Angular momentum of the particle rotating with a central force is constant due to

Point masses ${\mathrm{m}}_1$ and ${\mathrm{m}}_2$ are placed at the opposite ends of rigid rod of length $\text{L}$, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point $\text{L}$ on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ${\mathrm{\omega }}_0$ is minimum, is given by:

Persons $A$ and $B$ are standing on the opposite sides of a $3.5 \mathrm{m}$ wide water stream that they wish to cross. Each one of them has a rigid wooden plank whose mass can be neglected. However, each plank is only slightly longer than, $3 \mathrm{m}.$ So, they decide to arrange them together as shown in the figure schematically. With $B$ (the mass $17 \mathrm{kg}$ ) standing, the maximum mass of $A$, who can walk over the plank is close to,

A uniform rod of length ' ${\ell }^{\text{'}}$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$ the rod makes an angle $\theta$with it (see figure). To find $\theta$ equate the rate of change of angular momentum (direction going into the paper) $\frac{m{\ell }^2}{12}{\omega }^2\sin \theta$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces $F_H$and $F_v$ about the CM. The value of $\theta$ is then such that:

A rod of length $50 cm$ is pivoted at one end. It is raised such that it makes an angle of ${30}^o$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad s^{-1}$ ) will be $\left(g=10 m{s}^{-2}\right)$