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When you compute the energy of a rotating object you either consider the tangential velocity or the angular velocity. The two expressions are two different ways to look at the same observable, they both lead exactly to the same result and summing them is a mistake. For instance for a point like mass on a straight line we have: $$E_l = \frac{1}{2}m v^2$$ ...


4

No, it does not gain energy. The confusion arises because there's a force that does no work. If the car moves a distance $d \vec l = \vec v dt$, then the work done during that time $dt$ is $dW_{rope} = \vec F_{rope}\cdot d \vec l= 0$. This follows because $ \vec F_{rope}$ and $d \vec l$ are always perpendicular to each other (draw and check this!) which ...


1

The translational kinetic energy is simply $\frac{1}{2} m v^2$ where $v$ is the velocity of the center of mass. Rotational kinetic energy is $E_r = \frac{1}{2} I \omega^2$. To solve the problem, we must write the velocity of the rod as function of $\omega$ (or vice versa). Consider the above image. (Note that my convention for $\theta$ is different from ...


1

The potential energy $ P $ of the car doesn't change (the car stays on the ground the whole time), and because it moves uniformly in a circle, its speed $\left | \mathbf{V} \right |$ doesn't change. But the kinetic energy $ K $ of the car is $\frac {1}{2} m\left | \mathbf{V} \right |^2 $, so it doesn't change aswell. Hence, the total energy $ E= K + P $ ...


1

$\def\om{\omega}\def\vr{{\vec r}}\def\l{\left}\def\r{\right}\def\ve{{\vec e}}\def\vom{{\vec\omega}}\def\ds{\,'}$ Let the car move in the (x,y)-plane, let $m$ be the car's mass and let $J$ be the moment of inertia for the rotation about the axis through the center of mass aligned parallel to the z-direction. If you have a straight line and a circle with ...



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