A simple pendulum of mass 200, g and length 1 | vedclass

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(A) Given: Mass $m = 200\, g = 0.2\, kg$,Length $L = 100\, cm = 1\, m$,$g = 10\, m/s^2$.Initial angle $	heta_i = 60^\circ$,Final angle $	heta_f = 30

Vedclass · Accepted Answer

$7.32 	imes 10^6\, erg, 2.68 	imes 10^6\, erg$

Vedclass · Answer

(A) Given: Mass $m = 200\, g = 0.2\, kg$,Length $L = 100\, cm = 1\, m$,$g = 10\, m/s^2$.Initial angle $	heta_i = 60^\circ$,Final angle $	heta_f = 30^\circ$.Taking the lowest point of the swing as the reference level for potential energy ($PE = 0$ at $	heta = 0^\circ$):Potential energy at angle $	heta$ is $PE = mgL(1 - \cos	heta)$.At $	heta_f = 30^\circ$,$PE_f = mgL(1 - \cos 30^\circ) = 0.2 	imes 10 	imes 1 	imes (1 - \sqrt{3}/2) = 2 	imes (1 - 0.866) = 2 	imes 0.134 = 0.268\, J = 2.68 	imes 10^6\, erg$.By conservation of mechanical energy,Total Energy $E = mgL(1 - \cos 60^\circ) = 0.2 	imes 10 	imes 1 	imes (1 - 0.5) = 1\, J$.Kinetic Energy $KE_f = E - PE_f = 1 - 0.268 = 0.732\, J = 7.32 	imes 10^6\, erg$.

$A$ simple pendulum of mass $200\, g$ and length $100\, cm$ is moved aside until the string makes an angle of $60^\circ$ with the vertical. The kinetic and potential energies of the bob,when the string is inclined at $30^\circ$ to the vertical,are

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........$kJ$

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