What is the Maximum Angular Velocity of a Reverse Three-Speed Transmission?

[orangeincup]

I calculated a, b, and c, I just need help with d.

1. Homework Statement [/b]
The operation of a "reverse" three-speed automotive transmission is illustrated the in
figure. Each of the gears rotate about a fixed axis. Note that gears A and B, C and D, E
and F are in mesh. The radii of each of these gears are shown in the figure. The shaft G
starts from rest and reaches its maximum angular velocity of ωG = 60 rad/s. The mass
moment of inertia of the load shaft is given as 5.0 kgm^2
. The shaft G is driven by a 2 hp motor. Assume that the efficiency of the gear box is 80 %.

a) If gear A has 45 teeth and if B, C and D, E have the same diametral pitch
calculate the number of teeth in gears B, C, D, E, F.
b) Determine the angular velocity of the drive shaft H.
c) Calculate the torque on the output shaft.
d) How many turns does the load shaft make to reach its maximum angular velocity.

Homework Equations

P=ω*τ
τ=IgAc

The Attempt at a Solution

a) Used the ratios to solve, all end up having to be .5, so 45, 15, 15, 25, 35, 30

b) Multiplied ω*45/15, and repeated for every gear until I reduced it to 127ω
c)(127/.8) for efficiency, converted 2hp into 1491.5 W. 1491.5/(127/.8)=9.4
d) Not sure where to start for this... I assume I need something that has rpm in it, and I have to use the moment of inertia somehow.

 

[Simon Bridge]

(d) that would be correct - though it will help to think through the physics before looking for an equation.
What determines the mag angular velocity?
What is the initial angular velocity?
Does it have a uniform acceleration?

 

[orangeincup]

The rpm from the motor?
127ω? Or is that the max? I don't know how the acceleration of this system works
I think it is constant, but I don't know what it is

 

[orangeincup]

T=IgAc

9.4 N-m=5kg*m^2 * Ac
Ac=1.88

Solving for time: ω=ω0+Act
(126rad/s)/(1.88)=t
t=67.02s

126 rad/s * 60 / 2*∏ = 1203 rpm
1203 rpm *(67.02s/60)=1344 revolutions

Does this look correct?

 

[Simon Bridge]

T=IgAc

9.4 N-m=5kg*m^2 * Ac
Ac=1.88

Solving for time: ω=ω0+Act
(126rad/s)/(1.88)=t
t=67.02s

126 rad/s * 60 / 2*∏ = 1203 rpm
1203 rpm *(67.02s/60)=1344 revolutions

Does this look correct?

That looks likt the sort of reasoning I'd use yeah.
I knew you'd get there :)

 

[orangeincup]

According to my solution the answer is 667 rev, should I be using the 60 rad/s anywhere? Or is there any mistake you see?

 

[Simon Bridge]

According to my solution the answer is 667 rev, should I be using the 60 rad/s anywhere? Or is there any mistake you see?

Of course... That's the max speed.
I didn't go through your work in detail. Where did you get the figure you used?

 

[orangeincup]

Never mind

 

[orangeincup]

Never mind I figured it out.