Piston P has a velocity upwards of 68.95 mm/s. Find the angular velocity of arms AB and BD when q=67deg..

Question

https://imgur.com/a/yADcvhq

Answer 1

Hi Leo,

To find the angular velocities, this is a relative velocity problem commonly found in Dynamics.

For this analysis the following equation will be used: ^→v_x = ^→v_y + ^→v_x/y, which makes use of cross product.

Step #1 - Analyze Bar AB

^→v_A = 0 (mm/s)

^→v_B = ^→v_A + ^→v_B/A = ^→v_A +ω_AB x ^→r_A/B

(1) ^→v_B = 0-19.5ω_ABî +46 ω_ABĵ

Step #2 - Analyze Bar BD

^→v_D = 68.95ĵ (mm/s)

For ω_AB it will be assumed to be positive.

^→v_D = ^→v_B + ^→v_D/B = ^→v_B +ω_BD x ^→r_B/D

(2) 68.95ĵ = ^→v_B -142.8ω_BDî - 46ω_BDĵ

Step #3 - Sub equation (1) into equation (2)

(1) ^→v_B = 0-19.5ω_ABî +46 ω_ABĵ

(2) 68.95ĵ = ^→v_B -142.8ω_BDî - 46ω_BDĵ

(3) 68.95ĵ = -19.5ω_ABî +46 ω_ABĵ -142.8ω_BDî - 46ω_BDĵ

Step #4 - Equate Components

î

(4) 0 = -19.5ω_AB -142.8ω_BD

ĵ

(5) 68.95 = 46 ω_AB - 46ω_BD

Step #5 - Solve and Substitute

Using the î component and put in terms of ω_AB

0 = -19.5ω_AB -142.8ω_BD,

∴ (6) ω_{AB =} -142.8ω_BD /19.5

Sub equation (6) into equation (5), combine like terms, and solve:

68.95 = 46 (-142.8ω_BD /19.5) - 46ω_BD

ω_BD = -0.18 rad/s

Sub ω_BD into equation (4):

(4) 0 = -19.5ω_AB -142.8ω_BD = -19.5ω_AB -142.8(-0.18), ∴ ω_AB = 1.32 rad/s.

Step 6 - Do these answers make intuitive sense?

Using the right hand rule, bar AB is rotating counter clockwise and we notate the magnitude as positive. And bar DB is rotating clockwise, which we notate as negative. Knowing this and looking over the answers, then yes this makes both mathematical and intuitive sense.

Answers: ω_AB = 1.32 rad/s and ω_BD = -0.18 rad/s

Answer 2

I assume that the letter 'q' in your text refers to the angle 'θ' in the picture.

If I am wrong then my numeric result is wrong, but the approach is still applicable

to any other similar situation.

Let

a(t) be the length of AD at any time t,

b = 50 mm be the length of AB,

c = 150 mm be the length of BD,

θ(t) be the size of the marked angle at any time t,

θ'(t) = dθ/dt be the angular velocity of AB at any time t,

a'(t) = da/dt be the velocity of the piston at any time t,

t₁ be the time when a'(t₁) = 68.95 mm/s and θ(t₁) = 67°.

We are to find θ'(t₁).

When it will simplify notation, I will write

a instead of a(t)

a' instead of a'(t)

θ instead of θ(t)

θ' instead of θ'(t).

By the cosine theorem

c² = a² + b² - 2abcos(θ) (1)

Differentiate (1) with respect to t

0 = 2aa' - 2b(a'cos(θ) - asin(θ)θ')

Express

θ' = (ba'cos(θ) - aa')/absin(θ)

= a'(1/atan(θ) - 1/bsin(θ)) (2)

In (2) we know all the quantities at time t₁, except a(t₁).

That we can calculate from (1) as the quadratic equation

a² - 2abcos(θ) + b² - c² = 0

a = (2bcos(θ) ± √(4b²cos²(θ) - 4(b² - c²)))/2

= bcos(θ) ± √(b²cos²(θ) - b² + c²)

= bcos(θ) ± √(b²(cos²(θ) - 1) + c²)

= bcos(θ) ± √(c² - b²sin²(θ)) (3)

Substituting actual numbers

a(t₁) = 50 cos(67°) ± √(150² - 50²sin²(67°)) = 50 (cos(67°) ± √(3² - sin²(67°))) =

50 (0.39 ± 2.855)

Using only the positive root,

a(t₁) = 162.25 mm

Substitute into (2)

θ'(t₁) = 68.95×(1/(162.25×tan(67°)) - 1/(50×sin(67°))) = -1.3177 s^-1

The angular velocity is negative, because when the piston is going up, the angle θ is decreasing.

Secondly you are to calculate the angular velocity of the arm BD.

That is the derivative of the angle at the vertex D; call that angle η.

The approach is identical to calculating θ';

simply the role of θ is replaced by η, and the sides b and c switch roles.

Therefore we can reuse equations (1) and (2) after applying that substitution:

b² = a² + c² - 2accos(η) (4)

η' = a'(1/atan(η) - 1/csin(η)) (5)

Since we already know a(t₁), we can calculate cos(η) from (4):

cos(η) = (a² + c² - b²)/2ac

= (162.25² + 150² - 50²)/(2×162.25×150) = 0.95

Therefore

sin²(η) = 1 - 0.95² = 0.0975

sin(η) = √0.0975 = 0.31

tan(η) = sin(η)/cos(η) = 0.31/0.95 = 0.32

We can now substitute into (5)

η'(t₁) = 68.95×(1/(162.25×0.32) - 1/(150×0.31)) = -0.15 s^-1