Question

The length of hour hand of a wrist watch is 1.5 cm. Find magnitude linear acceleration of a particle on tip of hour hand.
(Ans : 3.175 x 10⁻¹⁰ m/s²)

Answer 1

linear acceleration is given by ,
$a=\sqrt{a_c^2+a_t^2}$

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

here, $a_c=3.175\times10^{-10}m/s^2$ [ it can be found by $a_c=\omega^2r$, where $\omega$ in case of hour hand is given by $\omega=\frac{2\pi}{T_H}$ , $T_H$ = 12 × 60 × 60 sec]
$a_t=0$

so, $a=\sqrt{(3.175\times10^{-10})^2+0^2}$
a = 3.175 × 10^-10 m/s²