Question

A rod of mass m and length L (= 50/3m) is hinged in plank of same mass m. The plank is kept at the corner of a smooth table and rod makes $\theta$ with the vertical. The system is released from the $\theta$ = 0°. Find the velocity (in m/s) of plank when the rod makes the $\theta$ = 180° (g=10m/s²)

Answer 1

10

Answer 2

Solution Explanation

1. Conservation of Horizontal Momentum:
    Let the plank’s velocity be $v$ to the right and the rod’s angular velocity be $\omega$.
    At the instant $\theta=180^\circ$ (rod vertical downward), the rod’s center of mass (located at $L/2$ from the hinge) has a relative horizontal speed $\frac{L}{2}\omega$ (to the left).
    Thus, the rod’s horizontal speed in the lab frame is $v - \frac{L}{2}\omega$.
    Since total horizontal momentum begins at 0:
    $m\,v + m\Bigl(v - \frac{L}{2}\omega\Bigr) = 0 \quad\Longrightarrow\quad 2v -\frac{L}{2}\omega = 0$  So,
    $\omega = \frac{4v}{L}.$

2. Conservation of Energy:
    Loss in gravitational potential energy of the rod’s center of mass as it falls a distance $L$ is $mgL$.
    The final kinetic energy consists of three parts:
    - Plank’s translational KE: $\frac{1}{2}m v^2$.
    - Rod’s translational KE (of its center of mass): $\frac{1}{2}m \Bigl(v - \frac{L}{2}\omega\Bigr)^2$.
    - Rod’s rotational KE about its center of mass: $\frac{1}{2}I_{\text{cm}}\omega^2$, with $I_{\text{cm}}=\frac{1}{12}mL^2$.
    At $\theta=180^\circ$ the rod is vertical downward so  $v - \frac{L}{2}\omega = v - \frac{L}{2}\left(\frac{4v}{L}\right) = v - 2v = -v,$  and $(v-2v)^2 = v^2$.

 Thus, the energy equation becomes:  $mgL = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{12}mL^2\right)\omega^2.$  Substitute $\omega = \frac{4v}{L}$:  $mgL = m v^2 + \frac{1}{2}\left(\frac{1}{12}mL^2\right) \left(\frac{16 v^2}{L^2}\right)  = m v^2 + \frac{8}{12}m v^2  = m v^2 + \frac{2}{3}m v^2  = \frac{5}{3}m v^2.$  Cancelling $m$ gives:  $\frac{5}{3}v^2 = gL \quad\Longrightarrow\quad v^2 = \frac{3gL}{5}.$

3. Substitute Given Values:
    $g = 10\,\text{m/s}^2$ and $L = \frac{50}{3}\,\text{m}$:  $v^2 = \frac{3 \times 10 \times \frac{50}{3}}{5}  = \frac{10 \times 50}{5}  = \frac{500}{5} = 100,$  so  $v = \sqrt{100} = 10\,\text{m/s}.$

---

Answer:
The velocity of the plank when the rod makes $\theta=180^\circ$ is 10 m/s.

---

Metadata:

• Subject: Physics

• Chapter: Mechanics

• Topic: Conservation of Energy & Momentum (Rotational Motion)

• Difficulty Level: Hard

• Question Type: single_choice (if options provided) / integer (value answer)