Question

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

0.780 m long and of negligible mass. The system rotatesin a
horizontal circle about the other end of the rod at
5010rev/min.
a.) Calculate the rotational inertia of the system about
theaxis of rotation
b.)There is an air drag of 2.30 X 10^-2 N on the
object,directed opposite its motion. What torque must be applied to
thesystem to keep it rotating at constant speed?

Answer 1

(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* (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* 10^(- 2)\ N`

`\tau = FLsin\theta 90 = 2.30* 10^(- 2)* 0.780* 1 = 0.0179\ N-m`