Question

A long, uniform rod of length  0.500 mm and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.42 rad/srad/s and a moment of inertia about the axis of 3.20×10⁻³ kg⋅m2kg⋅m2 . An insect initially standing on the rod at the axis of rotation decides to walk to the other end of the rod. When the insect has reached the end of the rod and sits there, its tangential speed is 0.173 m/sm/s . The insect can be treated as a point mass.

A) Find the mass of the rod?

B)  Find the mass of the insect?

Answer 1

A)Moment of inertia of rod =M/3

M=mass of rod=?

L=its length=0.5m

=3.2xkgm2

=M/3

M=3 /=3x3.2x/=0.0384kg=3.84xkg

B)Initial angular momentum of rod =

=initial angular velocity=0.42rad/s

Initial angular momentum of rod ==3.2xx0.42=1.344x

Final angular momentum (when the insect reach at other end) =''

'=final moment of inertia=+m

m=mass of insect=?

'=3.2x+(mx)

'=final angular velocity

tangential velocity v=0.173m/s

But v=L'

'=v/L=0.173/0.5=0.346rad/s

Final angular momentum =''=[3.2x+(mx)]0.346

By conservation of angular momentum,

Initial angular momentum=final angular momentum

1.344x=[3.2x+(mx)]0.346

1.344x/0.346=[3.2x+(mx)]

3.884x=[3.2x+(mx)]

3.884x-3.2x=(mx)

0.684x=(mx)

m=0.684x/=2.736xkg=mass of insect