Question

The length of hour hand of a wrist watch is 1.5 cm. Find magnitude angular acceleration.
(Ans : 0 rad/s²)

Answer 1

length of hour hand of a wrist watch is 1.5cm
initial angular velocity , $\omega_0=\frac{2\pi}{T}$
where T is time taken to complete one circle by hour hand.
$\omega_0=\frac{2\pi}{12\times60\times60}rad/s=\frac{\pi}{12\times1800}$
final angular velocity , $\omega_f=\frac{2\pi}{T}=\frac{2\pi}{12\times60\times60}=\frac{\pi}{12\times1800}$

actually, watch is example of uniform circular motion.
so, initial angular velocity = final angular velocity
now, use formula, $\omega_f=\omega_0+\alpha.t$
$\frac{\pi}{12\times1800}=\frac{\pi}{12\times1800}=\alpha.t$
$\alpha=0$ rad/s²

Answer 2

Answer: The correct answer is 0 rad/s².

Explanation:

Calculate the final angular velocity.

$\omega _{0}=\frac{2\pi }{T}$

Here, T is the time.

T is the time taken to complete one circle by hour hand.

Put T= 12 h and convert into seconds.

$\omega _{0}=\frac{2\pi }{(12)(60)(60)}$

Calculate the initial angular velocity.

$\omega _{i}=\frac{2\pi }{T}$

Put T= 12 h and convert into seconds.

$\omega _{i}=\frac{2\pi }{(12)(60)(60)}$

The expression for the rotational equation of motion.

$\omega _{f}=\omega _{i}+\alpha t$

Here, $\alpha t$ is the angular acceleration.

Put $\omega _{i}=\frac{2\pi }{(12)(60)(60)}$ and $\omega _{0}=\frac{2\pi }{(12)(60)(60)}$.

$\frac{2\pi}{(12)(60)(60)} =\frac{2\pi }{(12)(60)(60)}+\alpha t$

$0=\alpha t$

$\alpha = 0 rad/s^2$

Therefore, the magnitude of the angular acceleration is 0 rad/s².