# Rotation Mechanics [disk-axle mechanism]

1. Jan 17, 2008

### rohanprabhu

[SOLVED] Rotation Mechanics [disk-axle mechanism]

1. The problem statement, all variables and given/known data

Q. A disc is fixed at its center O and rotating with constant angular velocity $\omega$. There is a rod whose one end is connected at A on the disc and the other end is connected with a ring which can freely move along the fixed vertical smooth rod. At an instant when the rod is making an angle 30º with the vertical the ring is found to have a velocity $v$ in the upward direction. Find $\omega$ of the disk. Given that the point A is R/2 distance vertically above the point O and length of the rod AB is l.

2. Relevant equations

I don't know.. maybe:

$$v = r\omega$$
$$\tau = I^2\alpha$$

3. The attempt at a solution

The point 'A' moves in the inner circle of radius 'R/2'. I tried marking two positions in the inner rim, at 'A' and 'a' where the angle made by the axle with the vertical rod is $\theta$ and $\theta + d\theta$. The angle at the cross section of the two impressions of the rod comes out to be $d\theta$.

I somehow need to equate it with the angular displacement for the disk. The key idea here, however is that the length of the rod will remain constant. I'm just not able to understand as to how i go about using that...

also.. i think that if i can somehow prove that this is the case of rotation about a fixed point on the rod, then things will become much easier. how do i go about doing this?

any help is greatly appreciated. thanks...

Last edited by a moderator: May 3, 2017
2. Jan 18, 2008

### rl.bhat

Whether the disc and the vertical rod are in the same plane? Whether the point A on the rod is firmly fixed or pivot around A?

Last edited: Jan 18, 2008
3. Jan 18, 2008

### rohanprabhu

Point 'A' is exactly above the center of the circle when the axle makes 30º with the vertical rod. At other times, it is a point on a concentric circle on the disc of radius R/2.

And yes, they are in the same plane...

4. Jan 18, 2008

### Staff: Mentor

Take advantage of the fact that the length of the rod doesn't change. View its length as having vertical and horizontal components (Y & X, say). At the moment in question, the horizontal component is increasing at the rate of $\omega R$ (since end A is moving horizontally), so at what rate must the vertical component decrease to keep the length constant? (Think in terms of differentials.)

5. Jan 18, 2008

### rohanprabhu

ok.. so here's what i've assumed. Have a look at the figure:

http://img265.imageshack.us/img265/1679/solutiononeen5.jpg [Broken]

I've dropped a perpendicular on the vertical rod from the point 'A' at 'B'. I've taken the distance of this horizontal as 'y'. Also, the distance from the point of contact of the rod and the axle and 'B' is taken as 'x'.

So, we have:

$$x^2 + y^2 = l^2$$

diff. both sides w.r.t. time:

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$

Now, here $\frac{dx}{dt} = v$ and $\frac{dy}{dt} = \frac{R\omega}{2}$

on substituting these values, i get:

$$v = -\frac{y}{x} \frac{R\omega}{2}$$

here, $\frac{y}{x} = tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$

So,

$$\omega = -\frac{2\sqrt{3}v}{R}$$

is it right??

I have an option like that.. but they're all positive... what does my negative sign indicate? Although it does indicates the direction of axis of rotation.. but ofcourse it has to be relative to the assumptions i have taken for sign convention. So, relative to what exactly is the positive sign...

Last edited by a moderator: May 3, 2017
6. Jan 18, 2008

### Staff: Mentor

Looks good.
Why did you divide by 2? [Correction: Looks good!]
Except for that factor of 2. [Correction: Looks good!]

The vertical component of the rod length (your X value) decreases (thus is negative) at the same rate that end B rises (and is thus positive).

Edit: I was taking R as the distance from point A to the axis, but that distance is R/2 not R. Thanks to Shooting star for pointing out my error.

Last edited: Jan 18, 2008
7. Jan 18, 2008

### Shooting Star

The rotating point where one end of the rod is fixed is at a dist of R/2 from the centre.

8. Jan 18, 2008

### Staff: Mentor

D'oh! My bad. (I'll correct my error in the previous post. Thanks, Ss!)

9. Jan 18, 2008

### rohanprabhu

k. I get it now.

thanks a lot to everybody who helped me out. Doc Al, shooting star.. u guys r gr8 :D
thanks again..