Question

A sphere of radius $0.1m$ and mass $8\pi \text{ kg}$ is attached to the lower end of a steel wire of length $5.0m$ & diameter $10^{-3}m$ . The wire is suspended from $5.22m$ high ceiling of a room. When the sphere is made to swing like a simple pendulum, it just grazes the floor at its lowest point. Calculate the velocity of the sphere at the lowest position. $Y$ for steel = $1.994\times 10^{11}N{m}^{-2}$

• A. $7.7m{s}^{-1}$
• B. $4.4m{s}^{-1}$
• C. $2.2m{s}^{-1}$
• D. $8.8m{s}^{-1}$

Answer 1

Answer: (D)

As the length of the wire is $5m$ and diameter $2\times 0.1=0.2m$ and at the lowest point, it grazes the floor which is at a distance $5.22m$ from the roof, the increase in the length of the wire at lowest point

$\Delta \text{L}=5\text{.}22-\left(5+0\text{.}2\right)$
$=0\text{.}02\text{ m}$
So the tension in the wire (due to elasticity)
$T=\frac{YA}{L}\Delta L=\frac{1\text{.}994\times 10^{11}\times \pi {\left(5\times 10^{-4}\right)}^2\times 0\text{.}02}{5}=199\text{.}4\pi \text{N}$
and as the equation of circular - motion of a mass $m$ tied to a string in a vertical plane is
$\frac{m{v}^2}{r}=T-m\text{g cos}\theta$
So at the lowest point
$\frac{m{v}^2}{r}=T-m\text{g}$ $\left[\text{as }\theta =0\right]$
But here $r=5+0.02+0.1=5.12m$
So $\left(\frac{8\pi v^2}{5.12}\right)=199\text{.}4\pi -8\pi \times 9\text{.}8$
i.e., ${\left(\text{v}\right)}^2=\left(121\times 5\text{.}12/8\right)=77\text{.}44$
So $v=8.8m{s}^{-1}$