Question

The length of hour hand of a wrist watch is $1.5cm$. Find magnitude of angular acceleration?

• A. $0rad/se{c}^2$
• B. $2\pi rad/se{c}^2$
• C. $2rad/se{c}^2$
• D. $1rad/se{c}^2$

Answer 1

Answer: (A)

$0rad/se{c}^2$

Length of hour hand of a wrist watch is $1.5$cm

Initial angular velocity${\omega }_0=\frac{2\pi }{T}$

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

${\omega }_0=\frac{2\pi }{12\times 60\times 60}$rad/s$=\frac{\pi }{12\times 1800}$

final angular velocity,${\omega }_f=\frac{2\pi }{T}=\frac{2\pi }{12\times 60\times 60}$rad/s$=\frac{\pi }{12\times 1800}$ is an example of uniform circular motion.

So, initial angular velocity$=$final angular velocity

using the formula, ${\omega }_f={\omega }_0+\alpha .t$

$\Rightarrow \frac{\pi }{12\times 1800}=\frac{\pi }{12\times 1800}+\alpha .t$

$\Rightarrow \alpha =0$rad/$s^2$