Question

The length of the seconds hand in a wall clock is $10\text{cm}$. The magnitude of velocity and acceleration of its tip are

• A. ${10}^{-2}\text{m/s}$ and ${10}^{-3}{\text{m/s}}^2$
• B. ${10}^{-2}\text{m/s}$ and ${10}^{-4}{\text{m/s}}^2$
• C. ${10}^{-1}\text{m/s}$ and ${10}^{-2}{\text{m/s}}^2$
• D. ${10}^{-2}\text{m/s}$ and ${10}^{-2}{\text{m/s}}^2$

Answer 1

The correct option is A ${10}^{-2}\text{m/s}$ and ${10}^{-3}{\text{m/s}}^2$

The seconds hand of a clock completes one revolution in $60s$. Therefore, frequency $f=\frac{1}{60}Hz$

The angular velocity of the seconds hand is given by
$\omega =2\pi f=\frac{2\pi }{60}=\frac{\pi }{30}\text{rad/s}$


The tip of the seconds hand is at a distance $r=10\text{cm}=0.1\text{m}$ from the axis of rotation. Thus, linear velocity of the tip is given by
$v=\omega r=\left(\frac{\pi }{30}\right)\left(0.1\right)\approx 0.01={10}^{-2}\text{m/s}$

The direction of the velocity is tangential. The tip moves in a circle of radius $r=0.1\text{m}$ with a constant speed $v={10}^{-2}\text{m/s}$ .

Thus, tangential acceleration of the tip is zero and its radial (centripetal) acceleration is
$a_c=\frac{v^2}{r}=\frac{(0.01{)}^2}{0.1}=0.001={10}^{-3}{\text{m/s}}^2$

The direction of centripetal acceleration is radially inward.
Hence option A is the correct answer.