A bicycle of radius $0.3 m$ has a rim of mass $1.0 k g$ and $50$ spokes, each of mass $0.01 k g$ . What is its moment of inertia about its axis of rotation?

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Solution

Given that,
Radius of wheel $(R) = 0.3 m$
Mass of rim $(M) = 0.1 k g$
No. of spokes $(N) = 50$
Mass of spokes $(m) = 0.01 k g$
Lenght of spokes $(L = R) = 0.3 m$
Moment of inertia about its axis of rotation $(I) = ?$

we have,
$M o m e n t o f i n e r t i a o f r i m (I_{r i m});$

$I_{r i m} = M R^{2}$

$\Rightarrow I_{r i m} = 0.1^A (0.3)^{2}$

$\Rightarrow I_{r i m} = 0.09 K g - m^{2}$

$M o m e n t o f i n e r t i a o f s p o k e s (I_{s p o k e s}) :$

$I_{s p o k e s} = N \frac{m L^{2}}{3}$

$\Rightarrow I_{s p o k e s} = 50^A \frac{0.01^A (0.3)^{2}}{3}$

$\Rightarrow I_{s p o k e s} = 0.015 K g - m^{2}$

$T o t a l m o m e n t o f i n e r t i a (I) :$

$I = I_{r i m} + I_{s p o k e s}$

$\Rightarrow I = 0.09 + 0.015$

$\Rightarrow I = 0.105 K g - m^{2}$

$I = 105 K g - M^{2}$

Hence,
Total moment of inertia $(I)$ is $0.105 K g - M^{2}$

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