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:

$\frac{Ie}{lm} = 1.10*10^{-3}$

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

$\frac{Ie}{lm} = \frac{-N}{(nf-ni)}$

$\frac{Ie}{lm} = \frac{-(-4.2)}{(12125 - (8325))}$

$\frac{Ie}{lm} = 1.10*10^{-3}$

Answer 2

Answer:$1.105\times 10^{-3}$

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

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

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

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

$\frac{I_E}{I_M}=\frac{\omega _M_f}{\omega _E-\omega _E_f}$

$\frac{I_E}{I_M}=\frac{-4.2}{8325-12125}=0.0011052\approx 1.105\times 10^{-3}$