Question

A uniform rod of mass m and length $l$ pivoted at one of its top end is hanging freely in vertical plane. Another identical rod moving horizontally with velocity $v$ along a line passing through its lower end hits it and sticks to it. The two rods were perpendicular during the hit and later also they remain perpendicularly connected to each other. Find the maximum angle turned by the two-rod system after collision.

Answer 1

$\sin\theta_{max} - 3\cos\theta_{max} = \frac{3v^2}{5gl} - 3$

Answer 2

The problem involves two main parts: a collision and subsequent rotational motion.

Part 1: Collision and Conservation of Angular Momentum

Before the collision, we have a stationary rod (Rod 1) pivoted at its top end and an identical rod (Rod 2) moving horizontally. The collision is inelastic, and the two rods stick together, forming an 'L' shape, remaining perpendicular. We apply the conservation of angular momentum about the pivot point of Rod 1, as there are no external torques about this point during the collision.

1. Initial Angular Momentum ($L_{initial}$):

   • Rod 1 is at rest, so its initial angular momentum is 0.
   • Rod 2 has mass m and velocity v. It hits Rod 1 at its lower end. Let the pivot be the origin (0,0). Rod 1 hangs along the negative y-axis, so its lower end is at (0, -l). Rod 2 moves horizontally along the line y = -l.
   • The angular momentum of Rod 2 about the pivot is L_2 = r x p. Here, the position vector of the point of impact relative to the pivot is r = -l j_hat, and the momentum is p = m v i_hat.
   • L_2 = (-l j_hat) x (m v i_hat) = -m v l (j_hat x i_hat) = -m v l (-k_hat) = m v l k_hat.
   • So, the initial angular momentum of the system is L_{initial} = m v l.

2. Final Angular Momentum ($L_{final}$):

   • After the collision, the two rods stick together and rotate as a single system with angular velocity w.

   • L_{final} = I_{system} * w, where I_{system} is the moment of inertia of the combined system about the pivot.

   • Moment of inertia of Rod 1 ($I_1$): Rod 1 is pivoted at one end. I_1 = (1/3) m l^2.

   • Moment of inertia of Rod 2 ($I_2$): Rod 2 is attached perpendicularly to the bottom end of Rod 1. Its center of mass (CM) is at (l/2, -l) relative to the pivot. We use the parallel axis theorem: I_2 = I_{CM} + M d^2.

     • I_{CM} for a rod about its center is (1/12) m l^2.
     • d is the distance from the pivot to the CM of Rod 2: d^2 = (l/2)^2 + (-l)^2 = l^2/4 + l^2 = (5/4)l^2.
     • I_2 = (1/12) m l^2 + m (5/4) l^2 = (1/12) m l^2 + (15/12) m l^2 = (16/12) m l^2 = (4/3) m l^2.

   • Total moment of inertia ($I_{system}$): I_{system} = I_1 + I_2 = (1/3) m l^2 + (4/3) m l^2 = (5/3) m l^2.

3. Conservation of Angular Momentum: L_{initial} = L_{final}

   m v l = (5/3) m l^2 * w

   w = (m v l) / ((5/3) m l^2) = (3v) / (5l).

   This is the angular velocity of the system immediately after the collision.

Part 2: Rotational Motion and Conservation of Energy

After the collision, the system swings upwards. At the maximum angle theta_{max}, its kinetic energy will be momentarily zero, and all initial kinetic energy will be converted into potential energy.

1. Initial Potential Energy ($PE_{initial}$):

   • First, find the initial position of the center of mass (CM) of the combined system.
   • CM of Rod 1: C1 = (0, -l/2)
   • CM of Rod 2: C2 = (l/2, -l) (Rod 2 extends horizontally to the right)
   • CM of system C_{system} = (m*0 + m*l/2) / (m+m), (m*(-l/2) + m*(-l)) / (m+m)
   • C_{system} = (l/4, -3l/4).
   • Let's set the initial potential energy PE_{initial} = 0 at this initial height of the CM.

2. Initial Kinetic Energy ($KE_{initial}$): KE_{initial} = (1/2) I_{system} w^2 = (1/2) * (5/3) m l^2 * ((3v)/(5l))^2

   KE_{initial} = (1/2) * (5/3) m l^2 * (9v^2)/(25l^2) = (1/2) * (3/5) m v^2 = (3/10) m v^2.

3. Final Potential Energy ($PE_{final}$):

   • When the system swings to theta_{max}, the Rod 1 makes an angle theta_{max} with the initial vertical.
   • The initial coordinates of the CM are (x_0, y_0) = (l/4, -3l/4).
   • After rotating by theta_{max} (counter-clockwise, for example), the new coordinates of the CM (x_f, y_f) are: x_f = x_0 cos(theta_{max}) - y_0 sin(theta_{max}) = (l/4)cos(theta_{max}) - (-3l/4)sin(theta_{max}) = (l/4)(cos(theta_{max}) + 3sin(theta_{max})) y_f = x_0 sin(theta_{max}) + y_0 cos(theta_{max}) = (l/4)sin(theta_{max}) + (-3l/4)cos(theta_{max}) = (l/4)(sin(theta_{max}) - 3cos(theta_{max}))
   • The rise in height of the CM is h = y_f - y_0 = (l/4)(sin(theta_{max}) - 3cos(theta_{max})) - (-3l/4)
   • h = (l/4)(sin(theta_{max}) - 3cos(theta_{max}) + 3).
   • The total mass of the system is 2m.
   • PE_{final} = (2m) g h = (2m) g (l/4)(sin(theta_{max}) - 3cos(theta_{max}) + 3)
   • PE_{final} = (mgl/2)(sin(theta_{max}) - 3cos(theta_{max}) + 3).

4. Conservation of Energy: KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

   (3/10) m v^2 + 0 = 0 + (mgl/2)(sin(theta_{max}) - 3cos(theta_{max}) + 3)

   Cancel m from both sides:

   (3/10) v^2 = (gl/2)(sin(theta_{max}) - 3cos(theta_{max}) + 3)

   Multiply by 2:

   (3/5) v^2 = gl(sin(theta_{max}) - 3cos(theta_{max}) + 3)

   Rearrange to solve for the trigonometric part:

   sin(theta_{max}) - 3cos(theta_{max}) = (3v^2 / (5gl)) - 3

This is the equation for theta_{max}.

Solution:

1. Conservation of Angular Momentum during Collision:

   Initial angular momentum of the system about the pivot: $L_i = mvl$.

   Moment of inertia of Rod 1 about the pivot: $I_1 = \frac{1}{3}ml^2$.

   Moment of inertia of Rod 2 about the pivot (using parallel axis theorem): $I_2 = \frac{1}{12}ml^2 + m((\frac{l}{2})^2 + l^2) = \frac{1}{12}ml^2 + m\frac{5}{4}l^2 = \frac{16}{12}ml^2 = \frac{4}{3}ml^2$.

   Total moment of inertia of the combined system: $I_{sys} = I_1 + I_2 = \frac{1}{3}ml^2 + \frac{4}{3}ml^2 = \frac{5}{3}ml^2$.

   Final angular velocity just after collision: $w = \frac{L_i}{I_{sys}} = \frac{mvl}{\frac{5}{3}ml^2} = \frac{3v}{5l}$.

2. Conservation of Mechanical Energy after Collision:

   Initial kinetic energy of the system: $KE_i = \frac{1}{2}I_{sys}w^2 = \frac{1}{2}(\frac{5}{3}ml^2)(\frac{3v}{5l})^2 = \frac{1}{2}(\frac{5}{3}ml^2)(\frac{9v^2}{25l^2}) = \frac{3}{10}mv^2$.

   Initial position of the center of mass (CM) of the combined system: $C_{sys} = (\frac{m(0) + m(l/2)}{2m}, \frac{m(-l/2) + m(-l)}{2m}) = (\frac{l}{4}, -\frac{3l}{4})$.

   Let the initial potential energy be zero at this height.

   When the system rotates by an angle $\theta_{max}$, the new y-coordinate of the CM is $y_f = \frac{l}{4}\sin\theta_{max} - \frac{3l}{4}\cos\theta_{max}$.

   The change in height of the CM is $h = y_f - y_0 = (\frac{l}{4}\sin\theta_{max} - \frac{3l}{4}\cos\theta_{max}) - (-\frac{3l}{4}) = \frac{l}{4}(\sin\theta_{max} - 3\cos\theta_{max} + 3)$.

   Potential energy at maximum angle: $PE_f = (2m)gh = 2mg \frac{l}{4}(\sin\theta_{max} - 3\cos\theta_{max} + 3) = \frac{mgl}{2}(\sin\theta_{max} - 3\cos\theta_{max} + 3)$.

   At maximum angle, $KE_f = 0$.

   By conservation of energy: $KE_i + PE_i = KE_f + PE_f$.

   $\frac{3}{10}mv^2 + 0 = 0 + \frac{mgl}{2}(\sin\theta_{max} - 3\cos\theta_{max} + 3)$.

   $\frac{3}{5}v^2 = gl(\sin\theta_{max} - 3\cos\theta_{max} + 3)$.

   $\sin\theta_{max} - 3\cos\theta_{max} = \frac{3v^2}{5gl} - 3$.