Instantaneous center of rotation

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The discussion revolves around locating the instantaneous center of rotation (C) during the deceleration of a tire, where the tire's velocity is 40 ft/s and point A's velocity is 5 ft/s. Participants debate whether C can be positioned below A and explore the relationship between angular velocity and the motion of the tire. It is clarified that the center of rotation must be directly below the tire's center, as any deviation would introduce a vertical velocity component. The conversation emphasizes that the instantaneous center of rotation is not necessarily on the wheel itself but can be outside of it, depending on the velocities involved. Ultimately, the radius of the tire is confirmed to be 1 ft, which is crucial for further calculations.
edgarpokemon
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<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 shouldn't A be pointing to the left? help!

http://tinypic.com/r/2n7f7g2/9
 
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What's your definition of "instantaneous centre of rotation"?
 
PeroK said:
What's your definition of "instantaneous centre of rotation"?
as its name suggests?
 
edgarpokemon said:
as its name suggests?

Since you asked the following question:

edgarpokemon said:
Can the instantaneous center of rotation C be located below A?

It clearly isn't so obvious.
 
PeroK said:
Since you asked the following question:
It clearly isn't so obvious.

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.
 
edgarpokemon said:
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.

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?
 
PeroK said:
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?

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.
 
edgarpokemon said:
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.

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##?
 
would that be 1.143ft below point O?
 
  • #10
edgarpokemon said:
would that be 1.143ft below point O?

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

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