Velocity of a piston in a piston-shaft mechanism

[afaiyaz]

Homework Statement

In the figure, a piston P is connected to a cylinder. The piston is connected to a rotating wheel with two shafts AB and BC. The shaft AB is connected on the periphery of the wheel. The wheel is rotating with angular speed ω= 100 rad s^-1. At the moment A,C and the center of the wheel is collinear and θ=30° and dθ/dt = 500 rad s^-1 , what is the velocity of the piston? (AB = 1.5m, BC = 1m)

Homework Equations

a² = b²+c²-2abcosθ
Basic Trigonometry
Implicit DIfferentiation
dx/dt = dx/dθ × dθ/dt

The Attempt at a Solution

Since the linking rods are rigid, they must maintain their length throughout.

From the diagram, I assumed a pathway of shaft AB (denoted by y) and used it to find x, which I presumed is the horizontal motion of shaft BC. The rate of change of x with respect to time is, therefore, the velocity of the piston.

My working is as follows:
tan 60 = 1.5/y
∴y=√3/2 m.
Using sine rule,
sin C = y × (sin 120/1)
∴ C = 48.6°
sin (11.4) / x = sin (120) / 1
∴ x=0.228

Using cosine rule,
y²+x²+xy = 1
(2y × dy/dt) + (2x × dx/dt) + (x × dy/dt) + (y × dx/dt) = 0
∴ dx/dt = (- (2y+x) × dy/dt) / (2y+x)

dy/dt = 500 × 1.5 = 750 m/s
Plugging in values of y,x and dy/dt, I get -1111.9≈-1112 m/s.

I feel like this answers very unrealistic as it is more than 3.5 times the speed of sound. Could anyone point out where the mistake is? Could there be a completely different approach to the problem?

 

[arpon]

You forgot to upload the figure.

 

[afaiyaz]

 

[afaiyaz]

the original figure

 

[arpon]

Why didn't you consider $\omega$ in your solution?

 

[Nidum]

This is either a trick question or it was set by an academic with no knowledge of mechanisms .

There is no sensible answer .

 

[afaiyaz]

Do I have to? @arpon bro
If so, how do I use it in the solution?

 

[afaiyaz]

What do you mean by 'no sensible answer'? @Nidum
Could you please clarify?

 

[arpon]

Do I have to? @arpon bro
If so, how do I use it in the solution?

If $\frac{d\theta}{dt} = 0$ and $\omega$ is nonzero, will the piston move? What do you think?
Drawing diagrams may help.

 

[afaiyaz]

I think it should because, even if \frac{dθ}{dt}=0, the wheel is still rotating so there's ω.There may be no angualr velocity of the shaft AB along centre A, but there is the ω of the wheel. And if that's the case, I think ω has to be considered for calculating piston's velocity, although I'm not sure how to use it.

 

[haruspex]

What do you mean by 'no sensible answer'? @Nidum
Could you please clarify?

Because you could keep the position of C fixed and the wheel could still rotate. B will just move back and forth in an arc centred on C.
To make sense of the question there needs to beanother mechanical constraint.