Question

A 5.0 kg and 7.0 kg weight are placed on opposite ends of a massless 4.0 m rod (5.0 kg on the left, 7.0 kg on the right). Determine the total moment of inertia of the system and net torque if

The axis of rotation is halfway between the two masses
The axis of rotation is 0.50 m to the left of the 7.0 kg weight.
Which axis of rotation produces the most torque and why?

Answer 1

Axis of rotation in case 2 produces the most torque than case 1

Step-by-step explanation:

Moment of inertia is a quantity that determines the torque needed for a required angular acceleration about a rotational axis.

formula wise its mass multiplied by the square of distance of it from the axis of rotation.

⇒
`I=mr^(2)`

for n number of discreet bodies,
`I=\sum mr^(2)`

for a rigid body,
`I=\int\limits {\delta mr^(2)} \, dr`

And
`\tau=I\alpha`

where,
`\tau` is torque

`I` is moment of inertia and

`\alpha` is angular acceleration

Now coming to the problem,

Case 1(When the axis of rotation is halfway between the two masses):

`I=5*2^(2)+7*2^(2)`

`=20+28`

`=48kgm^(2)`

Case 2(When the axis of rotation is 0.50 m to the left of the 7.0 kg weight):

`I=5*3.5^(2)+7*0.5^(2)`

`=61.25+1.75`

`=63kgm^(2)`

From
`\tau=I\alpha` we know that
`\tau` is proportional to
`I` when angular acceleration is constant which means axis of rotation in case 2 produces the most torque than case 1 because moment of inertia in case 2 is more than that of case 1