Rotation of two cylinders inclined at an angle

[PhysicsBoi1908]

Homework Statement

A cylinder P of radius r[SUB]p[/SUB] is being rotated at a constant angular velocity ω[SUB]p[/SUB] with the help of a motor about its axis that is fixed. Another cylinder Q of radius r[SUB]q[/SUB] Free to rotate about its axis that is also fixed is touched with and pressed on P making angle θ between their axes. Soon after the cylinders are pressed against each other, a steady state is reached and cylinder Q acquires a constant angular velocity. Find the final angular velocity of Q and comment on frictional force.

Relevant Equations

v= ωr

A single pair of points will be in contact between P and Q. The frictional force will try to make the velocity of these points equal.
Say the final angular velocity of Q is ω_q.

The velocity of points in contact can never be equal because of difference in directions of ω_q and ω_p. If I break ω_qr_q into horizontal and vertical components though, then I can write that its horizontal component must be equal to ω_pr_p.

Thus

ω_qr_qcosθ=ω_pr_p​

And so we get ω_q to be:

ω_q=[tex]-ω_pr_p/r_qcosθ(sinθi+cosθj)​

Which is wrong. The correct answer is:

ω_q=-ω_pr_pcosθ/r_q(sinθi+cosθj)​

It seems that I was supposed to break ω_pr_p into components instead. But I don't see why I should've done that.

 

[anuttarasammyak]

As an extreme case of two cylinders are parallel position touching in line so $\theta=0$,
$$v=\omega_p r_p=\omega_q r_q$$
Another extreme case of perpendicular position, $\theta=\pi/2$,
$$\omega_q=0$$

In between
$$\omega_p r_p\ \cos\theta=\omega_q r_q$$
We get $\omega_q$ and to make it a vector
$$- cos\theta\ \ \mathbf{i} - sin\theta\ \ \mathbf{j}$$
should be multiplied.

 

[PhysicsBoi1908]

Your post is very helpful. Can you please comment if my thought process is right?
$$ω_p$$ is responsible for making Q rotate. And so we take component of $$ω_pr_p$$ and equate it to $$ω_qr_q$$
This is analogous to the scenario where force is applied on a box which is restricted to move along a straight line. If the force applied is inclined to the straight line, then only the component of the force along it will be responsible for the box's movement. Just like $$ω_pr_pcosθ$$ is responsible for $$ω_qr_q$$

 

[anuttarasammyak]

Though not force but velocity, it seems OK.
Rotation of P with tangential speed $\omega_p r_p$ would transmit
tangential speed of Q : $\omega_p r_p \cos\theta$
longitudinal speed of Q : $\omega_p r_p \sin\theta$
So if friction allows motion of the both direction, Q rotates around and proceed along the axis.
The latter is prohibited by some mechanical arrangement and the only former survives.

 

[PhysicsBoi1908]

This is really nice. Thanks a lot!