Question

A rod of length 10 m rotates about an axis perpendicular to its length and through its center. Two particles of masses m1 = 4.0 kg and m2 = 3.0 kg are connected to the ends of the rod. What is the angular momentum of the system if the speed of each particle is 2.5 m/s?

Answer #1

Angular momentum of a point mass is given by rmv sin, where r is position, m is mass, v is velocity , is angle between position and velocity.

Here, position of each particle=10/2= 5 m.

For first particle: r=5 m, m=4 kg, v=2.5 m/s,=90 degrees. So, angular momentum = 5*4*2.5 sin90 = 50 kg m2/s

Similarly,for second particle: r=5 m, m=3 kg, v=2.5 m/s,=90 degrees. So, angular momentum = 5*3*2.5 sin90 = 37.5 kg m2/s.

So,total angular momentum = 50+37.5 = 87.5 kgm2/s

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