Rotation 4 Question 8

9. A homogeneous rodAB of length L=1.8m and mass M is pivoted at the centre O in such a way that it can rotate freely in the vertical plane (figure).

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Answer:

Correct Answer: 9. (a) 12v7L

(b) 3.5ms1

Solution:

  1. In this problem we will write K for the angular momentum because L has been used for length of the rod.

(a) Angular momentum of the system (rod + insect) about the centre of the rodO will remain conserved just before collision and after collision, i.e. Ki=Kf.

 or MvL4=Iω=ML212+ML24ω or MvL4=748ML2ω i.e. ω=127vL

(b) Due to the torque of weight of insect about O, angular momentum of the system will not remain conserved (although angular velocity ω is constant). As the insect moves towards B, moment of inertia of the system increases, hence, the angular momentum of the system will increase.

Let at time t1 the insect be at a distance x from O and by then the rod has rotated through an angle θ. Then, angular momentum at that moment,

K=ML212+Mx2ω Hence, dKdt=2Mωxdxdt(ω= constant )τ=2MωxdxdtMgxcosθ=2Mωxdxdtdx=g2ωcosωtdt(θ=ωt)

At time t=0,x=L/4 and at time t=T/4 or π/2ω, x=L/2.

Substituting these limits, we get

L/4L/2dx=g2ω0π/2ω(cosωt)dt[x]L/4L/2=g2ω2[sinωt]0π/2ωL2L4=g2ω2sinπ2sin0L4=g2ω2 or ω=2gL

Substituting in Eq. (i), we get

2gL=127vL or v=7122gL=7122×10×1.8v=3.5m/s



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