# Tag Info

0

This is what I did and I think is simple and right : Assume $v$ linear speed of centre of mass downwards and $\omega$ angular speed around it. Use the fact the bottom point has no vertical speed to find relation between $v$ and $\omega$. And I am done.

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 $E= K + P$ ...

5

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: $$E_l = \frac{1}{2}m v^2$$ ...

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 ...

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 ...

0

Case a) Body with uniform motion (no rotation), with C the center of the disk and A a point on the edge (for example below, at a distance $R$).  \begin{aligned} \vec{v}_A & = (v,0,0) \\ \vec{\omega} & = (0,0,0) \\ \vec{v}_C & = \vec{v}_A + (0,-R,0) \times \vec{\omega} = (v,0,0) \\ \vec{L} & = m \vec{v}_C = (m v,0,0) \\ \vec{H}_A & = ...

0

From your description, I assume the disk only only translating, not rotating. Is this correct? If so, read on. If not, I'll delete. I'm uncomfortable with the first method that uses $L=I\omega$. In this equation, it is assumed that every point on the rigid body can be characterized by the same angular velocity $\omega$. From your description of the motion ...

0

The first method is correct. The second method is wrong because the equation you use only applies to point particles, not continuous masses with volume (such as a disk). You're incorrectly treating the disk as a point particle located at the disk center. If you want to use the second method, you'll need to use this equation for angular momentum of ...

0

Euler's rotation theorem states that every rigid motion that leaves some point fixed is equivalent to a rotation about an axis running through that point. It is easy to see that every rigid motion can therefore be decomposed in the prescribed manner (i.e., by first moving the centre of mass to its new location).

1

You have a differential equation that says $$a(x) = -0.01*w = \frac{d w}{d t}$$ What you did with the change of variables is correct, so $w$ cancels on either side. Otherwise you have a first order differential equation to solve.

Top 50 recent answers are included