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Is kinetic energy of center of mass of a system is equal to KE of of system?

by vkash
Tags: energy, equal, kinetic, mass
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vkash
#1
Mar5-12, 05:27 PM
P: 318
Does Kinetic energy(KE) of center of mass is equal to kinetic energy of system?
I think no...
Example take two identical particle moving with velocity v and -v. kinetic energy is [tex]mv^2[/tex] but KE of center of mass is zero>
have a look at this example.
A block of mass m is pushed towards towards a movable wedge of mass nm. and height h, with a velocity v. All the surfaces are smooth. Find the minimum value of u for which block will reach top of height.
one of the equation placed(in book) to solve this question is
Work done =change in Kinetic energy
[tex]-mgh=\frac{(m+nm)V^2}{2}-\frac{mu^2}{2}[/tex] (v=velocity of center of mass before collision)
(where this latex code is wrong)
I think this equation is incorrect. Because final Kinetic energy can't be written as kinetic energy of center of mass.
Am I correct??

Secondly i failed to solve this question. Any hint.,
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Syrianto
#2
Mar5-12, 08:40 PM
P: 2
Hey! First post here so maybe I can be of help and not give wrong info haha. To start with your initial question, no. The kinetic energy of the center of mass is not equal to the kinetic energy of the system. Kinetic energy of the system is the summation of all of the particles/objects in that system. So if you have two objects the total kinetic energy = 0.5(m1)(v1)^2 + 0.5(m2)(v2)^2

So yeah, for your actual problem. You have kinetic energy at the start and all of that is turning into potential energy when it goes up the block (if the initial speed is a minimum). Considering that energy is conserved, initial energy (kinetic) = final energy (potential)

Don't think I'm supposed to give away too much so maybe I should leave it at that and come back if you get it from there? Good luck!
ehild
#3
Mar6-12, 12:29 AM
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Quote Quote by vkash View Post
Does Kinetic energy(KE) of center of mass is equal to kinetic energy of system?
I think no...
Example take two identical particle moving with velocity v and -v. kinetic energy is [tex]mv^2[/tex] but KE of center of mass is zero>
have a look at this example.

one of the equation placed(in book) to solve this question is
Work done =change in Kinetic energy
[tex]-mgh=\frac{(m+nm)V^2}{2}-\frac{mu^2}{2}[/tex] (v=velocity of center of mass before collision)
You are right, the KE is the sum of the KE-s of all participants, but the whole KE is equal to the sum of the KE of the CM and the KE of all parts relative to the CM, that is Ʃ1/2 mi (v-VCM)2.

First the total KE is that of the block. If it goes up along the wedge it pushes it and slows down with respect to it. If it just reaches the top, its relative velocity with respect to the wedge is zero, the whole system moves together as a single body, with the velocity of the CM.
As there are internal forces only, the velocity of the CM does not change.

Edit: As there are only internal forces in the horizontal direction, the horizontal velocity of the CM does not change.

Find the velocity of the center of mass before collision and substitute for V into the equation. Solve for u.

(About Latex: Do not use "[bold]" inside it. )

ehild

ehild
#4
Mar6-12, 12:36 AM
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P: 10,673
Is kinetic energy of center of mass of a system is equal to KE of of system?

Hi Syrianto,

welcome at PF.

Quote Quote by Syrianto View Post
So yeah, for your actual problem. You have kinetic energy at the start and all of that is turning into potential energy when it goes up the block (if the initial speed is a minimum). Considering that energy is conserved, initial energy (kinetic) = final energy (potential)
That is not right. The energy of the whole system conserves. When the block slides upward along the wedge, it pushes it and the wedge starts to move. It accelerates till the block has velocity relative to it. At the end the whole thing moves together as a single body. The final KE is not zero.


ehild
vkash
#5
Mar6-12, 05:04 AM
P: 318
Quote Quote by ehild View Post
You are right, the KE is the sum of the KE-s of all participants, but the whole KE is equal to the sum of the KE of the CM and the KE of all parts relative to the CM, that is Ʃ1/2 mi (v-VCM)2.

First the total KE is that of the block. If it goes up along the wedge it pushes it and slows down with respect to it. If it just reaches the top, [highlight]its relative velocity with respect to the wedge is zero, the whole system moves together as a single body, with the velocity of the CM.[highlight]
As there are internal forces only, the velocity of the CM does not change.

Find the velocity of the center of mass before collision and substitute for V into the equation. Solve for u.

(About Latex: Do not use "[bold]" inside it. )

ehild
thanks!
since there is no horizontal force so there will no change in velocity of center of mass in horizontal direction.
vkash
#6
Mar6-12, 05:05 AM
P: 318
Quote Quote by Syrianto View Post
Hey! First post here so maybe I can be of help and not give wrong info haha. To start with your initial question, no. The kinetic energy of the center of mass is not equal to the kinetic energy of the system. Kinetic energy of the system is the summation of all of the particles/objects in that system. So if you have two objects the total kinetic energy = 0.5(m1)(v1)^2 + 0.5(m2)(v2)^2

So yeah, for your actual problem. You have kinetic energy at the start and all of that is turning into potential energy when it goes up the block (if the initial speed is a minimum). Considering that energy is conserved, initial energy (kinetic) = final energy (potential)

Don't think I'm supposed to give away too much so maybe I should leave it at that and come back if you get it from there? Good luck!
welcome to PF!

You got it little wrong actually wedge is also movable so it is required to calculate the change in KE of that wedge....
ehild
#7
Mar6-12, 06:40 AM
HW Helper
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P: 10,673
Quote Quote by vkash View Post
thanks!
since there is no horizontal force so there will no change in velocity of center of mass in horizontal direction.
You are right, I meant that there are no horizontal external forces.

ehild
Syrianto
#8
Mar6-12, 11:09 AM
P: 2
Ohh, sorry for putting out bad info then D=

When he said all surfaces were smooth I assumed that it meant the block + floor, and the ramp part of the wedge. If the bottom of the wedge was firmly planted to the ground would my statement be correct?


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