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

1.65×10⁻⁶ N

Step-by-step explanation:

m = Mass of ant =
`3* 10^(-6)\ kg`

r = Radius = 50 m

t = Time taken to complete on rotation = 60 seconds

Angular velocity

`\omega=(2\pi)/(t)\\\Rightarrow \omega=(2\pi)/(60)`

Centripetal acceleration is force that is acting outward

`F_c=m\omega^2r\\\Rightarrow F_c=3* 10^(-6)* \left((2\pi)/(60)\right)^2* 50\\\Rightarrow F_c=1.65* 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