find the kinetic energy of the bar

Question

A 30 cm long metal bar with mass 12 grams is attached to a pivot with a massless string 70 cm long. Find the kinetic energy of the bar in ergs when it is rotated at a rate of 3 revolutions per second.

Answer 1

MOI of Bar about its Center of mass = Ml²/12 = 12*10^-3 *( 0.3) ² /12 = 9 * 10^-5 Kg m² =I

Shifting the axis to pivot = I + Md² = I + Md² , d= 85 cm = 85*10^-2 m

8.67 *10^{- 3} = 8.67*10 ^-3 Kg m²

KE = 1/2( I w² ) = 0.5 * 8.67 *10^-3 * 2* pi * f , where f = frequency = 3

KE = 0.5 *8.67 *10 ^-3 *2 * 3.14 * 3 J

Now change in ergs

1J = 10 ⁷ ergs