Part A
A '29er' mounbtain bike wheel has a diameter of 29.0 in . What is the moment of inertia of this wheel (expressed in standard units)? The rim and tire have a combined mass of 0.850 kg . Remember that 1in = 2.54cm.
Express your answer to two significant figures and include the appropriate units.


I = 
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Part B
The mass of the hub (at the center) can be ignored (why?).
The mass of the hub (at the center) can be ignored (why?).
The hub mass can be ignored because its mass is very small, and so it has a very small rotational inertia. 
The hub mass can be ignored because its distance from the axis of rotation is very small, and so it has a very small rotational inertia. 
Part A.
Moment of inertia of wheel is given by:
I = M*R^2
M = Mass of combined rim and tire = 0.850 kg
R = radius of wheel = 29.0 in/2 = 14.5 in = 14.5*2.54 = 36.83 cm = 0.3683 m
So,
I = 0.850*0.3683^2
I = 0.1153 kg.m^2
In two significant figures
I = 0.12 kg.m^2
Part B.
Since axis of rotation is very small for hub mass, So it's moment of inertia will be very very low compared to rim and tire.
Correct option is B.
The hub mass can be ignored because its distance from the axis of rotation is very small, and so it has a very small rotational inertia.
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