please help, finding the kinetic energy of point mass

Question

A point-mass of 5 grams at the end of a massless 20 centimeter long string is rotated at a rate of 2 revolutions per second. Find the kinetic energy in ergs of the point mass.

Please show me the detailed step by steps as I am lost.

Answer 1

Solution:

The kinetic energy an object moving with a velocity V is KE = 1/2 mV²

In case of a rotating mass KE = 1/2 mV_t² where V_t = tangential velocity = ω r , where ω is the angular velocity and r is the radius of rotation given as 20 cm or the length of the string

The angular velocity is given as 2 rev/s knowing that one rev is equivalent to 2π in radians

Then ω = 2 ( 2π rad)/ s= 4π rad/s and V_t = 4π rad/s ( 20 cm) = 80 π cm / s = 251.3 cm/s

or V_t = 4π rad/s ( 0.2 m) = 2.513 m /s

KE = 1/2 ( 5g)(251.3 cm/s)² = 157879.2 g .cm²/s² = 157879 ergs, 1 erg = 1 g.cm²/s²

or KE = 1/2 ( 5 x 10^-3 kg)( 2.513 m/s)² = 0.015789 kg.m²/ s² = 0.015789 J ( joules)

Note that 1 J = 1 N. m = 1 (kg .m/s²).m

also note that 1 erg = 10^-7 J or 1 J = 10⁷ ergs , so 157879 erg ( 10^-7 J/erg) = 0.015789 J