Question

A sphere of radius $0.1$ m and mass $8\pi$ kg is attached to the lower end of a steel wire of length $5.0$m and diameter ${10}^{-3}$m. The wire is suspended from $5.22m$ 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 point.
(Y for steel = $1.994\times {10}^{11}N/m^2)$

• A. $7.5m{s}^{-1}$
• B. $8.2m{s}^{-1}$
• C. $8.8m{s}^{-1}$
• D. $6.5m{s}^{-1}$

Answer 1

The correct option is C $8.8m{s}^{-1}$

$\text{Given:-}$ Radius of sphere = 0.1 m

Mass of sphere = $8\pi$ Kg

Length of the steel wire = 0.5 m

Diameter of the steel wire = $10m^{-3}$

Young's modulus of steel = $1.994\times {10}^{11}N/m^2$

$\text{Solution:-}$

Let \Delta l be the extension of wire when the sphere is at mean position . Then, we have

$l+\Delta l+2r=5.22$ (or)

$\Delta l=5.22-l-2r$

$\Delta l=5.22-5-(2\times 0.1)$

$\Delta l=0.02m$

Let T be the tension in the wire at mean position during oscillations, then

$Y=\frac{T/A}{\left(\Delta l\right)/\left(l\right)}$

$\therefore T=\frac{YA\Delta l}{l}=\frac{Y\pi r^2\Delta l}{l}$

Substituting the values, we have

$T=\frac{(1.994\times {10}^{11})\times \pi \times {(0.5\times {10}^{-3})}^2\times 0.02}{5}$

$T=626.43N$

The equation of motion at mean position is,

$T-mg=\frac{m{V}^2}{R}..........\left(1\right)$

Here, $R=5.22-r=5.22-0.1=5.12m$

And $m=8\pi Kg=25.13kg$

Substituting the proper values in $Eq.\left(i\right)$ , we have

$\left(626.43\right)-(25.13\times 9.8)=\frac{\left(25.13\right)v^2}{5.12}$

Solving this equation, we get

$v=8.8m/sec.$

$\text{Hence the correct option is C}$