Question

A pendulum consists of a stone with mass m swinging on a string of length L and negligible mass. The stone has a speed of v0 when it passes its lowest point. (a) Write an expression for the speed of the stone when the string is at some angle theta with the vertical. (b) What is the greatest angle with the vertical that the string will reach during the stone's motion? (c) If the potential energy of the pendulum-Earth system is taken to be zero at the stone's lowest point, what is the total mechanical energy of the system? State all your answers in terms of the given variables and g.

Answer 1

Answer:

(a) $v = \sqrt{ mv_0^2 - 2gL(1-\cos(\theta))}$

(b) $\theta = \arccos(\frac{2gL-v_0^2}{2gL})$

(c) $E_{total} = \frac{1}{2}mv_0^2$

Explanation:

(a) The total mechanical energy of the system is conserved.

$K_A + U_A = K_B + U_B\\\frac{1}{2}mv_0^2 + 0 = \frac{1}{2}mv^2 + mgh \\h = L - L\cos(\theta) = L(1 - \cos\theta)\\\frac{1}{2}mv^2 = \frac{1}{2}mv_0^2 - mgL(1-\cos(\theta))\\v^2 = mv_0^2 - 2gL(1-\cos(\theta))\\v = \sqrt{ mv_0^2 - 2gL(1-\cos(\theta))}$

(b) The conservation of energy states

$K_A + U_A = K_M + U_M\\\frac{1}{2}mv_0^2 + 0 = 0 + mgL(1-\cos(\theta))\\1-\cos(\theta) = \frac{v_0^2}{2gL}\\\cos(\theta) = 1-\frac{v_0^2}{2gL} = \frac{2gL-v_0^2}{2gL}\\\theta = \arccos(\frac{2gL-v_0^2}{2gL})$

(c) As explained in part (a) the total mechanical energy of the system is equal to the initial kinetic energy, since the potential energy of the system at that point is zero.

$E_{total}= K_A + U_A = K_A + 0\\E_{total} = \frac{1}{2}mv_0^2$