Question

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

• A. $2.2\text{m/s}$
• B. $4.4\text{m/s}$
• C. $8.8\text{m/s}$
• D. $12.2\text{m/s}$

Answer 1

The correct option is C $8.8\text{m/s}$

Image drwan for the given problem:


Given that,
Radius of sphere, $r=0.1\text{m}$
Mass of the sphere, $m=8\pi \text{kg}$
Length of steel wire, $l=5\text{m}$
Diameter of steel wire, $d={10}^{-3}\text{m}$
Young's modulus of steel. $Y=1.994\times {10}^{11}{\text{N/m}}^2$

Let $\Delta l$ be the extension of wire when the sphere is at mean position. Then, we have
$l+\Delta l+2r=5.22$
$\Rightarrow \Delta l=5.22-l-2r$
Substuting the values,
$\Rightarrow \Delta l=5.22-5-2\times 0.1$
$\Rightarrow \Delta l=0.02\text{m}$

Let, $T$ be the tension in the wire at mean position during oscillations, then
$Y=\frac{\frac{T}{A}}{\frac{\Delta l}{l}}$

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

Substituting the values, we have
$T=\frac{(1.994\times {10}^{11})\times \pi \times (0.5\times {10}^{-3}{)}^2\times 0.02}{5}$

$\Rightarrow T=626.43\text{N}$

Let the velocity of the ball at the lowest position is $v$.
The equation of motion at mean position is,
$T-mg=\frac{m{v}^2}{R}\dots \dots \left(1\right)$

Here, $R=5.22-r=5.22-0.1=5.12\text{m}$

Substituting the values in Equation ($1$), we have
$\left(626.43\right)-(8\pi \times 9.8)=\frac{\left(8\pi \right)v^2}{5.12}$

Solving this equation, we get
$v=8.8\text{m/s}$

Hence, option (d) is correct answer.