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--- 183_notes:examples:the_moment_of_inertia_of_a_bicycle_wheel [2014/10/31 14:55] – created pwirving
+++ 183_notes:examples:the_moment_of_inertia_of_a_bicycle_wheel [2014/11/05 20:49] (current) – pwirving
@@ Line 6: / Line 6: @@
 === Facts ===
+Bicycle wheel has a mass of M
+Outer rim of bicycle wheel is at a radius R
 === Assumptions and Approximations ===
+Assume the mass of the spokes is negligible
+Thickness of the rim sufficiently thin so all of the mass is located at radius R
 === Lacking ===
+A representation for the moment of inertia of the bicycle wheel about its center of mass
 === Representations ===
+{{course_planning:course_notes:mi3e_09-019.jpg|}}
+$I = m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2}$
 === Solution ===
-Let m represent the mass of one atom in the rim. The moment of inertia is
+Let m represent the mass of one atom in the rim. The moment of inertia is therefore
-$I = m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2}$ + $m_{3}r^{3}_{\perp3}$ + $m_{4}r^{4}_{\perp4}$ + \cdot \cdot \cdot$
+$I = m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2}$ + $m_{3}r^{3}_{\perp3}$ + $m_{4}r^{4}_{\perp4} + \cdot \cdot \cdot$
+Substitute R for $r_{\perp1}$ and so forth as this the outer rim radius.
 $I = m_{1}R^{2} + m_{2}R^{2} + m_{3}R^{2} + m_{4}R^{2} + \cdot \cdot \cdot$
-$I = [m_{1} + m_{1} + m_{1} + m_{1} + \cdot \cdot \cdot]R^2$
+Gather the R's
+$I = [m_{1} + m_{1} + m_{1} + m_{1} + \cdot \cdot \cdot]R^2$
+We've assumed that the mass of the spokes is negligible compared to the mass of the rim, so that the total mass os just the mass of the atoms in the rim. Therefore the moment of inertia of a Bicycle Wheel is:
-For two masses, $I = m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2}$. The distance between masses is d, so the distance of each object from the center of mass is $r_{\perp1} = r_{\perp2} = d/2$.
+$I = MR^2$
-Therefore:
-$I = M(d/2)^{2} + M(d/2)^{2} = 2M(d/2)^{2}$
-$I = 2 \cdot (2.3$ x $10^{-26}kg)(0.75$ x $10^{-10}m)^2$
-$I = 2.6$ x $10^{-46} kg \cdot m^2$