1. The problem statement, all variables and given/known data
A thin-walled pipe with mass M and radius R (moment of inertia MR2) rolls without slipping on a horizontal surface. Inside the pipe, a solid cylinder of mass m and radius r (moment of inertia (1/2)mr2) also rolls without slipping under gravity. ##\alpha## and ##\phi## are, respectively, the angles through which the pipe and cylinder rotate about their centres, and ##θ## is the angular displacement of the centre of the cylinder from the downward vertical through the centre of the pipe. All three angles should be chosen so that they are zero for a configuration in which the cylinder is at the bottom of the pipe, and all three should increase in the same sense, i.e., either all clockwise, or all anticlockwise.

The questions and an illustration are given in the attachment.

2. Relevant equations
Lagrange Equations, kinetic and potential energies.

3. The attempt at a solution
I’ll repeat the questions here.

a)Find the holonomic constraint relating the angles
I got that the holonomic constraint between the angles was ##\alpha – \theta = \phi/2## after some geometry.

b)Show that, in the C.O.M frame of the cylinder, the cylinder rotates through an angle ##\theta + \phi##.
That is obvious, but I am not sure what the question wants me to say. In the frame co-moving with the centre of the cylinder, it sees the centre of mass of the cylinder sketch out an arc of a circle of radius R-r. From the vertical through the pipe, the angle to a point on the cylinder is then ##\theta + \phi##. (##\theta## takes you to the vertical through the cylinder and ##\phi## then takes you to a point on the cylinder, so relative to the zero angle, it is the sum.)

c)Find the kinetic energies of the pipe and cylinder.
I managed to obtain the correct expression for the pipe, but the cylinder I did not.
The general expression for ##T = T’ + V \cdot P’ + \frac{1}{2}MV^2##, where T the kinetic energy in frame S, T’ in frame S’ and V the relative speed of the frames.

Let S’ be the frame co-moving with the centre of the pipe. Then, ##T’ = \frac{1}{2}m((R-r)(\dot{\theta}+\dot{\phi})^2) + \frac{1}{2} (\frac{1}{2}mr^2) (\dot{\theta}+\dot{\phi})^2##. Since there is no slip of the pipe, ##\mathbf{V} = R\dot{\alpha}\underline{e}_x## and ##P’ = P – (M+m)V##, where ##P = MR\dot{\alpha} + m2r(\dot{\theta}+\dot{\phi})## is the momentum of the C.O.M of the pipe and cylinder respectively, relative to S.

Many thanks.

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