What Is the Velocity of the Piston in a Slider-Crank Mechanism?

[bobmarly12345]

Homework Statement

The instantaneous configuration of a slider crank mechanism has a crank GH 10cm long, the connecting rod HP is 50cm. The crank makes an angle of 60 degree with the inner dead centre position and is rotating at 110 rev/min. Determine the velocity of the piston P and the angular velocity of the link HP.

what i need is the steps how to do it up until the answer(i don't need the answer calculated just need to see how you would do it), I've changed all the values so it won't give me the answer to my own question. really been struggling with this because I've read 101 ways to do it and they all are over complicated and produc different results. need urgent help

Homework Equations

The Attempt at a Solution

right what I've done so far is found the length of the base of the diagram "PG" using x=0.1cos(60)+sqrt(0.5²-0.1²sin(60)² which comes out to around 0.542m
and found the velocity at H by using v=w X gh which comes out to 1.047Rad/s
Also would wHP the angular velocity part of the question would the formula for that be ((HG.w)/PH) X ((cos(60)/(sqrt(1-(HG²/PH²)sin²(60)))

 

[tiny-tim]

hi bobmarly12345!

if r_P and r_H are two points in a rigid body with angular velocity ω, then …

v_P - v_H = ω × (r_P - r_H)

so (in two dimensions) ω = |v_P - v_H| / PH ​

 

[SammyS]

Homework Statement

The instantaneous configuration of a slider crank mechanism has a crank GH 10cm long, the connecting rod HP is 50cm. The crank makes an angle of 60 degree with the inner dead centre position and is rotating at 110 rev/min. Determine the velocity of the piston P and the angular velocity of the link HP.

what i need is the steps how to do it up until the answer(i don't need the answer calculated just need to see how you would do it), I've changed all the values so it won't give me the answer to my own question. really been struggling with this because I've read 101 ways to do it and they all are over complicated and produce different results. need urgent help

Homework Equations

The Attempt at a Solution

right what I've done so far is found the length of the base of the diagram "PG" using x=0.1cos(60)+sqrt(0.5²-0.1²sin(60)² which comes out to around 0.542m
and found the velocity at H by using v=w X gh which comes out to 1.047Rad/s
Also would wHP the angular velocity part of the question would the formula for that be ((HG.w)/PH) X ((cos(60)/(sqrt(1-(HG²/PH²)sin²(60)))

.

I suggest calling the measure of $\angle GPH$ something like θ, and call the measure of $\angle GPH$ ϕ .

You can express the altitude of △$GPH$ as an expression in θ, and as an expression in ϕ . Equate these expressions, then take the derivative w.r.t. time, t.

$\displaystyle \omega=\frac{d\phi}{dt}=\text{ 110 rev/min .}$