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Chapter 1: Q46E (page 462)

Question: An \({\bf{85}}.{\bf{0}} - {\bf{kg}}\) mountain climber plans to swing down, starting from rest, from a ledge using a light rope \({\bf{6}}.{\bf{50}}\,{\bf{m}}\) long. He holds one end of the rope, and the other end is tied higher up on a rock face. Since the ledge is not very far from the rock face, the rope makes a small angle with the vertical. At the lowest point of his swing, he plans to let go and drop a short distance to the ground. (a) How long after he begins his swing will the climber first reach his lowest point? (b) If he missed the first chance to drop off, how long after first beginning his swing will the climber reach his lowest point for the second time?

Short Answer

Expert verified

(a) \(1.27\,{\rm{s}}\)

Step by step solution

01

Identification of given data

Length of rope \(L = 6.50\,{\rm{m}}\)

Mass of the mountain climber \(m = 85\,{\rm{Kg}}\)

02

Significance of time period of simple pendulum

The letter "\(T\)" stands for the period of time needed for the pendulum to complete one complete oscillation.

\(T = 2\pi \sqrt {\frac{L}{g}} \) 鈥�(i)

Where, \(L\) is the length of pendulum and \(g\) is the acceleration due to gravity (\(9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}\))

03

(a) Determining time taken by mountain climber to swing from initial point to lowest point

Total time taken by climber to one complete one oscillation is \(T\)

So, the time taken by climber to move from initial position to lowest position will be \(\frac{T}{4}\)

By using equation (i)

\(T = 2\pi \sqrt {\frac{L}{g}} \)

Substitute all the values in above equation

\(\begin{array}{c}T = 2\pi \sqrt {\frac{{6.50\,{\rm{m}}}}{{9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \\ = 2 \times 3.14 \times 0.814\,{\rm{s}}\\ = 5.11\,{\rm{s}}\end{array}\)

Hence the time taken by climber to move from initial position to lowest position is

\(\begin{array}{l} = \frac{T}{4}\\ = \frac{{5.11\,{\rm{s}}}}{4}\\ = 1.27\,{\rm{s}}\end{array}\)

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