Question

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance ?

Answer 1

Given Data

• Pendulum length: \(L = 1.5\) m
• Released from horizontal (90°)
• Energy dissipated: 5% against air resistance
• \(g = 9.8\) m/s²

● Start (Horizontal)
/ 
/ L = 1.5 m
/
● Bottom (v = ?)

Energy Analysis

Initial PE = \(mgL\) (reference at bottom)

95% converts to KE at bottom:

$$KE_\text{final} = 0.95 \times mgL$$ $$\frac{1}{2}mv^2 = 0.95 \, mgL$$ $$v = \sqrt{2 \times 0.95 \times gL}$$

Calculation

$$v_\text{ideal} = \sqrt{2gL} = \sqrt{2 \times 9.8 \times 1.5} = \sqrt{29.4} = 5.42 \, \text{m/s}$$

$$v_\text{actual} = \sqrt{0.95} \times 5.42 = 0.975 \times 5.42 = 5.28 \, \text{m/s}$$

Speed at Lowermost Point

\(v = \textbf{5.28 m/s}\)

Verification Table

Condition	PE Initial	KE Final	Speed
No resistance	100%	100%	5.42 m/s
5% loss	100%	95%	5.28 m/s

Key Insight

• Height drop = full radius \(L\) (horizontal start)
• \(\sqrt{0.95} \approx 0.975\) → only 2.5% speed reduction
• Energy loss minor effect due to square-root relationship