Question

The length of hour hand of a wrist watch is $1.5cm$. Find magnitude of:
a) Angular velocity
b) Linear velocity
c) Angular acceleration
d) Radial acceleration
e) Tangential acceleration
f) Linear acceleration of a particle on tip of hour hand.

Answer 1

Length$l=1.5cm=1.5\times {10}^{-2}m$

Time period of hour hand $T_h=12hours=12\times 60\times 60s=43200s$

Angular velocity$\omega =\frac{2\pi }{T}=\frac{2\pi }{43200}=1.454\times {10}^{-4}rad/s$

Linear velocity$v=l\omega =1.5\times {10}^{-2}\times 1.454\times {10}^{-4}=2.182\times {10}^{-6}m/s$

Angular acceleration$\alpha =\frac{d\omega }{dt}$

but $\omega$ is constant

$\therefore \alpha =0$

Radial acceleration$a_r=l{\omega }^2=1.5\times {10}^{-2}\times {(1.454\times {10}^{-4})}^2=3.17\times {10}^{-10}m/s^2$

Tagential acceleration$a_t=\frac{dv}{dt}$

but $v$ is constant

$\therefore a_t=0$

Linear acceleration of the particle on tip of an hour hand will be equal to the radial acceleration of the hour hand $=3.17\times {10}^{-10}m/s^2$