Question

Two points of a rod have velocities 3v and 2v as shown in figure. The angular speed of the rod will be $\frac{xv}{4l}$.

The value of x is ______.

Answer 1

20

Answer 2

Let the rod have a translational velocity V of its center of mass and an angular speed ω. For a vertical rod (length l) rotating about its center, the top end (at +l/2) and bottom end (at –l/2) have velocities:

$$v_{\text{top}} = V + \omega \frac{l}{2},\quad v_{\text{bottom}} = V - \omega \frac{l}{2}.$$

Given:

• Top end velocity = $3v$ (to the right)
• Bottom end velocity = $2v$ (to the left), which we take as $-2v$ (since left is negative).

Hence, we have:

$$V + \omega \frac{l}{2} = 3v \quad \text{(1)},$$ $$V - \omega \frac{l}{2} = -2v \quad \text{(2)}.$$

Subtract (2) from (1):

$$\left(V + \omega \frac{l}{2}\right) - \left(V - \omega \frac{l}{2}\right) = 3v - (-2v),$$ $$\omega l = 5v \quad \Longrightarrow \quad \omega = \frac{5v}{l}.$$

The problem states that the angular speed is given as:

$$\omega = \frac{xv}{4l}.$$

Setting the two expressions equal:

$$\frac{5v}{l} = \frac{xv}{4l}.$$

Cancel $\frac{v}{l}$:

$$5 = \frac{x}{4} \quad \Longrightarrow \quad x = 20.$$

Decompose the rigid body motion into translation (V) and rotation (ω). For endpoints, write $V \pm \omega \frac{l}{2}$ equal to the given velocities. Subtract the equations to solve for ω, then equate it with the given expression to find x.