Question

Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 8325 rev/min. The motorcycle rider forgets to throttle back, so the engine's angular speed increases to 12125 rev/min. As a result, the rest of the motorcycle (including the rider) begins to rotate clockwise about the engine at 4.2 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.

Answer 1

Answer:
`1.105* 10^(-3)`

Step-by-step explanation:

Given

Initial angular speed of engine(
`\omega _E`)=8325 rpm

Final angular speed of engine(
`\omega _E_f`)=12125 rpm

Initial angular speed of Motorcycle(
`\omega _M`)=0 rpm

Final angular speed of engine(
`\omega _M_f`)=4.2 rpm

as there is no external torque therefore angular momentum remains conserved

`I_E\omega _E+I_M\omega _M=I_E\omega _E_f+I_M\omega _M_f`

`I_E\omega _E+=I_E\omega _E_f+I_M\omega _M_f`

`I_E\left ( \omega _E-\omega _E_f\right )=I_M\omega _M_f`

`(I_E)/(I_M)=(\omega _M_f)/(\omega _E-\omega _E_f)`

`(I_E)/(I_M)=(-4.2)/(8325-12125)=0.0011052\approx 1.105* 10^(-3)`

Answer 2

`(Ie)/(lm) = 1.10*10^(-3)`

Step-by-step explanation:

GIVEN DATA:

Engine operating speed nf = 8325 rev/min

engine angular speed ni= 12125 rev/min

motorcycle angular speed N_m= - 4.2 rev/min

ratio of moment of inertia of engine to motorcycle is given as

`(Ie)/(lm) = (-N)/((nf-ni))`

`(Ie)/(lm) = (-(-4.2))/((12125 - (8325)))`

`(Ie)/(lm) = 1.10*10^(-3)`