Question

The rear wheel of a bicycle at rest is rotating about its axis with $n$ revolutions per second. The radius of the wheel is $R$ and mass $M$. The number of spokes is $N$ and mass of each spoke is $m$. Then the kinetic energy of the wheel is:

• A. $[{MR}^2+\frac{Nm{R}^2}{3}].2{\pi }^2{n}^2$
• B. $2{\pi }^2{n}^2[\frac{{MR}^2}{2}+\frac{Nm{R}^2}{12}]$
• C. ${\pi }^2{n}^2[\frac{{MR}^2}{2}+\frac{Nm{R}^2}{3}]$
• D. ${\pi }^2{n}^2[{MR}^2+\frac{Nm{R}^2}{12}]$

Answer 1

The correct option is A $[{MR}^2+\frac{Nm{R}^2}{3}].2{\pi }^2{n}^2$

$\begin{array}{c}\text{Given -}\ast \text{Mass of the wheel}=M\\ \ast \text{Mass of spoke}=m\\ \text{* Total num of spokes}=N\\ \text{* Angular velocity}=\omega =n\text{Rev per second}\\ \text{OR}w=2\pi n\text{Rad}18\text{.}\end{array}$

$\begin{array}{c}\text{We know that Rotational Kinetic energy is}\\ KE=\frac{1}{2}I{\omega }^2=2{\pi }^2{n}^2.I\\ I\to \text{Moment of Inertia}\\ \text{calculating}I\to \end{array}$

$\begin{array}{cc}I=& M{R}^2+N\times \left(\frac{m{R}^2}{3}\right)\text{(Spokes)}\\ & \uparrow \text{heel}\\ \Rightarrow I& =M{R}^2+\frac{Nm{R}^2}{3}\\ \Rightarrow KE=& 2{\pi }^2{n}^2[M{R}^2+\frac{Nm{R}^2}{3}]\\ \text{Hence, option (A) is correct.}& \end{array}$