Question

A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5100 rev/min.
(a) Calculate the rotational inertia of the system about the axis of rotation.
(b) There is an air drag of 2.60 x 10.

Answer 1

Answer:

0.88752 kgm²

0.02236 Nm

Explanation:

m = Mass of ball = 1.2 kg

L = Length of rod = 0.86 m

$\theta$ = Angle = 90°

Rotational inertia is given by

$I=mL^2\\\Rightarrow I=1.2\times 0.86^2\\\Rightarrow I=0.88752\ kgm^2$

The rotational inertia is 0.88752 kgm²

Torque is given by

$\tau=FLsin\theta\\\Rightarrow \tau=2.6\times 10^{-2}\times 0.86sin90\\\Rightarrow \tau=0.02236\ Nm$

The torque is 0.02236 Nm