Question

The length of hour hand of a wrist watch is $1.5cm$. Find magnitude of linear acceleration of a particle on tip of hour hand?

• A. $3.175\times {10}^{-9}m/s^2$
• B. $3.175\times {10}^{-7}m/s^2$
• C. $3.175\times {10}^{-12}m/s^2$
• D. $3.175\times {10}^{-10}m/s^2$

Answer 1

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

Linear acceleration is given by

$a=\sqrt{a_c^2+a_t^2}$

where $a_c$ denotes centripetal acceleration and $a_t$ denotes the tangential acceleration and $a$ denotes linear acceleration of a particle on top of hour hand.

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$a_c={\omega }^{2}r$

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

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

The length of hour hand of a wrist watch is $1.5$cm , $r=1.5$cm $=0.015$m

In case of uniform, circular motion, centripetal or angular acceleration $=0$.

we know, tangential acceleration $=$ radius of circular path $\times$ angular acceleration

$=0.015\times 0=0$

Hence, tangential acceleration $a_t=0$

So, $a=\sqrt{{(3.175\times {10}^{-10})}^2+0}$

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