Question

Four equal masses M are spaced at equal intervals (each of length d) along a horizontal, straight rod whose mass can be ignored. The system is to be rotated about a vertical axis passing through the mass at the left end of the rod and perpendicular to it.

(a) What is the moment of inertia of the system about this axis? (Assume that each mass is a particle, with no finite size.)
(b) What minimum force, applied to the farthest mass, will give the system an angular acceleration?

Answer 1

a) 14M
`d^(2)`

Step-by-step explanation:

a)The inertia of a particle moving in a circular axis is given by,

`I=Mr^(2) \\`

I = Moment of inertia

M = mass of the particle

r = perpendicular distance from axis of rotation.

And by adding moment of inertia of each particle we can come to the moment of inertia of the system.

I = M
`(0d)^(2)`+M
`d^(2)` +M
`(2d)^(2)`+M
`(3d)^(2)`

= 14M
`d^(2)`

b) Your question is incomplete but I'll write how to find the minimum force required to give a system given angular acceleration.

Minimum force is found when applied from the furthest point to the axis of rotation in the system.

, by τ = Fr, whereτ = torque , F = Force , = perpendicular distance from axis of rotation.

For minimum force r = 3d

And also τ = Iα where I = Moment of inertia and α = angular acceleration

By combining the two equations you get minimum force as,

F = Iα/r

F' = 14M
`d^(2)`α/3d

= 14Mαd/3