Question

A sphere of radius 0.1m and mass 8 pi kg is attached to the lower end of a steel wire of length 5.0 m and diameter 10^(-3). The wire is suspended from 5.22 m high ceiling of a room . When the sphere is made to swing as a simple pendulum, it just grazes the floor at its lowest point. Calculate the velocity of the sphere at the lowest position . Young's modulus of steel is (1.994xx10^(11) N//m^(2)).

Answer 1

Given:

A sphere of radius R = 0.1m and mass 8 pi kg .

It is attached to the lower end of a steel wire of length, l = 5.0 m and diameter d = 10^(-3).

The wire is suspended from a height h = 5.22 m .

Young's modulus of steel is Y = 1.994 x 10^(11) N//m^(2))

To Find:

The velocity of the sphere at the lowest position.

Solution:

Let Δl be the extension in the string.

Then h = l + Δl + 2r

• 5 + Δl + 2 x 0.1 = 5.22
• Δl = 0.02m

Let T be the tension on the string,

Then Youngs modulus,

• Y = Tl/AΔl
• T = Yπr² Δl / l = 1.994 x 10^(11) x π x  (0.5 x$10^{-3}$)² x 2 x $10^{-2}$ / 5
• T =  626.43 N

Applying force equation at mean postion,

• T - mg  = mV²/R
• R = 5.22  - 0.1 = 5.12m
• m = 8π = 25.13Kg
• 626.43 - 25.13x9.8 = 25.13xV²/5.12
• 380.24 x  5.12 / 25.13 = V²
• V =  8.8m/s

The velocity of the sphere at the lowest position is 8.8m/s.