Question

A uniform spherical shell of mass $\mathbf{\text{M}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{kg}}\mathbf{}$and radius $\mathbf{\text{R}}\mathbf{=}\mathbf{}\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{}\mathbf{\text{cm}}$can r otate about a vertical axis on frictionless bearings (Fig.$\mathbf{10}\mathbf{}\mathbf{-}\mathbf{}\mathbf{47}$). A massless cord passes around the equator of the shell, over a pulley of rotational inertia $\mathit{l}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{\text{kgm}}}^{\mathbf{\text{2}}}$and radius,$\mathbf{\text{r}}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathbf{\text{cm}}$and is attached to a small object of mass .$\mathbf{\text{m}}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{60}\mathbf{}\mathbf{}\mathbf{\text{kg}}$There is no friction on the pulley’s axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen $\mathbf{82}\mathbf{}\mathbf{}\mathbf{\text{cm}}\mathbf{}$after being released from rest? Use energy considerations.

Short answer

The speed of the object after being released from rest is$1.4\text{m}/\text{s}.$

Step-by-step solution

Given

1. Mass of the spherical shell is,$\text{M}=4.5\text{kg}$
2. Radius of the spherical shell is,$R=8.5cm=0.085\text{m}$
3. Rotational inertia of the pulley is,$\text{I}=3.0\times {10}^{-3}{\text{kg.m}}^2.$
4. Radius of the pulley is,$r=5.0\text{cm}=0.05\text{m}.$
5. Mass of the small object is,$\text{m}=0.60\text{kg}$
6. $\text{h}=42\text{cm}=0.82\text{m}.$
   

Understanding the concept

Applying the law of conservation of energy to the given system,we can write an equation. Inserting angular speed in terms of speed of the system and given values in it, we can find the speed of the object after being released from rest.

Formulae:

The M.I of the spherical shell is

$\text{I}=\frac{2}{3}{\text{MR}}^2$

${\text{E}}_i={\text{E}}_f$

$\omega =\frac{\text{v}}{\text{r}}{\omega }^2$

Calculate the speed of the object when it has fallen  after being released from rest

M.I of the spherical shell is

${\text{I}}_s=\frac{2}{3}{\text{MR}}^2=\frac{2}{3}\left(4.5\right){\left(0.085\right)}^2=0.02167{\text{kg.m}}^2$

An object is released from rest. So its initial K.E is zero.

According to conservation of energy,

$$\begin{gathered} {\text{E}}_i={\text{E}}_f \\ {\text{K.E}}_i+{\text{P.E}}_i={\text{K.E}}_f+{\text{P.E}}_f \end{gathered}$$

In this case,

$$\begin{gathered} 0+{\text{P.E}}_{\text{i}}={\text{K.E}}_{sphere}+{\text{K.E}}_{pulley}+{\text{K.E}}_{object} \\ \Rightarrow \text{mgh}=\frac{1}{2}{\omega }_s^2+\frac{1}{2}\text{I}{\omega }_p^2+\frac{1}{2}{\text{mv}}^2 \end{gathered}$$

But,$\omega =\frac{\text{v}}{\text{r}}$

Hence,

$\left(0.60\right)\left(9.8\right)\left(0.82\right)=\frac{1}{2}\left(0.02167\right)\left(\frac{v}{0.085}\right){)}^2+\frac{1}{2}\left(3.0\times {10}^{-3}\right){\left(\frac{v}{0.05}\right)}^2+\frac{1}{2}\left(0.60\right){\text{v}}^2$

$\Rightarrow 4.8216=1.499{\text{v}}^2\text{+}0.6{\text{v}}^2{\text{+0.3vv}}^2$

$\Rightarrow {\text{v}}^2=2$

$\therefore \text{v}=1.4\text{m}/\text{s}$

Therefore, the speed of the object after being released from rest is$1.4\text{m}/\text{s}$