A small 100 gram block is attached to a light hand-held cord, which passes throu

Question

A small 100 gram block is attached to a light hand-held cord, which passes through a smooth hole, drilled through the center of the smooth table top shown below. The table is horizontally oriented. as shown below. The block rotates with initial tangential velocity, v = 20 cm/s, with a radius of revolution, r =40 cm. If the radius of revolution is increased to 50 cm, calculate the final angular frequency, omega f, of the block's motion Select one a. 0.125 rad/s b, 0.25 rad/s G.0.5 rad/s d. 0.625 rad/s problem #8 is the problem above. Calculate the difference in the total orbital angular momentum for the two differing orbits of the block in problem #8. Select one: a. 1.90625 x 10e-3 kgm2/sec b, 3.8125 x 10e-3 kg·m2/sec c. 7.625 x10e-3 kg'm2/sec d, 9.53 x 100-3 kg-m2/sec

Explanation / Answer

1] Initial angular frequency = v/r = 20cm/s/40cm = 0.50 rad/s

final angular frequency = 0.50*50/40 = 0.625 rad/s

2]difference in angular momentum = mv1r1 - mr2^2w2

= 0.100*0.20*0.40 - 0.100*0.50^2*0.625

= 0.007625 kgm^2/s

= 7.625*10^-3 kgm^2/s option c is correct. Note: only d option in first part gives an answer that matches with an option of second part.

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