Question

A bead of mass 2kg is fitted onto a rod with a length of 8m and can move on it with friction having the coefficient of friction 0.5. At the initial moment the bead is in the middle of the rod. The rod moves translationally in a horizontal plane with an acceleration 1m/sec$^2$ in the direction forming an angle 37° with the rod in gravity free region. Find the time (in sec) when the bead will leave the rod.

Answer 1

4

Answer 2

The problem describes a bead on a rod in a gravity-free region, with the rod accelerating translationally. We need to find the time it takes for the bead to leave the rod.

We analyze the motion of the bead in the non-inertial frame of the rod. In this frame, a pseudo force $\vec{F}_{\text{pseudo}} = -m \vec{a}_0$ acts on the bead, where $m$ is the mass of the bead and $\vec{a}_0$ is the acceleration of the rod. Given $m = 2 \, \text{kg}$ and $a_0 = 1 \, \text{m/s}^2$, the magnitude of the pseudo force is $F_{\text{pseudo}} = m a_0 = 2 \times 1 = 2 \, \text{N}$. The acceleration $\vec{a}_0$ is in a direction forming an angle of $37^\circ$ with the rod. Let's align the rod with the x-axis. The acceleration $\vec{a}_0$ is in the horizontal plane at an angle of $37^\circ$ with the rod. We can assume the angle is with the positive x-axis without loss of generality for the magnitude calculation of the components. The pseudo force $\vec{F}_{\text{pseudo}}$ is in the opposite direction, making an angle of $180^\circ - 37^\circ = 143^\circ$ with the positive x-axis, or equivalently, an angle of $37^\circ$ with the negative x-axis.

Let's resolve the pseudo force into components parallel and perpendicular to the rod (x-axis and y-axis, respectively). The component of the pseudo force parallel to the rod is $F_x = F_{\text{pseudo}} \cos(180^\circ - 37^\circ) = F_{\text{pseudo}} (-\cos 37^\circ)$. Using $\cos 37^\circ \approx 0.8$, $F_x = 2 \times (-0.8) = -1.6 \, \text{N}$. The component of the pseudo force perpendicular to the rod is $F_y = F_{\text{pseudo}} \sin(180^\circ - 37^\circ) = F_{\text{pseudo}} \sin 37^\circ$. Using $\sin 37^\circ \approx 0.6$, $F_y = 2 \times 0.6 = 1.2 \, \text{N}$.

The forces acting on the bead in the frame of the rod are the pseudo force, the normal force from the rod, and the friction force. The normal force $N$ from the rod acts perpendicular to the rod. In the direction perpendicular to the rod, the net force must be zero for the bead to stay on the rod. The normal force balances the component of the pseudo force perpendicular to the rod. $N = |F_y| = 1.2 \, \text{N}$.

The friction force acts parallel to the rod and opposes the relative motion of the bead along the rod. The coefficient of kinetic friction is given as $\mu = 0.5$. The magnitude of the kinetic friction force is $f_k = \mu N = 0.5 \times 1.2 = 0.6 \, \text{N}$.

The component of the pseudo force along the rod is $F_x = -1.6 \, \text{N}$. This force tends to move the bead in the negative x direction. The magnitude of this force component is $|F_x| = 1.6 \, \text{N}$. Since $|F_x| > f_k$ (1.6 N > 0.6 N), the bead will move along the rod. The direction of motion will be in the direction of the net force along the rod. The pseudo force component is -1.6 N. Since the bead starts from rest, it will initially move in the direction of the pseudo force component along the rod, which is the negative x direction. When the bead moves in the negative x direction, the kinetic friction force acts in the positive x direction. The net force on the bead along the rod is $F_{\text{net}, x} = F_x + f_k = -1.6 + 0.6 = -1.0 \, \text{N}$.

The acceleration of the bead relative to the rod is $a_{\text{rel}, x} = \frac{F_{\text{net}, x}}{m} = \frac{-1.0}{2} = -0.5 \, \text{m/s}^2$. This acceleration is constant and directed along the rod in the negative x direction.

The rod has a length of 8m. The bead starts in the middle of the rod. Let's set the origin at the middle of the rod, with the x-axis along the rod. The ends of the rod are at x = -4m and x = +4m. The initial position of the bead is $x(0) = 0$. The initial velocity of the bead relative to the rod is $v_{\text{rel}, x}(0) = 0$. The acceleration is $a_{\text{rel}, x} = -0.5 \, \text{m/s}^2$.

The position of the bead relative to the middle of the rod at time $t$ is given by the equation of motion: $x(t) = x(0) + v_{\text{rel}, x}(0) t + \frac{1}{2} a_{\text{rel}, x} t^2$ $x(t) = 0 + 0 \times t + \frac{1}{2} (-0.5) t^2 = -0.25 t^2$.

The bead will leave the rod when its position reaches either end, i.e., when $x(t) = +4$m or $x(t) = -4$m. Since the acceleration is in the negative x direction ($a_{\text{rel}, x} = -0.5 \, \text{m/s}^2$), the bead will move towards the end at x = -4m. We need to find the time $t$ when $x(t) = -4$. $-4 = -0.25 t^2$ $t^2 = \frac{-4}{-0.25} = \frac{4}{0.25} = \frac{4}{1/4} = 16$. $t = \sqrt{16} = 4 \, \text{s}$. (We take the positive value for time).

The time when the bead will leave the rod is 4 seconds.