Question

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

• A. $3.175\times {10}^{-4}m/s^2$
• B. $4.175\times {10}^{-4}m/s^2$
• C. $6.754\times {10}^{-4}m/s^2$
• D. $4.754\times {10}^{-4}m/s^2$

Answer 1

The correct option is A $3.175\times {10}^{-4}m/s^2$

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

$r=1.5$cm

For hour hand$T_H=12$hr$=12\times 60\times 60=43200$sec

We know that angular velocity,$\omega =\frac{2\pi }{T_H}$

$=\frac{2\pi }{43200}$

$=\frac{2\times 22}{7\times 43200}=1.455\times {10}^{-4}$rad/s.

Radial acceleration$\alpha ={\omega }^{2}r$

$={(1.455\times {10}^{-4})}^2\times 0.015$m/$s^2$

$=3.175\times {10}^{-10}$m/$s^2$