Finding Angle of Twist at Gear A Relative to Motor

[Jonski]

Homework Statement

What is the angle of twist at gear A relative to the motor? (Round to the closest 6th decimal)

Homework Equations

I know to find the angle its θ = ∑ (TL/JG)
but it can also be θ = Lτ/Gr

The Attempt at a Solution

So i used both these equations, and I got (-29.18*0.052)/(1.57x10^-8 * 83x10^9) = -1.163833x10^-3 rad, which was the same result for the other one too. However, this was wrong and I'm not sure what I am doing wrong?

 

[SteamKing]

Homework Statement

View attachment 98919
What is the angle of twist at gear A relative to the motor? (Round to the closest 6th decimal)

Homework Equations

I know to find the angle its θ = ∑ (TL/JG)
but it can also be θ = Lτ/Gr

The Attempt at a Solution

So i used both these equations, and I got (-29.18*0.052)/(1.57x10^-8 * 83x10^-9) = -1.163833 rad, which was the same result for the other one too. However, this was wrong and I'm not sure what I am doing wrong?

G = 83×10⁹ Pa

You should check your arithmetic. θ seems to be missing a couple of factors of 10.

When you isolate gear A, you should draw a free body diagram to make sure you have the correct net torque acting at that location. I don't think you can say that the angle of twist is caused only by the torque acting from the left of the gear, and that the torque acting to the right has no effect.

 

[Jonski]

G = 83×10⁹ Pa

You should check your arithmetic. θ seems to be missing a couple of factors of 10.

When you isolate gear A, you should draw a free body diagram to make sure you have the correct net torque acting at that location. I don't think you can say that the angle of twist is caused only by the torque acting from the left of the gear, and that the torque acting to the right has no effect.

But the -29.18 is the internal torque between them, so doesn't that account for torque both sides of the gear

 

[SteamKing]

But the -29.18 is the internal torque between them, so doesn't that account for torque both sides of the gear

If you do a check, that 29 N-m is the torque produced by the 11 kW motor operating at 60 Hz.

Looking at the rest if the gears on this shaft, there are much greater torque loads being applied than this piddly 29 N-m, and these torques are turning in different directions.

 

[Jonski]

If you do a check, that 29 N-m is the torque produced by the 11 kW motor operating at 60 Hz.

Looking at the rest if the gears on this shaft, there are much greater torque loads being applied than this piddly 29 N-m, and these torques are turning in different directions.

So would it be more like:
(-29.18*0.052)/(1.57x10^-8 * 83x10^9) + (1100.82*0.092)/(J * 83x10^9) + (-249.18*0.111)/(J * 83x10^9) + (-779.18*0.138)/(J* 83x10^9)

 

[SteamKing]

So would it be more like:
(-29.18*0.052)/(1.57x10^-8 * 83x10^9) + (1100.82*0.092)/(J * 83x10^9) + (-249.18*0.111)/(J * 83x10^9) + (-779.18*0.138)/(J* 83x10^9)

Something like that.

You've already calculated J for the 20 mm dia. shaft. You need to calculate J for the 38 mm shaft.