Work Done by the Gravitational Force — Question 5

Question 5

A fixed pendulum support does no work while gravity changes speed

Take a uniform gravitational field with $g=10$ m/s$^2$, upward y positive, and neglect air resistance. A 0.50 kg bob hangs from a massless inextensible cord of length 2 m at a fixed pivot. It is released from rest with the cord $60^\circ$ from vertical. The cord remains taut.

1. Find the vertical drop to the lowest point and gravity work.

2. Find speed at the lowest point and explain why tension does no work.

3. Find tension at the lowest point using the radial acceleration.

Question 5 - Solution

A fixed pendulum support does no work while gravity changes speed

(a) Use the pendulum geometry.

The governing relation is $$\boxed{h=L-L\cos\theta,\qquad W_g=mgh}$$

$$h=L(1-\cos60^\circ)=1\ \mathrm m,$$ so $$W_g=mgh=5\ \mathrm J.$$

(b) Apply the work theorem.

Tension is radial and displacement tangent to the fixed circle, so its work is zero. Thus $$\tfrac12(0.50)v^2=5,$$ giving $$v=\sqrt{20}=2\sqrt5\ \mathrm{m/s}.$$

(c) Include the inward acceleration.

At the bottom, $$T-mg=mv^2/L.$$ Hence $$T=5+0.50(20)/2=10\ \mathrm N,$$ larger than weight even though tension did no work.