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I'm working through mechanics questions and came across the following:

Two identical particles P and Q, each of mass m, are attached to the ends of a diameter of a light thin circular hoop of radius a. The hoop rolls without slipping along a straight line on a horizontal table with the plane of the hoop vertical. Initially, P is in contact with the table. At time t, the hoop has rotated through an angle θ. Write down the position at time t of P, relative to its starting point, in cartesian coordinates.

I have no idea where to start. Any help would be appreciated, I feel like I'm missing something obvious.

redknightofthesun
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1 Answers1

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A quick sketch will confirm the following:

When the hoop has rotated through angle $\theta$, the point of contact of the hoop with the straight line ia at a distance $a\theta$ from the starting point (since the hoop rolls without slipping).

Therefore taking the initial position of $P$ as the origin, and the horizontal straight line as the $x$ axis, the coordinates of $P$ at angle $\theta$ are:

$$(x,y)=(a\theta-a\sin\theta, a\cos \theta)$$

Due to symmetry the potential energy remains constant and therefore, due to conservation of energy, the kinetic energy remains constant (assuming no air resistance). Therefore the angular speed remains constant and equal to the initial value $\omega$, say, in which case we can write:

$$(x,y)=(a\omega t-a\sin\omega t, a\cos \omega t)$$

David Quinn
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