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 $L = 100\, cm = 1\, m$,$g = 10\, m/s^2$.Let the lowest point be the reference level for potential ener

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 $L = 100\, cm = 1\, m$,$g = 10\, m/s^2$.Let the lowest point be the reference level for potential energy $(PE = 0)$.The height of the bob at an angle $	heta$ with the vertical is $h = L(1 - \cos	heta)$.At $	heta_1 = 60^\circ$,the initial potential energy is $PE_i = mgL(1 - \cos 60^\circ) = mgL(1 - 0.5) = 0.5 mgL$.At $	heta_2 = 30^\circ$,the potential energy is $PE_f = mgL(1 - \cos 30^\circ) = mgL(1 - \frac{\sqrt{3}}{2}) \approx mgL(1 - 0.866) = 0.134 mgL$.Calculating $PE_f$: $PE_f = 0.2 	imes 10 	imes 1 	imes (1 - 0.866) = 2 	imes 0.134 = 0.268\, J = 2.68 	imes 10^6\, erg$.By the law of conservation of mechanical energy,$TE_i = TE_f$.$PE_i + KE_i = PE_f + KE_f$.Since the pendulum is released from rest,$KE_i = 0$.$KE_f = PE_i - PE_f = mgL(\cos 30^\circ - \cos 60^\circ) = 0.2 	imes 10 	imes 1 	imes (0.866 - 0.5) = 2 	imes 0.366 = 0.732\, J = 7.32 	imes 10^6\, erg$.

$A$ 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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