Question

A bicycle wheel has mass of the rim $1kg$ and $50$ spokes each of mass $5g$ if the radius of the wheel is $40cm$ then find the moment of inertia of the wheel.
a. $0.160kg{m^2}$
b. $0.174kg{m^2}$
c. $0.18kg{m^2}$
d. $0.196kg{m^2}$

Answer 1

Hint: We want to find the moment of inertia of the wheel along its centre. Here we clearly see that we have to find a moment of inertia of two things: one is the circumference of the wheel which is like a ring. And another thing is spokes which act like a rod and find its moment of inertia along one end of spokes.

Complete step by step answer:
We can solve this question in two parts first we calculate the moment of inertia of outer ring of wheel and second part is spokes of wheel which rotating along there one end as seen in figure

As we can clearly see in figure the outer part of wheel (rim) act as a ring so the moment of inertia of ring along its centre is given as
$ \Rightarrow {I_{ring}} = M{R^2}$
Where $M - $ mass of the ring
$R - $ Radius of ring
Mass of wheel rim is given $M = 1kg$
Radius of wheel is $R = 40cm = 40 \times {10^{ - 2}}m$
$ \Rightarrow {I_{rim}} = 1 \times {\left( {40 \times {{10}^{ - 2}}} \right)^2}$
Solving it
${I_{rim}} = 0.16kg{m^2}$ ........ (1)

Now we calculate the moment of inertia of spoke along one end of spoke we know the moment of inertia of rod along its end is ${I_{rod}} = \dfrac{1}{3}m{l^2}$
Mass of spoke is given $m = 5g = 5 \times {10^{ - 3}}kg$
Length of spoke is $l = 40cm = 40 \times {10^{ - 2}}m$
So the moment of inertia of spoke along its end is
$ \Rightarrow {I_{spoke}} = \dfrac{1}{3} \times 5 \times {10^{ - 3}} \times {(40 \times {10^{ - 2}})^2}$

Solving this,
$
   \Rightarrow {I_{spoke}} = \dfrac{{8000}}{3} \times {10^{ - 7}} \\
   \Rightarrow {I_{spoke}} = 0.00026kg{m^2} \\
 $
This is moment of inertia of one spoke but there are 50 spokes so moment of inertia of 50 spokes is
$ \Rightarrow {I_{spokes}} = 50 \times 0.00026$
$ \Rightarrow {I_{spokes}} = 0.013kg{m^2}$

So the total moment of inertia along centre of wheel is $I = {I_{rim}} + {I_{spoke}}$
$ \Rightarrow I = 0.16 + 0.013$
$ \Rightarrow I = 0.173Kg{m^2}$
$\therefore I = 0.174kg{m^2}$

Hence, the correct answer is option (B).

Note: Here we use the moment of inertia of spoke along its end is $I = \dfrac{1}{3}m{l^2}$ some time we learned formula of inertia of rod along its centre of mass is ${I_{com}} = \dfrac{{m{l^2}}}{{12}}$
To find its moment of inertia along one end we apply parallel axis theorem which is $I = {I_{com}} + m{a^2}$
Here $a = \dfrac{l}{2}$ distance between com of rod and one end of rod
$I = \dfrac{{m{l^2}}}{{12}} + \dfrac{{m{l^2}}}{4}$
Solve it
$I = \dfrac{{m{l^2}}}{3}$
So we use this formula in the above question.