PLEASE EXPLAIN THOROUGHLYSolutionB In 1 revolution it covers

Solution

B)

In 1 revolution, it covers 2*pi radians.

Hence, in 200 revolutions, it\'ll cover 200*2*pi = 1256.64 radians

Work done = Torque*angles turned in radians

= 120 * 1256.64

= 150796.4 J

= 150.8 kJ ~ 151 kJ

Hence, choice (ii) is correct.

D)

MOI of table = mr² / 2

= 15 * 0.7² / 2

= 3.675 kg-m²

Total MOI of table + performer = 10 + 3.675 = 13.675 kg-m²

By angular momentum conservation before and after the jump,

I₁ * w₁ = I₂ * w₂

3.675 * 1.05 = 13.675 * w₂

w₂ = 0.28 rad/s