A puck of mass m1-67.0 g and radius r1 4.10 cm glides across an air table at a s

Question

A puck of mass m1-67.0 g and radius r1 4.10 cm glides across an air table at a speed of v = 1.50 m/s as shown in Figure a It makes a glancing collision with a second puck of radius r2 = 6.00 cm and mass m2 = 115 g (initially at rest) such that their rims just touch. Because their rims are coated with instant-acting glue, the pucks stick together and rotate after the collision (Figure b) rig (a) What is the magnitude of the angular momentum of the system relative to the center of mass? Kg m2/s (b) What is the angular speed about the center of mass? rad/s Need Help?

Explanation / Answer

Calculate the location of the of mass from first puck just after collision .

R = m2*(r1+r2)/(m1+m2)

    = 115*(4.1+6.00)/67+115=6.381 cm

Thus, the angular momentum of the system is:

L = m1v*R

=(0.067)(1.5)(0.06381) =0.0064 kgm²/s

this is the final angular momentum.

Total moment of inertia of the system is:

Now I = I1+I2+(M*r)

           =0.5*0.067*0.041² +0.5*0.115*0.06² + 0.067*0.06381² + 0.115*(0.101 – 0.06381)² =

           = 6.95*10-4 kgm²

wf = L/I = 9.1 rad/sec

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