Question

A sphere of radius 10cm and mass ${\mathbf{25}}{\text{ }}{\mathbf{kg}}$is attached to the lower end of steel wire of length ${\mathbf{5}}{\text{ }}{\mathbf{m}}$and diameter of ${\mathbf{4}}{\text{ }}{\mathbf{mm}}$which is suspended from the ceiling of a room. The point of support is ${\mathbf{521}}{\text{ }}{\mathbf{cm}}$above the floor. When the sphere is set swinging as a simple pendulum, its lowest point just grazes the floor. Calculate the velocity of the ball at its lowest position.

Answer 1

As we know that when we apply a force on anybody there is some elongation and compression depending upon the situation. In this condition due to gravitational force there will be an elongation in the system due to which it will just graze the floor and continue its oscillatory motion. Formula for young’s modulus of elasticity $(Y) = \dfrac{{FL}}{{A\Delta L}}$ where $F$ is the applied force , $L$ is the length of wire, $A$ is the area of wire and $\Delta L$ is the elongation in the wire.

Complete step by step solution:
We can get the following data clearly from the above question. Therefore, we can write:
Unstretched length of Steel wire $\left( L \right){\text{ }} = 500cm$
Steel wire diameter $\left( d \right){\text{ }} = {\text{ }}4mm{\text{ }} = {\text{ }}0.4cm$
Radius of the steel wire$\left( R \right){\text{ }} = 0.02cm$
Mass of sphere $= {\text{ }}25{\text{ }}kg$
Radius of sphere $= 10{\text{ }}cm$
Distance of floor from ceiling $= {\text{ }}521{\text{ }}cm$
Young’s modulus of elasticity for steel $= 2 \times {10^{11}}$ $N/{{\text{m}}^2}$
Now, clearly from above we can write:
Total length of system before release $= {\text{ }}500 + {\text{ }}\left( {10 + 10} \right){\text{ }} = 520cm$
Due to gravitational force the elongation in wire $(){\text{ }} = {\text{ }}521cm{\text{ }}-{\text{ }}520{\text{ }}cm{\text{ }} = {\text{ }}1{\text{ }}cm{\text{ }} = {\text{ }}0.01m$
On balancing the forces, we can write:
$T = (m{v^2}/R) + mg$ (here R refers to the radius of arc that the system will follow i.e. $5m$of string
+ $10{\text{ }}cm$radius of sphere)
$T = \dfrac{{25{v^2}}}{{5.1}} + 25 \times 9.8 \\\ T = 4.9{v^2} + 245 \\\$
Now,

$Y = \dfrac{{FL}}{{A\Delta L}} \\\ \Rightarrow \Delta L = \dfrac{{FL}}{{AY}} \\\ \Rightarrow \Delta L = \dfrac{{FT}}{{AY}} \\\$
On putting the values the equation reduces as:
${10^{ - 2}} = \dfrac{{4.9{v^2} + 245) \times 5}}{{\dfrac{{\pi {d^2}}}{4} \times 2 \times {{10}^{11}}}} \\\ \Rightarrow {10^{ - 2}} = \dfrac{{(4.9{v^2} + 245) \times 5}}{{\dfrac{{3.14 \times {{(0.02)}^2}}}{4} \times 2 \times {{10}^{11}}}} \\\$

$v = 31m/s$

Note: At first visualize the problem and then proceed because many times the crux of the problem lies in its visualization.
Calculate tension and hence equate it to $m{v^2} + {\text{ }}mg$.