Question

A smooth horizontal rod of length $\ell$ = 1 m and mass M = 3 kg with a sleeve of mass m = 1 kg initially getting at end A is rotated with an initial angular velocity $\omega_0$ = 10 rad/s. When the sleeve reaches the other end B, which of the following options is/are correct. (Take g = 10 m/s²)

• A. The velocity of sleeve relative to the rod is $5\sqrt{2}$ m/s.
• B. The velocity of sleeve relative to the ground is $5\sqrt{3}$ m/s
• C. The normal force acting on the sleeve is $25\sqrt{2}$ newton
• D. The normal force acting on the sleeve is $15\sqrt{6}$ newton.

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The system consists of a horizontal rod of length $\ell$ and mass M, rotating about a vertical axis at one end A, and a sleeve of mass m sliding along the rod. Initially, the sleeve is at end A, and the rod rotates with an angular velocity $\omega_0$. The rod is smooth, so there is no friction between the sleeve and the rod. Since the rod is horizontal, gravity does not affect the motion in the horizontal plane. We assume that there is no external torque applied about the axis of rotation, so the angular momentum of the system about the axis of rotation is conserved.

Let r be the distance of the sleeve from the axis of rotation. The moment of inertia of the rod about the axis is $I_{rod} = \frac{1}{3}M\ell^2$. The moment of inertia of the sleeve about the axis is $I_{sleeve} = mr^2$. The total moment of inertia of the system is $I = I_{rod} + I_{sleeve} = \frac{1}{3}M\ell^2 + mr^2$. The angular momentum of the system is $L = I\omega = (\frac{1}{3}M\ell^2 + mr^2)\omega$. Initially, at end A, r=0 and $\omega = \omega_0$. The initial angular momentum is $L_0 = (\frac{1}{3}M\ell^2 + m(0)^2)\omega_0 = \frac{1}{3}M\ell^2\omega_0$. When the sleeve reaches end B, r=$\ell$. Let the angular velocity be $\omega_f$. The final angular momentum is $L_f = (\frac{1}{3}M\ell^2 + m\ell^2)\omega_f$. By conservation of angular momentum, $L_f = L_0$. $(\frac{1}{3}M\ell^2 + m\ell^2)\omega_f = \frac{1}{3}M\ell^2\omega_0$ $\ell^2(\frac{1}{3}M + m)\omega_f = \frac{1}{3}M\ell^2\omega_0$ $\omega_f = \frac{\frac{1}{3}M}{\frac{1}{3}M + m}\omega_0 = \frac{M}{M + 3m}\omega_0$. Given M = 3 kg, m = 1 kg, $\omega_0$ = 10 rad/s. $\omega_f = \frac{3}{3 + 3(1)} \times 10 = \frac{3}{6} \times 10 = 5$ rad/s.

Now, we use conservation of mechanical energy. The initial kinetic energy is the rotational kinetic energy of the rod. The sleeve is initially at rest relative to the rod at r=0, so its initial kinetic energy is zero. $K_0 = \frac{1}{2}I_0\omega_0^2 = \frac{1}{2}(\frac{1}{3}M\ell^2)\omega_0^2$. When the sleeve reaches end B, its kinetic energy consists of rotational kinetic energy and translational kinetic energy relative to the rod. The kinetic energy of the rod is $\frac{1}{2}I_{rod}\omega_f^2 = \frac{1}{2}(\frac{1}{3}M\ell^2)\omega_f^2$. The kinetic energy of the sleeve is $\frac{1}{2}I_{sleeve,f}\omega_f^2 + \frac{1}{2}mv_{r,f}^2 = \frac{1}{2}m\ell^2\omega_f^2 + \frac{1}{2}mv_{r,f}^2$, where $v_{r,f}$ is the radial velocity of the sleeve relative to the rod at end B. The total final kinetic energy is $K_f = \frac{1}{2}(\frac{1}{3}M\ell^2 + m\ell^2)\omega_f^2 + \frac{1}{2}mv_{r,f}^2$. By conservation of energy, $K_f = K_0$. $\frac{1}{2}(\frac{1}{3}M\ell^2 + m\ell^2)\omega_f^2 + \frac{1}{2}mv_{r,f}^2 = \frac{1}{2}(\frac{1}{3}M\ell^2)\omega_0^2$. $(\frac{1}{3}M + m)\ell^2\omega_f^2 + mv_{r,f}^2 = \frac{1}{3}M\ell^2\omega_0^2$. Substitute $\omega_f = \frac{M}{M+3m}\omega_0$. $(\frac{M+3m}{3})\ell^2(\frac{M}{M+3m}\omega_0)^2 + mv_{r,f}^2 = \frac{1}{3}M\ell^2\omega_0^2$. $\frac{M^2}{3(M+3m)}\ell^2\omega_0^2 + mv_{r,f}^2 = \frac{1}{3}M\ell^2\omega_0^2$. $mv_{r,f}^2 = \frac{1}{3}M\ell^2\omega_0^2 - \frac{M^2}{3(M+3m)}\ell^2\omega_0^2 = \frac{M(M+3m) - M^2}{3(M+3m)}\ell^2\omega_0^2 = \frac{3Mm}{3(M+3m)}\ell^2\omega_0^2 = \frac{Mm}{M+3m}\ell^2\omega_0^2$. $v_{r,f}^2 = \frac{M}{M+3m}\ell^2\omega_0^2$. $v_{r,f} = \sqrt{\frac{M}{M+3m}}\ell\omega_0$. Substitute the given values: M = 3 kg, m = 1 kg, $\ell$ = 1 m, $\omega_0$ = 10 rad/s. $v_{r,f} = \sqrt{\frac{3}{3+3(1)}} \times 1 \times 10 = \sqrt{\frac{3}{6}} \times 10 = \sqrt{\frac{1}{2}} \times 10 = \frac{10}{\sqrt{2}} = 5\sqrt{2}$ m/s. (A) The velocity of sleeve relative to the rod is $5\sqrt{2}$ m/s. This is correct.

The velocity of the sleeve relative to the ground has two components: the radial velocity relative to the rod $v_{r,f}$ and the tangential velocity due to the rotation $v_{t,f} = \ell\omega_f$. These two components are perpendicular. $v_{t,f} = \ell\omega_f = 1 \times 5 = 5$ m/s. The magnitude of the velocity of the sleeve relative to the ground is $v_{ground} = \sqrt{v_{r,f}^2 + v_{t,f}^2} = \sqrt{(5\sqrt{2})^2 + 5^2} = \sqrt{50 + 25} = \sqrt{75} = 5\sqrt{3}$ m/s. (B) The velocity of sleeve relative to the ground is $5\sqrt{3}$ m/s. This is correct.

The normal force acting on the sleeve is the force exerted by the side of the rod on the sleeve. This force is perpendicular to the rod and the axis of rotation, and it provides the centripetal force for the sleeve's tangential acceleration in the inertial frame. In the rotating frame, this force is in the tangential direction. The tangential equation of motion for the sleeve in the rotating frame is $N = m(r\dot{\omega} + 2\dot{r}\omega)$, where N is the normal force from the side of the rod. From angular momentum conservation, $(\frac{1}{3}M\ell^2 + mr^2)\omega = L_0$. Differentiating with respect to time, $(2mr\dot{r})\omega + (\frac{1}{3}M\ell^2 + mr^2)\dot{\omega} = 0$. $\dot{\omega} = -\frac{2mr\dot{r}\omega}{\frac{1}{3}M\ell^2 + mr^2}$. At end B, r=$\ell$, $\dot{r} = v_{r,f}$, $\omega = \omega_f$. $\dot{\omega}_f = -\frac{2m\ell v_{r,f}\omega_f}{\frac{1}{3}M\ell^2 + m\ell^2} = -\frac{2m\ell v_{r,f}\omega_f}{\ell^2(\frac{1}{3}M + m)} = -\frac{2mv_{r,f}\omega_f}{\ell(\frac{1}{3}M + m)}$. The normal force at end B is $N_f = m(\ell\dot{\omega}_f + 2v_{r,f}\omega_f)$. $N_f = m(\ell(-\frac{2mv_{r,f}\omega_f}{\ell(\frac{1}{3}M + m)}) + 2v_{r,f}\omega_f) = m(-\frac{2mv_{r,f}\omega_f}{\frac{1}{3}M + m} + 2v_{r,f}\omega_f) = m v_{r,f}\omega_f (2 - \frac{2m}{\frac{1}{3}M + m}) = m v_{r,f}\omega_f (2 - \frac{6m}{M+3m})$. Substitute the values: m=1 kg, $v_{r,f} = 5\sqrt{2}$ m/s, $\omega_f = 5$ rad/s, M=3 kg. $N_f = 1 \times 5\sqrt{2} \times 5 (2 - \frac{6 \times 1}{3 + 3 \times 1}) = 25\sqrt{2} (2 - \frac{6}{6}) = 25\sqrt{2} (2 - 1) = 25\sqrt{2} \times 1 = 25\sqrt{2}$ newton. (C) The normal force acting on the sleeve is $25\sqrt{2}$ newton. This is correct. (D) The normal force acting on the sleeve is $15\sqrt{6}$ newton. This is incorrect.

Final check: Option A: $v_{r,f} = 5\sqrt{2}$ m/s. Correct. Option B: $v_{ground} = 5\sqrt{3}$ m/s. Correct. Option C: $N_f = 25\sqrt{2}$ N. Correct. Option D: $N_f = 15\sqrt{6}$ N. Incorrect.