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 celling 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.

$(Y_{\text{steel}}=1.994\times {10}^{11}{\text{N/m}}^2)$

• A. $4.4{\text{ms}}^{-1}$
• B. $2.2{\text{ms}}^{-1}$
• C. $6.6{\text{ms}}^{-1}$
• D. $8.8{\text{ms}}^{-1}$

Answer 1

The correct option is D $8.8{\text{ms}}^{-1}$

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

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

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

$\Rightarrow \Delta l=5.22-5-2\times 0.1=0.02\text{m}$


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

$Y=\frac{T/A}{\Delta l/l}$

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

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

The equation of motion at mean position is,
$T-mg=\frac{m{v}^2}{R}...\left(i\right)$

Here,
$R=5.22-r=5.22-0.1=5.12\text{m}$
$m=8\pi \text{kg}=25.13\text{kg}$

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, $v=8.8\text{m/s}$.