Question

A small object with mass 1.30 kg is mounted on one end of arod

Answer 1

Answer:

(a) $I_{system} = 1.014\ kg.m^{2}$

(b) $\tau = 0.0179\ N-m$

Solution:

As per the question:

Mass of the object, m = 1.30 kg

Length of the rod, L = 0.780 m

Angular speed, $\omega = 5010\ rev/min$

Now,

(a) To calculate the rotational inertia of the system about the axis of rotation:

Since, the rod is mass less, the moment of inertia of the rotating system and that of the object about the rotation axis will be equal:

$I_{system} = ML^{2} = 1.30\times (0.780)^{2} = 0.791\ kg.m^{2}$

(b) To calculate the applied torque required for the system to rotate at constant speed:

Drag Force, F = $2.30\times 10^{- 2}\ N$

$\tau = FLsin\theta 90 = 2.30\times 10^{- 2}\times 0.780\times 1 = 0.0179\ N-m$