Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 7560 rev/min. The motorcycle rider forgets to throttle back, so the engine's angular speed increases to 12300 rev/min. As a result, the rest of the motorcycle (including the rider) begins to rotate clockwise about the engine at 3.40 rev/min. Calculate the ratio IE/IM of the moment of inertia of the engine to the moment of inertia of the rest of the motorcycle (and the rider). Ignore torques due to gravity and air resistance.

Explanation:

IEWEf+ IMWmf = IEWEo + IMWmo

IEWEf + IMWmf = IEWEo + IM

IE ( Wef - WEo) = - IMWmf

IE/IM = - WMf/(WEf - WEo)

Clockwise rotation is negative

Counterclockwise is positive.

Given:

Wmf = 3.40

Wef = 12300rev/min

WEo =7560rev/min

The ratio of the moment of inertia will be:

Substituting into the equation gives:

IE/ IM

= -(-3.40)/((+12300) -( +7560))

IE/IM = 3.40/ 4740

IE/IM = 7.17×10^-4

Im/Ie = 1394.11, Ie/Im= 7.17×10-⁴.

Explanation:

This problem involves the principle of conservation of angular momentum. Initially when the rider gets of the total angular momentum as a result of the counterclockwise turning of the engine of the motorcycle. Shortly after the the angular speed of the engine increases and as result the rider and every other part of the motorcycle gain angular momentum.

Let ωe1 = initial angular speed of the engine,

ωe2 = final angular speed of the engine, ωm1 = = initial angular speed of the motorcycle and rider

ωm2 = final angular speed of the motorcycle and rider.

Ie = moment of inertia of the engine.

Im = moment of inertia of the motorcycle

Taking counterclockwise rotation as positive and clockwise as negative,

ωe1 = 7560rev/min

ωe2 = 12300rev/min

ωm1 = 0 motorcycle and rider were not initially rotating.

ωm2 = -3.40rev/min.

Sum of angular momentum is constant but before and after the action of the throttle.

Ie×ωe1 + Im×ωm1 = Ie×ωe2 lm×ωm2

Substituting the given values into the equation above, we have that

Ie×7560+ Im×0 = Ie×12300 + lm×(-3.40)

7560Ie = 12300Ie -3.4Im

3.4Im = 12300Ie – 7560Ie

3.4lm = 4740Ie

Im/Ie = 4740/3.40

Im/Ie = 1394.11

The ratio of the moment of inertia of the engine is large. No wonder speed of rotation of the motorcycle and rider did not change by much. A similar example of a scenario like this is when a small car collides with a huge truck. The car would experience a rapid change in velocity but the truck won't change its velocity by much. This is because of the mass (inertia) of the huge truck vs the small car.

a straight line sloped