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1

The force due to sliding friction causes the center of mass to decelerate - since it is the only external force acting on the ring, and we know the center of mass moves as though all forces act there. You might find this earlier answer of mine and the links therein helpful. As for the "in the real world the ring will stop completely" part of your question - ...


3

For most of the trials the racket head reached peak speed just at the time of impact. The racket head showed great acceleration just before impact where the racket head speed went from around 10 m/s to its peak value [35 m/s] in less than 0.1 seconds Probably tou have misinterpreted the text, it nowhere says that the acceleration gives more power, ...


1

Several things here. First - how efficiently you transfer momentum depends in part on whether you hit the "sweet spot". The location of the sweet spot is a function of the instantaneous center of rotation - an almighty swing from the shoulder puts the center of rotation further back so the same velocity of the racket might be less efficient. That depends on ...


1

They are the same thing. Parts of the racket though produce more force than others, suppose I completely missed the strings and hit the metal part (that's why without technique is so important). Usually, when you play you hit in what is called the sweet spot which is a vibration node and although it may not give you the biggest pop or bounce it gives you ...


8

Use the spatial inertia to relate linear/angular momentum to changes in linear/angular speed $$\begin{pmatrix} \vec{L}\\ \vec{H}_A \end{pmatrix} = \begin{bmatrix} m {\bf 1}_{3×3} & -m [\vec{c}\times] \\ m [\vec{c}\times] & I_{cm}-m [\vec{c}\times][\vec{c}\times] \end{bmatrix} \begin{pmatrix} \Delta\vec{v}_A\\ \Delta \vec{\omega} \end{pmatrix}$$ ...


0

you can use GOM playef pres F to count revolutions if you count 6 frames per revolution 6x5= 30 frames then you spin is 5 rev per second thats is tha same 5x60 second= 300 rpms, or use iphone Spinorama free app


0

Assuming it is an elastic collision, you will have, right after it, a ball with the same rolling motion (anticlockwise) but translating in the opposite direction at $v_0$. Now there is friction because the ball is "sliding", and the new equilibrium movement will be when both are moving without sliding again at $v_k$. Let me know if you do not know how to ...


2

When you fix a reference point (take it to be the origin of your reference frame) you can write the position as $$\vec{r} = r \hat{r}$$ where $\hat{r}$ is the unit vector pointing toward the particle. Deriving you obtain $$\vec{v} = \frac{dr}{dt} \hat{r}+ r \frac{d \hat{r}}{dt}$$ The first term is the radial component of the velocity, the second one is ...


4

I'm not an engineer, but this is how id do it. According to your rules, we can use a computer, and Audacity. You get a pair of headphones. You plug the headphones into the microphone jack on your computer. You open Audacity. You get a very small magnet. You glue the very small magnet to one of the fan blades. You turn the fan on. You hold one of the ...


0

This is a problem that involves only calculation of velocities from other velocities, no influence (forces) needs to be considered. Imagine the system as it appears in the inertial system where the point of contact of the two wheels is at rest. Since the contact point is at rest, the mass points of both wheels that touch each other are at rest. Since any ...



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