find the kinetic energy in ergs, please help

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 in ergs of the bar when it is rotated at a rate of 3 revolutions per second.

Please show me step by steps also, as I'm still extremely confused with answering this.

Answer 1

Solution:

This is an example of physical body rotation where ω = √( m g r / I) with r being the distance from the pivot of rotation to the center of gravity of the rotating object, In this case and for a bar attached to a massless string it will be the length of the string plus the length from the end of the string to the center of gravity of the rotating bar. r = 70 cm + ( 30 cm/2) = 85 cm = 0.85 m

I is the moment of inertia of a long bar = 1/3 m L² where L is the bar length = 30 cm =0.3 m

then ω = √( m g r / 1/3 m L²) simplifying ω = √ ( 3 g r / L²) = √ ( 3 . 9.8 m/s² . 0.85 m / ( 0.3 m)² ) =

√ ( 25 m²/ s² / 0.09 m²) = 16.7 /s

= 16.7 s^-1 = 16.7 rad/s

The Kinetic energy is KE = 1/2 m V_t² where V_t is the tangential velocity and

V_t = ω r = 16.7 rad / s ( 0.85 m) = 14.2 m/s/

KE = 1/2 ( 12 g x kg/ 1000 g)( 14.2 m/s)² = 1.21 kg.m²/ s² = 1.21 J ( joules) = 1.21 X 10⁷ ergs

note that 1 J = 10⁷ ergs or 1 erg = 10^-7 J

or KE = 1/2 ( 12 g) ( 14.2 m/s x 100 cm/m) = 1/2 ( 12 g) ( 1420 cm/s)² = 12,100,000 g.cm²/s² = 1.21 x 10⁷ ergs