doc Al
#6
Jan1-13, 09:03 AM
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Quote Quote by duplaimp View Post
I also tried that

v = [itex]\sqrt{a*r}[/itex]

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
#7
Jan1-13, 09:12 AM
P: 33
Quote Quote by Doc Al View Post
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
#8
Jan1-13, 09:22 AM
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Quote Quote by duplaimp View Post
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
#9
Jan1-13, 09:41 AM
P: 33
Quote Quote by Doc Al View Post
Show what you did with it.
a = [itex]\frac{v^{2}}{r}[/itex] and then F = m*a <=> F = m*[itex]\frac{v^{2}}{r}[/itex]
<=> 16 = 1.63 * [itex]\frac{v^{2}}{0.5}[/itex]

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

Quote Quote by Doc Al View Post
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=[itex]\frac{1}{2}[/itex]m[itex]v^{2}[/itex] so I need the mass
Doc Al
#10
Jan1-13, 09:46 AM
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Quote Quote by duplaimp View Post
a = [itex]\frac{v^{2}}{r}[/itex] and then F = m*a <=> F = m*[itex]\frac{v^{2}}{r}[/itex]
Perfect so far.
<=> 16 = 1.63 * [itex]\frac{v^{2}}{0.5}[/itex]

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=[itex]\frac{1}{2}[/itex]m[itex]v^{2}[/itex] 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
#11
Jan1-13, 10:01 AM
P: 33
Quote Quote by Doc Al View Post
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 = [itex]\frac{F}{m}[/itex]

a = [itex]\frac{v^{2}}{r}[/itex] <=> [itex]v^{2}=a*r[/itex] <=> [itex]v^{2}=\frac{F}{m}*r[/itex]

K = [itex]\frac{1}{2}*m*\frac{F*r}{m} = \frac{1}{2}*F*r[/itex]

Thanks!
Doc Al
#12
Jan1-13, 10:06 AM
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Quote Quote by duplaimp View Post
Ok, finally I figured that :P

So, i did this:
F = ma <=> a = [itex]\frac{F}{m}[/itex]

a = [itex]\frac{v^{2}}{r}[/itex] <=> [itex]v^{2}=a*r[/itex] <=> [itex]v^{2}=\frac{F}{m}*r[/itex]

K = [itex]\frac{1}{2}*m*\frac{F*r}{m} = \frac{1}{2}*F*r[/itex]

Thanks!
Cool.

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

[tex]F = \frac{m v^2}{r}[/tex]
Thus:
[tex]m v^2 = F r[/tex]
[tex]\frac{1}{2}m v^2 = \frac{1}{2}F r[/tex]


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