Answer to Question #165179 in Optics for Jason

Question

A 2-meter-long pendulum with a 1 kg bob is pulled back from its equilibrium position 25 degrees. 

a. What is its potential energy at that point? 

b. What is its kinetic energy as it passes through the equilibrium position? 

c. What is the pendulum’s maximum velocity?

Answer 1

(a)

$$PE=mgh=mg(L-Lcos\theta)=mgL(1-cos\theta).$$ $$PE=1\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot2\ m\cdot(1-cos25^{\circ})=1.84\ J.$$

(b) The potential energy of the pendulum when the bob is pulled back from its equilibrium position 25 degrees converted into the kinetic energy at the equilibrium position:

$$PE=KE=1.84\ J.$$

(c) The pendulum has maximum velocity at the equilibrium position. Then, we can find the pendulum’s maximum velocity from the law of conservation of energy:

$$KE=\dfrac{1}{2}mv_{max}^2,$$ $$v_{max}=\sqrt{\dfrac{2KE}{m}}=\sqrt{\dfrac{2\cdot 1.84\ J}{1\ kg}}=1.92\ \dfrac{m}{s}.$$