Find maximum kinect energy

In summary, the conversation discusses a problem involving an object moving in a circular path with a radius of 0.5m on a horizontal surface. The rope has a maximum tension of 16N before breaking. The question is asking for the maximum kinetic energy of the object. The conversation goes on to discuss finding the mass and velocity of the object, but it is ultimately determined that the mass is not needed to solve for kinetic energy. The final formula for calculating kinetic energy is K = 1/2 * F * r.
  • #1
duplaimp
33
0
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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  • #2
Hint: Make use of the fact that the motion is circular.
 
  • #3
I also tried that

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

But how to find a? Will it be g?
 
  • #4
yes..
 
  • #5
It is solved :) Thanks!
 
  • #6
duplaimp said:
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.
 
  • #7
Doc Al said:
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?
 
  • #8
duplaimp said:
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.
 
  • #9
Doc Al said:
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

Doc Al said:
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
 
  • #10
duplaimp said:
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.
 
  • #11
Doc Al said:
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!
 
  • #12
duplaimp said:
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]
 

1. What is maximum kinect energy?

Maximum kinect energy refers to the highest amount of energy that can be generated and detected by a kinect sensor. This energy is typically in the form of infrared light and is used to track movements and gestures in a three-dimensional space.

2. How is maximum kinect energy measured?

The maximum kinect energy is measured in watts (W) or kilowatts (kW) and can be calculated by multiplying the voltage and current of the energy source. It can also be measured using specialized equipment such as a light meter or oscilloscope.

3. What factors affect the maximum kinect energy?

Several factors can affect the maximum kinect energy, such as the type and quality of the energy source, the distance between the sensor and the energy source, and any obstructions or interference in the environment that may affect the energy transmission.

4. Can the maximum kinect energy be increased?

Yes, the maximum kinect energy can be increased by using a more powerful energy source, optimizing the placement and alignment of the sensor, and minimizing any factors that may affect the energy transmission. However, there is a limit to how much the maximum kinect energy can be increased.

5. Why is finding the maximum kinect energy important?

Finding the maximum kinect energy is important as it determines the range and accuracy of the kinect sensor. A higher maximum energy allows for a larger tracking area and more precise detection of movements and gestures, which is crucial in applications such as motion capture, virtual reality, and gaming.

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