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A homogeneous solid circular cylinder of radius \(r\), length \(L\) and mass \(m\) rolls without slipping inside a rough circular track of radius \(R\), as shown in %FigBelow{-20,-12}{80} Choose the centre of the track as the origin of axes, with \(Z\)- axis pointing vertically upwards.

• Obtain expressions for kinetic energy and angular momentum of the cylinder in terms of \(\pmb{\dot{\theta}}\).
• If the cylinder is released from rest at point \(\theta =\pi/2\), determine the angular velocity \(\pmb{\dot{\theta}}\) of the cylinder when in position \(\theta=0\).

• The total kinetic energy, and angular momentum, will be sum of two parts. The first part is translation of centre of mass and the second part is rotation about the centre of mass. The centre of mass rotates about the origin with angul The kinetic energy of the centre of mass at any instant of time is \begin{equation} T_\text{CM} = \frac{1}{2} M \vec{v}^2 = \frac{1}{2}M (R-r)^2 \dot{\theta}^2 \end{equation} where \(|\vec{v}|\) is the absolute value of the velocity given by \(|\vec{v}|= (R-r) \dot{\theta}\). Let \(\omega\)be the angular velocity of rotation about the centre of mass. The kinetic energy of rotational motion about the centre of mass is \begin{equation} T_\text{rf} = \frac{1}{2} I \omega^2 = \frac{1}{4} Mr^2 \omega^2. \end{equation} We now need to find a relation between \(\omega\) and \(\dot{\theta}\). The velocity of any point on the is cylinder is vector sum of translational velocity of the centre of mass and velocity due to rotation about the centre of mass. As the cylinder rolls without slipping the instantaneous velocity of the point of contact is zero. This give a relation between \(\dot{\theta}\) and \(\omega\). \[ (R-r) \dot{\theta} = \omega r\] Thus we get the total kinetic energy as \begin{eqnarray} T&=& \frac{1}{2} I \omega^2 = \frac{1}{4} Mr^2 \omega^2 + \frac{1}{2}M (R-r)^2 \dot{\theta}^2\\ &=& \frac{3}{8} M (R-r)^2 \dot{\theta}^2 \end{eqnarray} The angular momentum angular momentum about the origin is obtained by vector addition of angular momentum of rotation about the centre of the cylinder and the angular momentum of the centre of mass about the center of the track. Thus angular momentum is along the normal out of the plane of paper and has magnitude given by \begin{eqnarray} L&=& M (R-r)^2 \dot{\theta} - I \omega \\ &=& M (R-r) r \dot{\omega} - \frac{1}{4}M r^2 \dot{\omega}\\ &=& M r(\omega) \Big(R-r-\frac{1}{4} r \Big)\\ &=& M r(R-\frac{5r}{4})\omega.\\ &=& M \frac{4R-5r}{4 (R-r)} r \omega. \end{eqnarray}
• The total energy of the system as a function of \(\theta\) is \begin{equation} E= \frac{3}{8} M (R-r)^2 \dot{\theta}^2 + MgR(1- \cos\theta). \end{equation} The bottom point,\(\theta=0\) , has been chosen as a reference point for zero potential energy. At the initial point \(\theta=\pi/2\), \(\dot{\theta}=0\). Using energy conservation and equating the total energy at \(\theta=0\) and \(\theta=\pi/2\), we get \begin{equation} \frac{3}{8} M (R-r)^2 \dot{\theta}^2 = MgR. \end{equation} Thus the angular velocity when the cylinder reaches the bottom point is given by \begin{equation} \dot{\theta} =\sqrt{\frac{8gR}{3(R-r)^2}}. \end{equation}

Show that the principal moments of inertia of a linear chain consisting of two kinds of atoms \(a,b\), with origin chosen to coincide with the centre of mass, is given by \[I_1=I_2=\frac{1}{M}\sum_{i\ne j}m_im_j \,d_{ij}^2, \qquad I_3=0.\] where the summation includes each pair of \(i,j\) atoms once and \(d_{ij}\)is the distance between atoms in the pair and \(M\) is the total mass. Verify that this gives correct answer for a triatomic molecule.

Let the positions of the masses be \(x_i\) measured from some origin on the line joining the atoms. The moment of inertia about the origin, \(I_0\) is given by \[ I_0 = \sum_i m_i x_i^2.\] Letting \(I_\text{cm}\) denote the moment of inertia about the centre of mass. Using the parallel axes theorem we have \[I_0= I_\text{cm} + M X^2 , \] where \[ X = \frac{1}{M} \sum_i m_i x_i\] is the position of the centre of mass and \(M\) is the total mass. Therefore we get \begin{eqnarray} I_\text{cm} &=& I_0 - M X^2 = \frac{1}{M}\Big\{M \sum_i x_i^2 - \Big(\sum_im_ix_i\Big)^2\Big\} \\ &=& \frac{1}{M}\Big\{\Big(\sum_jm_j\Big) \sum_i (m_i x_i^2) - \Big(\sum_im_ix_i\Big)\Big(\sum_j m_j x_j\Big) \Big\}\\ &=& \frac{1}{M}\sum_{ij}\Big\{ m_jm_i (x_i^2 -x_ix_j) \Big\}\\ &=& \frac{1}{M}\sum_{i\ne j}\Big\{ m_jm_i (x_i^2 -x_ix_j) \Big\}\\ &=& \frac{1}{2M}\sum_{i\ne j}\Big\{ m_jm_i (x_i^2 -2x_ix_j + x_j)^2\Big\}\label{EQ01} \end{eqnarray} In the last step, for \(S_{ij}\) symmetric under exchange \(i\leftrightarrow j\), we have used \[\sum_{ij} S_{ij} T_{ij} =\frac{1}{2}\sum_{ij} S_{ij} \big(T_{ij} + T_{ji}\big)\] In the sum in \eqRef{EQ01}, each pair {ij} is counted twice, hence we get \begin{eqnarray} I_\text{cm} &=& \frac{1}{M}\sum_\text{pairs}\Big\{ m_jm_i (x_i^2 -2x_ix_j + x_j)^2\Big\}\\ &=& \frac{1}{M} \sum_\text{pairs} (x_i-x_j)^2. \end{eqnarray} where now each pair is counted once.