Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 7530 rev/min. The motorcycle rider forgets to throttle back, so the engine's angular speed increases to 12600 rev/min. As a result, the rest of the motorcycle (including the rider) begins to rotate clockwise about the engine at 3.50 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.
To solve this problem, we need to use the conservation of angular momentum. Angular momentum is given by the product of the moment of inertia (I) and the angular velocity (ω).
Angular momentum (L) = I * ω
According to the problem, just after the motorcycle launches into the air, its engine is turning counterclockwise at 7530 rev/min, which can be converted to ω1 = (7530 rev/min) * (2π rad/rev) * (1 min/60 sec) = 788.58 rad/s.
Then, the angular velocity of the engine increases to 12600 rev/min, which can be converted to ω2 = (12600 rev/min) * (2π rad/rev) * (1 min/60 sec) = 1319.91 rad/s.
The rest of the motorcycle (including the rider) rotates clockwise about the engine at 3.50 rev/min, which can be converted to ω3 = (3.50 rev/min) * (2π rad/rev) * (1 min/60 sec) = 0.3665 rad/s.
Using the conservation of angular momentum, we can equate the initial angular momentum (L1) with the final angular momentum (L2).
L1 = L2
(Ie * ω1) = (Ie * ω2) + (Im * ω3)
Where:
Ie is the moment of inertia of the engine
Im is the moment of inertia of the rest of the motorcycle (and the rider)
ω1 is the initial angular velocity of the engine
ω2 is the final angular velocity of the engine
ω3 is the angular velocity of the rest of the motorcycle (and the rider)
We can rearrange the equation to solve for the ratio IE/IM:
(Ie * ω1) - (Ie * ω2) = Im * ω3
(Ie * (ω1 - ω2)) = Im * ω3
IE/IM = ω3 / (ω1 - ω2)
Now we can substitute the values:
IE/IM = 0.3665 rad/s / (788.58 rad/s - 1319.91 rad/s)
Calculating this ratio will give us the answer.
To solve for the ratio of moment of inertia, we need to use the principles of conservation of angular momentum.
The initial angular momentum (L1) of the system (motorcycle + rider) is given by the sum of the angular momentum of the engine (LE) and the angular momentum of the rest of the motorcycle and rider (LM):
L1 = LE + LM
The final angular momentum (L2) of the system is also given by the sum of the angular momentum of the engine (LE) and the angular momentum of the rest of the motorcycle and rider (LM):
L2 = LE + LM
Since there is no net external torque acting on the system, the initial and final angular momenta are equal:
L1 = L2
Now let's calculate each term:
The initial angular momentum of the engine (LE) is given by:
LE = Ie * ωe
where Ie is the moment of inertia of the engine and ωe is the initial angular speed of the engine.
The final angular momentum of the engine (LE) is given by:
LE' = Ie * ωe'
where ωe' is the final angular speed of the engine.
The initial angular momentum of the rest of the motorcycle and rider (LM) is given by:
LM = IM * ωm
where IM is the moment of inertia of the rest of the motorcycle and rider and ωm is the initial angular speed of the rest of the motorcycle and rider.
The final angular momentum of the rest of the motorcycle and rider (LM) is given by:
LM' = IM * ωm'
where ωm' is the final angular speed of the rest of the motorcycle and rider.
Now, substituting these values into the conservation of angular momentum equation:
LE + LM = LE' + LM'
Ie * ωe + IM * ωm = Ie * ωe' + IM * ωm'
Given:
ωe = 7530 rev/min
ωe' = 12600 rev/min
ωm' = -3.50 rev/min (negative sign indicates clockwise rotation)
Substituting the values:
Ie * 7530 + IM * ωm = Ie * 12600 + IM * (-3.50)
Simplifying further:
7530Ie + IM * ωm = 12600Ie - 3.50IM
Rearranging the terms:
7395Ie + 3.50IM = IM * ωm
Let's call this equation (1).
To calculate the ratio of moment of inertia (IE/IM), we need another equation relating the moments of inertia. We know that the rest of the motorcycle and rider rotate about the engine, so:
IM * ωm = Ie * ωe'
Substituting this equation into equation (1):
7395Ie + 3.50 * (Ie * ωe') = Ie * ωe'
7395Ie + 3.50Ie * ωe' = Ie * ωe'
Simplifying further:
7395Ie = Ie * ωe' - 3.50Ie * ωe'
7395 = ωe' - 3.50 * ωe'
Now, dividing both sides of the equation by ωe:
7395/ωe = ωe' - 3.50
Finally, solving for the desired ratio:
IE/IM = (7395/ωe) / (ωe' - 3.50)