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

0.88752 kgm²

0.02236 Nm

Step-by-step 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* 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* 10^(-2)* 0.86sin90\\\Rightarrow \tau=0.02236\ Nm`

The torque is 0.02236 Nm