Question

At 2 'o'clock in the afternoon, the hour hand of the clock has a mechanism failure because of which its hour hand (length =0.5m) falls under gravity with its one end being attached to the clock centre, calculate the angular velocity of the hand as it goes past the 6 ‘o clock mark.
A) \[\sqrt{2}\] rad/s
B) \[10\sqrt{2}\] rad/s
C) \[10\sqrt{3}\] rad/s
D) \[10\sqrt{5}\] rad/s

Answer 1

Hint: The hour hand falls under the influence of gravity, so, the only force acting is the gravitational force.
Thus the acceleration of the body must be equal to the acceleration due to gravity.

Complete step by step answer:
It was resting at 2’ o'clock. We can find out the gravitational potential energy,
\[U=mgl\sin \xi \], where \[\xi \] is angle between 2 'O clock and horizontal = 30°
$
\implies U=mgl\sin 30 \\
\implies U=\dfrac{mgl}{2} \\
$
Now, when the hour hand is at 6’ o'clock, since it was moving it will have both kinetic energy and potential energy, so,
\[\Rightarrow U=-mgl\] and \[K=\dfrac{m{{v}^{2}}}{2}\]
By using the law of conservation of energy which says total energy must be constant,
\[\Rightarrow \] \[\dfrac{mgl}{2}=-mgl+\dfrac{m{{v}^{2}}}{2}\]
= \[v=\sqrt{3gl}\]
But we want to find out the angular velocity, so by using \[v=l\omega \]
\[\omega =\dfrac{\sqrt{3gl}}{l}=\sqrt{\dfrac{3g}{l}}=\sqrt{\dfrac{30}{0.5}}=10\sqrt{3}\]
\[\Rightarrow \omega = 10\sqrt{3}\]rad/s

So, the correct answer is “Option C”.

Note:
Here when the hand falls due to the gravitational force acts on the two sides and thus a moment is created and thus the hand starts rotating under the influence of gravity, thus, here torque is provided by the gravity itself. The rotation circle radius is the length of the hour hand. No another force acts on the system, so energy remains conserved.