# Is the velocity of a string that is being rotated around a central point the same at any point on the string?

Say that you have a weight attached to a 2 metre long string, and you are rotating the weight at 5 m/s. Is every point on the string going to be rotating at that same velocity of 5 m/s, or is the velocity of the string going to change according to how far away you are from the centre of rotation?

I'm looking at the linear velocity.

• It is the angular velocity being same on every point on the string not velocity . Jul 1, 2019 at 8:40

For different points of the string to have same velocities, their direction of movement and speeds would to be the same as well. If the string is taut then all points are moving in the same direction at any given time, but it's easy to show that the speeds are not the same: If you have your 2 meter long string, the endpoint will move the total distance of 2·π·(2 m)≈12,6 meters during one full revolution. If you specify the speed of 5 m/s, one revolution would then take 4π/5 s ≈ 2,5 seconds.

Clearly one full revolution takes the same time for all points on the string (if it's considered taut, rigid). But where the endpoint travels more than 12 meters in one revolution, points closer to the rotational axis end up moving a smaller distance as any point closer will trace a circle that is smaller than the one drawn by the endpoint. Thus the points closer in travel a smaller distance in the exact same time and hence their speed is less.

(Linear) Velocities therefore can't be the same.

For an object rotating about an axis, every point on the object has the same angular velocity. The tangential velocity of any point is proportional to its distance from the axis of rotation, i.e. $$\mathbf{v_{\perp}} = \boldsymbol{\omega} \times \mathbf{r}.$$ So it depends what you mean by "the velocity of the string."

Depending on the 'type' of velocity.

• Angular velocity will be constant for every particle / unit length of the string i.e. 5/2 revolutions s-1

• Linear velocity will change according to the distance from the point of rotation V= r. w, where V is the linear velocity, r is the length of the string from the point of rotation and w is the angular velocity

• You've got it the wrong way around.
– nasu
Jul 1, 2019 at 14:14
• Thank you for the correction Jul 1, 2019 at 14:16
• Angular velocity is not 5 m/s. This the magnitude of the linear velocity at the end of the string.
– nasu
Jul 1, 2019 at 14:23