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Question

A rod of mass 2 kg and length 1 m is lying in the horizontal plane and pivoted about its one end. Initially, it is rotating about its pivoted end (axis perpendicular to horizontal plane) with an angular velocity 2 rad/s. Suddenly, an angular impulse →J is given to the rod, because of which its angular velocity becomes 10 rad/s. Find the magnitude of angular impulse →J.

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Solution

The correct option is **D** 163 kg m2/s

Given,

Mass of rod =M=2 kg

Lemgth of rod =L=1 m

MOI of rod about its end I=ML23=2×123

I=2/3 kg m2

As we know from impulse-angular momentum theorem,

→J=t2∫t1d→L=−−→ΔL=→L2−→L1

where, L2= final angular momentum

L1= initial angular momentum =Iω1

L1=23×2=43 kg m2/s

L2= Final angular momentum =Iω2=23×10=203 kg m2/s

Therefore,

|→J|=|→L2−→L1|=203−43=163 kg m2/s

Given,

Mass of rod =M=2 kg

Lemgth of rod =L=1 m

MOI of rod about its end I=ML23=2×123

I=2/3 kg m2

As we know from impulse-angular momentum theorem,

→J=t2∫t1d→L=−−→ΔL=→L2−→L1

where, L2= final angular momentum

L1= initial angular momentum =Iω1

L1=23×2=43 kg m2/s

L2= Final angular momentum =Iω2=23×10=203 kg m2/s

Therefore,

|→J|=|→L2−→L1|=203−43=163 kg m2/s

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