JEE Main

Numerical 1: Conservation of Angular Momentum

• Initial angular momentum of the system: $$L_i = I_i\omega_i$$ $$I_i = MR^2 + mR^2 = (M + m)R^2$$ $$\omega_i = \omega$$

• Final angular momentum of the system: $$L_f = I_f\omega_f$$ $$I_f = (M + m)R^2$$ $$\omega_f = \frac{I_i\omega_i}{I_f}=\frac{\omega (M+m)R^2}{(M+m)R^2}=\boxed{\omega}$$

Numerical 2: Simple Pendulum with Added Mass

• Moment of inertia of the rod about the pivot: $$I = \frac{1}{3}ML^2$$
• Torque acting on the system: $$\tau = mgl\sin\theta$$
• Using the equation of motion: $$\tau = I\alpha$$ $$\alpha = \frac{\tau}{I} = \frac{mgl\sin\theta}{\frac{1}{3}ML^2}$$
• Period of oscillation: $$T = 2\pi\sqrt{\frac{I}{mgL\sin\theta}}$$ $$T = 2\pi\sqrt{\frac{\frac{1}{3}ML^2}{mgl\sin\theta}}$$

$$T = 2\pi\sqrt{\frac{L}{3g\sin\theta}}$$

Numerical 3: Rolling Sphere

• Translational kinetic energy of the sphere: $$K_t = \frac{1}{2}Mv^2$$
• Rotational kinetic energy of the sphere: $$K_r = \frac{1}{2}I\omega^2$$
• Total kinetic energy of the sphere: $$K = K_t + K_r$$
• Angular speed of the sphere: $$\omega = \frac{v}{R}$$
• Moment of inertia of the sphere: $$I = \frac{2}{5}MR^2$$
• Substituting these values into the equation for total kinetic energy, we get: $$K = \frac{1}{2}Mv^2 + \frac{1}{2}\left(\frac{2}{5}MR^2\right)\left(\frac{v^2}{R^2}\right)$$ $$K = \frac{7}{10}Mv^2$$

Therefore, the angular speed of the sphere is $$\omega = \frac{v}{R}$$

CBSE Board Exam

Numerical 1: Moment of Inertia of a Uniform Disk

• Moment of inertia of the disk about the axis of rotation: $$I = \frac{1}{2}MR^2$$
• Substituting the given values, we get: $$I = \frac{1}{2}(10\text{ kg})(0.2\text{ m})^2 = 1 \text{ kg m}^2$$

Therefore, the moment of inertia of the disk about the axis of rotation is $$I = 1 \text{ kg m}^2$$

Numerical 2: Kinetic Energy of a Uniform Disk

• Kinetic energy of the disk: $$K = \frac{1}{2}I\omega^2$$
• Substituting the given values, we get: $$K = \frac{1}{2}(20\text{ kg})(0.5\text{ m})^2(30\text{ rad/s})^2 = 900 \text{ J}$$

Therefore, the kinetic energy of the disk is $$K = 900 \text{ J}$$

Numerical 3: Period of Oscillation of a Uniform Rod

• Moment of inertia of the rod about the pivot: $$I = \frac{1}{3}ML^2$$
• Torque acting on the system: $$\tau = mgl\sin\theta$$
• Using the equation of motion: $$\tau = I\alpha$$ $$\alpha = \frac{\tau}{I} = \frac{mgl\sin\theta}{\frac{1}{3}ML^2}$$
• Period of oscillation: $$T = 2\pi\sqrt{\frac{I}{mgL\sin\theta}}$$ $$T = 2\pi\sqrt{\frac{\frac{1}{3}ML^2}{mgl\sin\theta}}$$ $$T = 2\pi\sqrt{\frac{L}{3g\sin\theta}}$$