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Suppose a bus with mass $m_1$ moving horizontally with velocity $\dot{x}_1$, and a person with mass $m_2$ running with velocity $\dot{x}_2$ inside the vehicle. What is the total kinetic energy of the system?

I know that kinetic energy is given by

$$T = \frac{mv^2}{2}$$

and I know that velocity (and therefore kinetic energy) depends on a reference frame. If we take two generalized (independent) reference frames $x_1$ and $x_2$ that are fixed outside the bus and at the same position, is the total kinetic energy $a$, $b$, other?

$$a) \quad T = \frac{m_1\dot{x}_1^2}{2} + \frac{m_2\dot{x}_2^2}{2}$$ $$b) \quad T = \frac{m_1\dot{x}_1^2}{2} + \frac{m_2(\dot{x}_1 + \dot{x}_2)^2}{2}$$

Obs: this is the very first part of a question where I will later apply Lagrange's equation. The system is a little bit more complex, but I am only having difficulties to interpret this two bodies.

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  • $\begingroup$ using the reference outside the bus it's just the sum of the two KE's. Mass of bus, bus velocity, mass of person, bus + running speed velocity. I don't know what your equations mean. $\endgroup$
    – Tiger Guy
    Feb 12, 2022 at 0:53

1 Answer 1

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first part

For this part of the question:

Suppose a bus with mass $m_1$ moving horizontally with velocity $\dot{x}_1$, and a person with mass $m_2$ running with velocity $\dot{x}_2$ inside the vehicle. What is the total kinetic energy of the system?

the answer is b) i.e. $$T = \frac{m_1\dot{x}_1^2}{2} + \frac{m_2(\dot{x}_1 + \dot{x}_2)^2}{2}$$

also, I would suggest to write this as:

$$T = \frac{m_1\dot{x}_{1|O}^2}{2} + \frac{m_2(\dot{x}_{1|O} + \dot{x}_{2|1})^2}{2}$$

where:

  • $\dot{x}_{1|O} $ : is the velocity of the bus with respect to the inertial frame
  • $\dot{x}_{2|1} $ : is the velocity of the person (2) with respect to the bus frame (1)

Second part

However, this part

If we take two generalized (independent) reference frames $x1$ and $x2$ that are fixed outside the bus and at the same position, is the total kinetic energy a, b, other?

shows a bit of confusion.

I assume that you mean to say that x1 is the position vector for the bus and x2 is the position vector for the person inside, and both of them refer to the inertial system outside the bus. In that case, to avoid confusion I would use the symbols $x_{1|O}$, x_{2|O}, O being the origin of the axis in the inertial reference system. In that case, the answer would be:

$$T = \frac{m_1\dot{x}_{1|O}^2}{2} + \frac{m_2\dot{x}_{1|O}^2}{2}$$


Difference in the two.

The velocity of the bus is the same in both cases i.e. $\dot{x}_{1|O}$.

The difference is in the velocity of the person. I.e.

$$ \dot{x}_{2|O}= \dot{x}_{1|O} + \dot{x}_{2|1}$$

i.e. the velocity of the person on the bus from an observer fixed on the inertial system O ( $\dot{x}_{2|O}$), would be equal to the velocity of the bus on the inertial system ( $\dot{x}_{1|O}$) plus the velocity of person with respect to the bus frame $\dot{x}_{2|1}$

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  • $\begingroup$ Maybe I have write down poorly my question. The velocity $\dot{x}_2$ is with respect to $O$, i.e., it is already $\dot{x}_2$ plus the relative speed of the person to the buss. Then, should the answer be $a)$ instead? I will rewrite the question to avoid confusion. $\endgroup$
    – Iuri
    Feb 12, 2022 at 1:37
  • $\begingroup$ I think the comment is also poorly written. ”The velocity $\dot {x}_2$ is with respect to O, i.e., it is already $\dot {x}_2$ plus the relative speed of the person to the buss". What is the difference then between the first and the second $\dot {x}_2$? $\endgroup$
    – NMech
    Feb 12, 2022 at 1:56
  • $\begingroup$ Let me try again. $\dot{x}_{1|O}$ is the speed of $m_1$ w.r.t. $O$, the fixed frame of reference outside the bus. $\dot{x}_{2|O}$ is the speed of $m_2$ w.r.t. $O$. $\dot{x}_{2|1}$ is the difference between $\dot{x}_{2|O}$ and $\dot{x}_{2|O}$. The thing is, in order to apply Lagrange's equation (to get an equation of motion) we need generalized coordinates, which are independent of each other. That is why I think the kinetic energy is $a)$, but I do not understand quite well why. $\endgroup$
    – Iuri
    Feb 12, 2022 at 6:40
  • $\begingroup$ What is your interpretation/understanding of generalized coordidates? Maybe the problem is there. Also, without trying to be patronizing (because I am known way too ofter to be as hasty when writing replies), a friendly advice is read the sentences after you type them. The following was meant to be a clarification but it still is carelessly typed "$\dot{x}_{2|O}$ is the speed of m2 w.r.t. O. $\dot{x}_{2|1} $is the difference between $\dot{x}_{2|O}$ and $\dot{x}_{2|O}$." $\endgroup$
    – NMech
    Feb 12, 2022 at 7:22

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