Question

A 3.00 × 10^-6 kg ant rides the second hand of an enormous vertical wall clock. The ant smoothly sweeps a circle of radius 50.0 m. What is the magnitude of the upward force felt by the ant due to the second hand when it is exactly at 6 AM?

Answer 1

Answer:

1.65×10⁻⁶ N

Explanation:

m = Mass of ant = $3\times 10^{-6}\ kg$

r = Radius = 50 m

t = Time taken to complete on rotation = 60 seconds

Angular velocity

$\omega=\frac{2\pi}{t}\\\Rightarrow \omega=\frac{2\pi}{60}$

Centripetal acceleration is force that is acting outward

$F_c=m\omega^2r\\\Rightarrow F_c=3\times 10^{-6}\times \left(\frac{2\pi}{60}\right)^2\times 50\\\Rightarrow F_c=1.65\times 10^{-6}\ N$

The magnitude of the upward force felt by the ant due to the second hand would be 1.65×10⁻⁶ N