Calculate Horizontal Distance of Particle at Linear Speed Rolling

[sachin123]

Homework Statement

A point on the rim of the wheel of radius R under pure rolling has absolute ground speed of (v root 2) or v*(2)^(1/2).the angular speed of rotation is ω.
At what horizontal distance from the instantaneous axis of the rim is the particle located?

Homework Equations

absolute ground speed means linear speed right?

The Attempt at a Solution

I really don't know how absolute ground speed varies with height.
Can someone please tell me how I must start with the problem?What is the idea?
Thank You.

 

[Doc Al]

How fast does a point on the rim move with respect to the center of the wheel? How fast does the center of the wheel move with respect to the ground? Hint: Add those vectors.

 

[sachin123]

I tried it with your method like this.
As seen from the diagram I uploaded,
the horizontal component of Rω (speed of particle)is taken in terms of theta.
Then v + Rω sin (theta) =v*(2)^(1/2).
I found theta and then found the required distance.
But that doesn't match with the answer given.
Am I wrong somewhere?(actual answer is R *root 2).

 

[Doc Al]

I tried it with your method like this.
As seen from the diagram I uploaded,
the horizontal component of Rω (speed of particle)is taken in terms of theta.
Then v + Rω sin (theta) =v*(2)^(1/2).

Why are you only considering the horizontal component? It's the total velocity that must equal v*(2)^(1/2).

 

[sachin123]

Okay.
I took the whole vector and added it.I get the answer as R.It seems right as the particle at the side will have a velocity v upward and v sideways.
But still,the answer is supposed to be R root 2,which confuses me even further,as the radius of the rim is only R!Am I missing something big?

 

[Doc Al]

Okay.
I took the whole vector and added it.I get the answer as R.It seems right as the particle at the side will have a velocity v upward and v sideways.

I would agree with that.

But still,the answer is supposed to be R root 2,which confuses me even further,as the radius of the rim is only R!Am I missing something big?

Hmm... Did you post the question word for word exactly as given? (If it's from a textbook, which one?)

 

[sachin123]

Here it is word for word:

A point on the rim of the wheel of radius R under pure rolling has absolute ground speed of (v root 2) or v*(2)^(1/2).the angular speed of rotation is ω.
It is 'x' distance away from the instantaneous axis .Find x.

yes I made some errors while writing it here,BUT what of it?
It shouldn't make any difference no?

(from an assignment,no textbook)

 

[Doc Al]

It is 'x' distance away from the instantaneous axis .Find x.

Where's the instantaneous axis? Find the distance from that axis to the point on the rim. (Nothing to do with 'horizontal', as your original post stated.)

 

[sachin123]

Yes,nothing got to do with horizontal,I'm very sorry.
Isn't the instantaneous axis the one passing through the centre of the rim,into the plane of paper (of my attached diagram)?

 

[Doc Al]

Isn't the instantaneous axis the one passing through the centre of the rim,into the plane of paper (of my attached diagram)?

No. The instantaneous axis is the axis about which the wheel can be considered to be in pure rotation. The center won't do, since it is translating.

 

[sachin123]

Okay.Thank You Doc Al.I understood.