Can Somebody check my center of mass answer?

In summary, a man standing on a railroad flatcar can change the velocity of the car by running to the left and jumping off at the left end. By using the formula v_{cf} = \frac{v_{ci}(m_c+m_m) - m_mv_{mcf}}{m_c+m_m}, we can find the final velocity of the car after the man jumps off. This is due to the fact that the center of mass velocity is always constant when there are no external forces.
  • #1
toesockshoe
265
2

Homework Statement


A railroad flatcar of mass Mc can roll without friciton along a straight horizontal track as shown in the diagram. Initially a man of mass Mm is standing on the car which is moving to the right relative to the ground with a speed of vi. What is the change in velocity of the car relative to the ground if the man runs to the left so that his speed relative to the car is given as vmc just before he jumps off at the left end?

the diagram is here: #5: http://nebula.deanza.edu/~Newton/4A/4AHWSet7.html

Homework Equations



[itex] x_{icm}=x_{fcm} [/itex]

The Attempt at a Solution


Because there are no external forces, we can say that the center of mass velocity is always constant correct?[/B]
so [itex] v_{icm}=v_{fcm} [/itex]

[itex] \frac{m_cv_{ci}+m_m+v_{ci}}{m_c+m+m} + \frac{m_cv_{cf}+m_mv_{mf}}{m_c+m_m} [/itex]

we know that the initial velocity of the car and man are teh same. the unknowns in that equation are final velocity of the car and the final velocity of the man... we are given the final velocity of the man respect to the car, so we need to convert it to the final velocity of the man respect to the ground using the following mnemonic:

[itex] v_{mc} = v_{mg} + v_{gc} [/itex] those should be vectors but idk how to do latex vectors
[itex] v_{mc} = v_{mg} - v_{cg} [/itex] i swapped the subscripts
we need Vmg... so
[itex] v_{mg} = v_{mc}+v_{cg} [/itex]
[itex] v_{cg} [/itex] is the same as [itex] v_{cfinal} [/itex] or [itex] v_{cf} [/itex] for short...
plug this back into formula i put on the 2nd line (the velocity equation)
also the masses in the denominator cancel out... so substituting v_cg (or v_cf) in for the equation gives us

[itex] v_{cf} = \frac{v_{ci}(m_c+m_m) - m_mv_{mcf}}{m_c+m_m} [/itex]

i know this isn't the change in velocity, but u can just subtract it from the given initial car speed. is this correct?
 
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  • #2
toesockshoe said:

The Attempt at a Solution


Because there are no external forces, we can say that the center of mass velocity is always constant correct?[/B]
so [itex] v_{icm}=v_{fcm} [/itex]

[itex] \frac{m_cv_{ci}+m_m+v_{ci}}{m_c+m+m} + \frac{m_cv_{cf}+m_mv_{mf}}{m_c+m_m} [/itex]

I'm sure you meant to have an equal sign instead of the middle plus sign.

Your work looks correct. I agree with your result.
 
  • #3
TSny said:
I'm sure you meant to have an equal sign instead of the middle plus sign.

Your work looks correct. I agree with your result.
yes yes, that should be an = sign. thanks!
 

Related to Can Somebody check my center of mass answer?

1. What is the center of mass?

The center of mass is the point in an object or system where the mass is evenly distributed in all directions. It can also be thought of as the point where the object would balance if placed on a fulcrum.

2. Why is it important to check the center of mass?

Checking the center of mass is important because it allows us to understand the stability and balance of an object or system. It is also crucial in determining the motion and behavior of an object under the influence of external forces.

3. How do you calculate the center of mass?

The center of mass can be calculated by finding the weighted average of the individual masses in an object or system. It is typically represented as a point in space with x, y, and z coordinates.

4. Can the center of mass change?

Yes, the center of mass can change depending on the distribution of mass within an object or system. If more mass is added or removed from a specific location, the center of mass will shift accordingly.

5. What happens if the center of mass is outside the object?

If the center of mass is outside the object, it means that the object is not balanced and will have a tendency to rotate or topple over. This can be seen in objects such as a leaning tower or a tilted book on a shelf.

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