# Instantaneous center of rotation

1. Apr 16, 2017

### edgarpokemon

<Mentor's note: ^Moved from a technical forum and therefore no template.>

at the instant shown during deceleration, the velocity of the tire is 40 ft/s to the right and the velocity of point A is 5ft/s to the right. locate the instantenous center of rotation.

Can the instantaneous center of rotation C be located below A? I also used Vo= Va + (w×ro/a) where w is the radial velocity and ro/a is the distance of o with respect to a. I get a value of 35 rad/s for w clockwise, but if it is clockwise, then shouldnt A be pointing to the left? help!

http://tinypic.com/r/2n7f7g2/9

Last edited by a moderator: Apr 16, 2017
2. Apr 16, 2017

### PeroK

What's your definition of "instantaneous centre of rotation"?

3. Apr 16, 2017

### edgarpokemon

as its name suggests?

4. Apr 16, 2017

### PeroK

Since you asked the following question:

It clearly isn't so obvious.

5. Apr 16, 2017

### edgarpokemon

so the center of rotation (C) will be located between points O and A? my books says that if two velocities are known and are parallel, it says to draw a perpendicular line from the tail of one of the vectors to the tail of the other velocity, and then draw a line from the tip of one vector to the tip of the other vector until it intersects with the perpendicular line, and that happens below point A.

6. Apr 16, 2017

### PeroK

That's not a definition. Anyway, I guess you mean:

https://en.wikipedia.org/wiki/Instant_centre_of_rotation#Pure_translation

In which case it is not necessarily part of the body. You need to imagine the whole plane as a rigid body and find a point on the plane which is instantaneously at rest.

One hint, although perhaps it's too late, is that you don't need to calculate the angular velocity. Think instead about the motion relative to the centre of the wheel. The wheel is moving at $40ft/s$, a point on the rim directly below the centre is moving at $5ft/s$, so that's $-35ft/s$ relative to the centre. So, which point is moving at $-40ft/s$ relative to the centre?

7. Apr 16, 2017

### edgarpokemon

would that be point A? It asks me to locate the instantaneous center of rotation and then calculate everything from there, which includes to find the velocities of point B and D, which i found by using the calculated angular velocity of 35. I think that the wheel is always moving, so the instantaneous center of rotation is not on the wheel but outside the wheel, but i am not sure where! i found the problem in google and they the velocity of A pointing to the left, even though the problem states it is pointing to the right.

8. Apr 16, 2017

### PeroK

One thing at a time:

The centre of rotation has to be on a point directly below the centre, otherwise the velocity has a y-component.

Hint: the speed is proportional to the distance from the centre. So, how far do you have to go to get to a speed of $40ft/s$?

9. Apr 16, 2017

### edgarpokemon

would that be 1.143ft below point O?

10. Apr 16, 2017

### PeroK

Who knows? You might know the radius of the wheel, but I certainly don't!

Do you mean $\frac{40}{35} R$, where $R$ is the radius of the wheel?

11. Apr 16, 2017

### edgarpokemon

oh right! sorry the tire has a diameter of 24 inches, so the radius is 1 ft.