# Tag Info

## New answers tagged rotational-kinematics

1

That line will get Coriolis acceleration $$\vec{a} = -2 \vec{\Omega} \times \vec{v}$$ ($\Omega$ is the angular speed of the earth's rotation, with a direction pointing into the ground from the view of the south pole). As it's going across the pole, there's a right angle between $\Omega$ and $v$ and the absolute value will be simply $$a = 2\Omega v$$ and the ...

2

Your angular velocity vector is $$\vec{\omega} = \Omega \frac{ \vec{r}_D - \vec{r}_A }{|\vec{r}_D - \vec{r}_A|}$$ where $\vec{r}_A = (0,0.2,0.12)$, $\vec{r}_D = (0.3,0,0)$, $\vec{r}_B = (0.3,0.2,0.12)$ in meters and $\Omega = 90\;{\rm rad/s}$. Your velocity kinematics is $$\vec{v}_B = \vec{\omega} \times ( \vec{r}_B - \vec{r}_A )$$ And acceleration ...

1

When the player hits the ball with top spin, it makes the ball, well, spin. By spinning, the ball will modify the airflow around itself and thus create an air pressure profile which will deflect the ball : this is the Magnus effect. So by applying top spin on the ball the way tennis players do, the ball is rotating in the direction of the trajectory. This ...

1

This is only true for engineering units which have $I$ in ${\rm lbf\,in^2}$. In the metric system the units of $I$ are ${\rm kg\, m^2}$. So to convert force ${\rm lbf}$ to mass you divide by $g$.

0

We want the condition angular speed to be zero .So use:$$\omega=d\theta/dt=0$$

0

If you define kinetic energy as $KE =\frac{1}{2} m (\vec{v}_G \cdot \vec{v}_G) + \frac{1}{2} I_G (\vec{\omega} \cdot \vec{\omega} )$ where $v_G$ is the linear velocity of the CG and $I_G$ is the mass moment of inertia about the CG, and then transform the quantities to the handle of the rigid body (i.e. the pin) then you will get what Wikipedia has.

0

Like linear momentum, angular moment is used to describe the degree of resistance to change in angular motion. There are two types of changes to consider. a) Change in rotation speed and b) Change in rotation direction. I think the first is kind of trivial to understand, but the second is a little more tricky. If you have vector that is riding on a rotating ...

Top 50 recent answers are included