Translational Kinetic energy, am i squaring vectors correctly?

In summary, the conversation discusses the calculation of the total kinetic energy of a system of three masses with their respective velocities. The equations and attempts at solving the problem are mentioned, with emphasis on finding the velocity of the center of mass and using it in the kinetic energy formula. However, it is noted that two different methods were attempted and both resulted in an incorrect solution. The expert summarizer suggests that the mistake may lie in not considering the individual kinetic energies of the masses and only focusing on the center of mass velocity.
  • #1
Kibbel
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Homework Statement


m1 = 5 kg, v1 = < 11, -4, 12 > m/s
m2 = 7 kg, v2 = < -14, 6, -6 > m/s
m3 = 5 kg, v3 = < -27, 35, 24 > m/s

What is the total kinetic energy of this system?

Homework Equations



Momentum(total) = Mass(total)*Velocty(of Center of Mass)

Ktranslation = (Mass(total)*(Velocity of Ceneter of mass)^2)/2

The Attempt at a Solution



I followed this as close to the textbook as possible and still two ways i did it i got it wrong.

I started off by finding the momentum of the entire system by doing m1v1+m2v2+m3v3. Then i divided that by the total mass (m1+m2+m3). this gave me the velocity of the center of mass.

One i had that i plugged that directly into the equaltion to find the total kinetic energy of the system. v << c, and there is no Kinetic vibrational or rotational energy, so its just translational.

so i plugged that into the K translational formula, but neither solutions i got worked out.

I know my velocity of center of mass is correct at <-10.47, 11.59, 8.118>

So i did the dot product multiplication of that to itsself, and got 309.85. squared that to get 96007.6, then mutiplied by 17/2, to get 816060 Joules, which is not accurate.

(17/2)(<-10.47,11.59,8.118>)^2 = 816060 Joules.

other solution i tried was finding the magnitude of Velocity of center of mass, and squaring that and multiplying it by 17/2. which ends up working out exactly the same.


can someone tell me what I am doing wrong? I've followed every rule and haven't ommitted everything and I've done the problem several times so i know I am not making a small mistake
 
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  • #2
Think about two masses with equal but opposite momenta. The velocity of the center of mass is zero, but the total kinetic energy of the system is obviously not since the masses are moving. You need to calculate the kinetic energy of the masses separately and then add them up.
 
  • #3
thanks vela, i should've looked at this more conceptually instead of just jumping in and plugging in numbers
 

1. What is Translational Kinetic Energy?

Translational Kinetic Energy is the energy possessed by an object due to its motion. It is a form of energy that an object has by virtue of its motion and is directly proportional to its mass and the square of its velocity.

2. How is Translational Kinetic Energy calculated?

The formula for calculating Translational Kinetic Energy is E = 1/2 * m * v^2, where E is the energy, m is the mass of the object, and v is the velocity of the object.

3. Why is the velocity squared in the formula for Translational Kinetic Energy?

The velocity is squared in the formula because kinetic energy is directly proportional to the square of the velocity. This means that as the velocity increases, the kinetic energy increases exponentially.

4. Am I squaring vectors correctly in the formula for Translational Kinetic Energy?

No, the formula for Translational Kinetic Energy does not involve squaring vectors. It only involves squaring the magnitude of the velocity, which is a scalar quantity. Vectors are not squared in the formula.

5. What are some real-life examples of Translational Kinetic Energy?

Some examples of Translational Kinetic Energy in everyday life include a moving car, a swinging pendulum, a rolling ball, and a falling object. Essentially, any object that is in motion possesses Translational Kinetic Energy.

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