Turntable - rotation?

**fardeen_gen** · May 19th 2009, 08:52 AM

A horizontal turntable in the form of a uniform disc of mass 150 kg is mounted on a light, frictionless vertical axis at its centre. Two men of mass 75 kg stand at opposite ends of a diameter, and they and the turntable are at rest. The men then move round the table in the same direction and at the same constant speed. Calculate the angle they have turned through in space when they have made one complete circuit of the table?

Answer:

Spoiler:

Any ideas?

**Kiwi_Dave** · May 21st 2009, 03:51 AM

Let $\phi _{p}$ be the angle through which the people turn with respect to the laboratory.

Let $\phi _{d}$ be the angle through which the disk turns with respect to the laboratory.

$v_{p}=\frac{r \phi_p}{t}$ because speed is constant

angular momentum of people $p_{p}=mrv_{p}=2*75*r*v_{p}$

angular momentum of disk = - angular momentum of people

$p_{disk}=I \omega = \frac 1 2 150 r^2 \frac {\phi_d}{t}=-mrv_{p}=-2*75*r*v_{p}=-2*75*r*\frac{r \phi_p}{t}$

$\therefore \phi_d=-2\phi_p$

Now at the completion of one revolution across the surface of the disk

$\phi_p-\phi_d=2\pi$

You now have two equations in two variables phi_d and phi_p so you can solve them. Though I get a different answer to your spoiler, perhaps I made a silly algebraic mistake above.

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