Question

A rigid semi-circular wire of radius r = 50 cm is supported on its vertical plane by a hinge at O and a smooth peg A. If the peg starts from O and moves with constant speed $v_0 = 2 \ cm/s$ along the horizontal, the angular velocity $\Omega$ (in rad/s) of the wire at the instant $\theta = 60^\circ$ is $\frac{x}{100} rad/s$, where x is _____.

Answer 1

x = 8

Answer 2

We note that the semi‐circular wire has one end fixed at O and the other end A sliding on a smooth peg that is moving horizontally with speed

$v_0=2~\text{cm/s}$.

Since the wire is rigid and rotates about O, point A (the free end) has a speed due to rotation given by

$v_A=\Omega \,r$,

but its direction is tangential to the circle. If we take θ as the angle between the line OA and the vertical (with O on the wall), then the x–coordinate of A is

$x_A=r\sin\theta$,

and the corresponding horizontal component of the tangential velocity is

$(v_A)_x=\Omega\, r\cos\theta$.

Because the peg forces A to move purely horizontally with speed $v_0$, we equate:

$\Omega\, r\cos\theta=v_0$.

Thus,

$\Omega=\frac{v_0}{r\cos\theta}$.

Substitute the given values $r=50~\text{cm}$, $v_0=2~\text{cm/s}$ and $\theta=60^\circ$ (with $\cos 60^\circ=0.5$):

$\Omega=\frac{2}{50\times0.5}=\frac{2}{25}~\text{rad/s}$.

Since the angular velocity is expressed in the problem as $\frac{x}{100}~\text{rad/s}$, we have:

$\frac{x}{100}=\frac{2}{25} \quad\Longrightarrow\quad x=100\times\frac{2}{25}=8$.

Minimal Explanation of the Solution:

1. Tangent speed at A: $v_A=\Omega r$; horizontal component is $\Omega\,r\cos\theta$.
2. Match to peg’s speed: $\Omega\,r\cos\theta=v_0$ so $\Omega=\frac{v_0}{r\cos\theta}$.
3. Substitute: $r=50$, $v_0=2$, $\cos 60^\circ=0.5$ to get $\Omega=\frac{2}{25}$ rad/s.
4. Compare: $\frac{2}{25}=\frac{8}{100}$ so $x=8$.