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  • Help me answer: A thin uniform bar of length L and mass 8 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively. The masses stick to th

A thin uniform bar of length L and mass 8 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively. The masses stick to the bar after collision at a distance L/3 and L/6 respectively from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be :

  • Option 1)  (v/5L)
  • Option 2) (6v/5L)
  • Option 3) (3v/5L)
  • Option 4) (v/6L)
 

Answers (2)

best_answer

As we have learnt

Moment of inertia of uniform rod of length(l) -

I= \frac{Ml^{2}}{12}

- wherein

About axis passing through its centre & perpendicular to its length.

 

Law of conservation of angular moment -

\vec{\tau }= \frac{\vec{dL}}

- wherein

If net torque is zero

i.e. \frac{\vec{dL}}= 0

\vec{L}= constant

angular momentum is conserved only when external torque is zero .

 

 Centre of mass from system from 0

=\frac{8m*0+m(L/3)-2m(L/6)}{8m+m+2m}=0

So,centre of mass is at zero.

From conservation of ang.momentum;L_{i}=L_{f}

L_{i}=m.(2v)*(L/3)+2mv*(L/6)=mvL

L_{f}=\left [ (8m).\frac{L^{2}}{12}+m.(L/3)^{2}+2m.(L/6)^{2} \right ]\omega

       =\left [ \frac{2}{3}mL^{2}+\frac{mL^{2}}{9}+\frac{mL^{2}}{18} \right ]\omega =\left ( \frac{12+2+1}{18} \right )mL^{2}\omega= \frac{5}{6}mL^{2}\omega

\frac{5}{6}mL^{2}\omega=mvl \therefore \omega =\frac{6v}{5L}


Option 1)

\frac{v}{5L}

Option 2)

\frac{6v}{5L}

Option 3)

\frac{3v}{5L}

Option 4)

\frac{v}{6L}

Posted by

Plabita

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