Answer to Question #111420 in Mechanics | Relativity for Brandon Salas

Question #111420
An ideal pendulum of length L = 1.2 m supports a mass of m = 0.5 kg. Initially the pendulum is lifted such that it makes an angle of θ = 21 degrees with respect to the vertical.

part(A) When the pendulum is released from rest, what is the maximum speed, vm in m/s, the mass reaches?
Expert's answer

When the pendulum is in lifted state, it has potential energy "mgh", where "h" is the height above the origin, and zero kinetic energy.

Pendulum reaches its maximum speed, when going through the equilibrium point ("\\theta = 0").

Lets choose the origin to be located at the equilibrium position of the mass.

Then, in lifted state, the height above the origin will be equal to "h = L - L \\cos \\theta" and in maximum speed state "h = 0".

Hence, total energy in lifted state is just potential energy "E_1 = mgh = mg (L - L\\cos \\theta)", and in maximum speed state it is just kinetic energy "E_2 = \\frac{m v^2}{2}".

By the law of conservation of energy, these two energies must be equal, therefore "m g L (1- \\cos \\theta) = \\frac{m v^2}{2}", from where "v = \\sqrt{ 2 g L (1 - \\cos \\theta)} \\approx 4.69\\frac{m}{s}".

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