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## Homework Statement

A man is walking at a constant velocity v along the circumference of a disc which is pivoted at a point on it's circumference. The man starts diametrically opposite to the pivot and the disc starts at rest. The man and disc have the same mass. When the man reaches back to his original place on the disc, what will the angular displacement of the disc be?

**2. Relevant stuff**

I've made a picture to illustrate the problem and (mostly) to illustrate my attempt at a solution. The top left corner of the picture shows the system at time zero, and the bigger part of the picture shows the system at some arbitrary later time (when the man has walked a distance θ, so I guess the time would be T=θR/v, but that's not important).

## The Attempt at a Solution

ß is the angular displacement of the disc (as shown in the picture)

θ is the angular displacement of the man relative to the disc (as shown in the picture)

ω is the angular velocity of the disc with respect to the pivot

Let's define L to be the distance between the pivot and the person (which depends on θ)

And for just a moment let's say L is a vector from the pivot to the person, then let's define V

_{⊥}as the component of the person's velocity which is perpendicular to L

From conservation of angular momentum:

[itex]mV_⊥L=Iω[/itex]

[itex]I[/itex], the moment of inertia of the disc, should be [itex]\frac{3}{2}mR^2[/itex] so we get:

[itex]V_⊥=\frac{3ωR^2}{2L}[/itex]

We can determine the radial component of the velocity V

_{r}from [itex]V_⊥^2+V_r^2=V^2[/itex]

I think the last constraint is that the direction of the velocity is tangent to the circle. From geometric considerations I get:

[itex]\tan(\frac{θ}{2})=\frac{V_r}{V_⊥}[/itex]

Also I forgot to mention [itex]L=\sqrt{2(1+cos(θ)}[/itex]

So from all this you get ω as a function of θ, but since v is constant, [itex]θ=\frac{vt}{R}[/itex] and so we have ω(t)

Since [itex]β=∫_0^Tω(t)dt[/itex] the answer should be [itex]∫_0^{\frac{2\pi R}{v}}ω(t)dt[/itex] but this does not give me the correct answer.