Question

A pendulum consists of a 2.0 kg stone swinging on a 4.0 m string of negligible mass. The stone has a speed of 8 m/s when it passes its lowest point. (a) What is the speed when the string is at 60° to 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?

Answer 1

(a) By using the law of conservation of energy,

(1/2)mv1^2 + mg*0 = (1/2)mv2^2 + mg*y2

Where y2 = L - Lcos theta = L*(1 - cos theta)

0.5mv1^2 = 0.5*mv2^2 + mg*L*(1 - cos theta)

v2 = sqrt(v1^2 - 2 g*L(1 - cos theta)

= sqrt(8^2 - 2*9.8*4*(1 - cos 60 degree)

= 4.98 m/s

(b) at greatest angle, v2 = 0

By law of conservation of energy,

mv1^2 /2 = mgy2

v1^2 / 2 = g*y2

Where, y2 = L(1 - cos theta)

v1^2 / 2 = g*L*(1 - cos theta)

cos theta = 1 - v1^2 /(2*g*L) = 1 - 8^2 /(2*9.8*4)

theta = arccos(1 - 8^2 /(2*9.8*4))

= 79.63 degree

(c) At the lowest point, total mechanical energy

= (1/2)mv1^2

= (1/2)*2*(8)^2

= 64 J