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| <title>Balls of mass m, each connected by a thread, move in a circle with a constant velocity v. The kinetic energy of each ball, equal to \frac{mv^2}{2}, does not change. If we switch to a reference frame in which the center of the thread moves in a straight line in the plane of rotation at a speed , the energy of each of the balls changes from zero to 4 \frac{mv^2}{2}. What causes this energy change? Does the total kinetic energy change in the specified reference frame?</title> | | <title>Balls of mass m, each connected by a thread, move in a circle with a constant velocity v. The kinetic energy of each ball, equal to \frac{mv^2}{2}, does not change. If we switch to a reference frame in which the center of the thread moves in a straight line in the plane of rotation at a speed , the energy of each of the balls changes from zero to 4 \frac{mv^2}{2}. What causes this energy change? Does the total kinetic energy change in the specified reference frame?</title> |
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| <h3 id="back-link"><a href="../../#2.4">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.4">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.4.1.$ Balls of mass $m$, each connected by a thread, move in a circle with a constant velocity $v$. The kinetic energy of each ball, equal to $\frac{mv^2}{2}$, does not change. If we switch to a reference frame in which the center of the thread moves in a straight line in the plane of rotation at a speed $,$ the energy of each of the balls changes from zero to $4 \frac{mv^2}{2}$. What causes this energy change? Does the total kinetic energy change in the specified reference frame? | | $2.4.1.$ Balls of mass $m$, each connected by a thread, move in a circle with a constant velocity $v$. The kinetic energy of each ball, equal to $\frac{mv^2}{2}$, does not change. If we switch to a reference frame in which the center of the thread moves in a straight line in the plane of rotation at a speed $,$ the energy of each of the balls changes from zero to $4 \frac{mv^2}{2}$. What causes this energy change? Does the total kinetic energy change in the specified reference frame? |
| </p> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
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| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.4.1_1.png" | | <img src="2.4.1_1.png" |
| loading="lazy" width="350" /> | | loading="lazy" width="350" /> |
| <figcaption> | | <figcaption> |
| Relative velocity in an inertial frame of reference | | Relative velocity in an inertial frame of reference |
| </figcaption> | | </figcaption> |
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| <p> | | <p> |
| | | |
| In the initial, stationary frame of reference, the energy of the system is | | In the initial, stationary frame of reference, the energy of the system is |
| $$E_{k1} = 2\frac{mv^2}{2}$$ | | $$E_{k1} = 2\frac{mv^2}{2}$$ |
| Velocity of the system | | Velocity of the system |
| $$v_s = \frac{mv+mv}{2m} = v$$ | | $$v_s = \frac{mv+mv}{2m} = v$$ |
| In the inertial frame of reference of the system moving with velocity $\vec{v}$ | | In the inertial frame of reference of the system moving with velocity $\vec{v}$ |
| $$E_{k2} = 0 +\frac{m(2v)^2}{2}=4 \frac{mv^2}{2}$$ | | $$E_{k2} = 0 +\frac{m(2v)^2}{2}=4 \frac{mv^2}{2}$$ |
| To go to the inertial frame of reference, it is necessary to give the system of total mass $2m$ a velocity $v$ | | To go to the inertial frame of reference, it is necessary to give the system of total mass $2m$ a velocity $v$ |
| $$\Delta E_k = \frac{2mv^2}{2}$$ | | $$\Delta E_k = \frac{2mv^2}{2}$$ |
| Thus, the law of conservation of energy is fulfilled and there is no change in total kinetic energy in the reference frame | | Thus, the law of conservation of energy is fulfilled and there is no change in total kinetic energy in the reference frame |
| $$\boxed{E_{k2} = E_{k1} + \Delta E_k}$$ | | $$\boxed{E_{k2} = E_{k1} + \Delta E_k}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| In a moving frame of reference, the tension force does work. No. | | In a moving frame of reference, the tension force does work. No. |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Almaskhan Arsen<br> | | Almaskhan Arsen<br> |
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