# Homework Help: Angular velocity of a connecting rod on a rotating wheel.

1. Jul 6, 2009

### Samuelb88

1. The problem statement, all variables and given/known data
A rotating wheel with radius 40 cm and a connecting rod PQ with length 1.2m. The pin P slides back and fourth along the x-axis as the wheel rotates counterclockwise at a rate of 360 revolutions per minute. Find the angular velocity of the connecting rod, da/dt, in radians per second when θ = Pi/3.

-Connecting Rod PQ = 120 cm.
-O(0,0)
-Q(40cosθ,40sinθ) - at Pi/3.
-P(x,0)
-θ = Angle QOP
-a = Angle OPQ
-Pin P slides along the x-axis at 360 rev./min.

2. Relevant equations
A circle, centered at the origin with radius 40. Three points have been given that form a scalene triangle, triangle OQP. Point O(0,0), point Q(40cosθ,40sinθ), and point P(x,0). The point Q that rotates on the curve of the cirlce. The point P slides back and fourth along the x-axis at 360 revolutions per minute. I am suppose to find the angular velocity, da/dt in rad./sec of the connecting rod where a = the angle OPQ.

3. The attempt at a solution

The following information was determined at θ = Pi/3.

From the diagram given, point Q(40cosθ,40sinθ) moves along the circle (rotating wheel). Thus, the coordinates of point Q(20,20sqrt(3)).

I've constructed two right triangles. Triangle OQS, where the line segment QS is perpendicular to the x-axis (therefore creating two right triangles) and triangle SPQ.

Triangle OQS: x1^2 + y1^2 = 40^2
where x1 = 20; and y1 = 20sqrt(3).

Triangle SPQ: x2^2 + y1^2 = 120^2
where x2 = sqrt(13,200) = 114.9; and y1 is the same as triangle OQS. (y1 = sqrt(3).)

The original triangle: Triangle OQP (which is not a right triangle).
I've assigned the following notations to each side.
-Line segment OQ = Side A = radius of circle = 40.
-Line segment QP = Side B = length of rod = 120.
-Line segment OP = Side C = x1 + x2 = 20 + 114.9 = 134.9.

It's asking to me find da/dt in rad./sec at θ = Pi/3. This is where I am confused about to proceed. Thus far, I've found it very difficult to find a relevant equation that relates both θ and a.

Here's what I got:
1) Examining triangle OQP, from the Law of Sines:
sin(a)/A = sin(θ)/B

2) From differentiating da/dt, I get:
(A(-cos(a))(a) - sin(a)(0))/A^2 = (B(-cos(θ))(θ) - sin(θ)(0))/B^2

-Which simplifies to...
a(-cos(a)/A) = θ(-cos(θ)/B)

3) Explicitly defining the equation in terms of a, I get:
a = (θ)((Acos(θ))/(Bcos(a)))

Which when the values of each variable in the equation does not equal the answer. I was aware when working through the problem using the law of Sines to express da/dt that I would get dθ/dt as a variable in the final answer which I was unsure if θ could be assigned the rate of 360rev./min. which simplifies to 6rev./sec. which I assumed was 12*Pi/sec.

Another problem using the law of sines, I get the term, cos(a) in my final expression, which I am almost positive shouldn't be there.

I've also tried expressing sides A and B in terms of the pythagorean theorem of A^2 + B^2 = C^2. Where A = sqrt(x1^2 +y1^2); and B = sqrt(x2^2 + y1^2). But when differentiated using the law of sines, leads to an expression with dx/dt and dy/dt terms which aren't given. I've also tried expressing the angle a in terms of arcsin which was found using the expression given from the law of sines. However, again when differentiated becomes very messy...

Any ideas?

Last edited: Jul 6, 2009
2. Jul 6, 2009

### rl.bhat

Projection of Q on the y-axis is
y = Asinωt, where A = 40 cm and ω = 12π/ s.
y = Rsinα where R = 120 cm
So Rsinα = Αsinωt. Take the derivative with respect to t. You get
Rcosα*dα/dt = Aωcosωt
Rcosα = R( 1 - sin^2α)^2 = R[ 1- (y/R)^2]
Put y = Asinθ and find dα/dt when θ = π/3.

Last edited: Jul 7, 2009
3. Jul 7, 2009

### Samuelb88

Hey so i'm a bit confused about your explanation. I understand for the most part the steps, except for...

y=Asinωt graphs the displacement (?) of point A as it moves along the circle with amplitude A and ω = θ + α ???

secondly... i am bit confused by cosα = (1-sin^2(α))^2

R(cosα) = R(1 - (sinα)^2)^2

correct me if i'm wrong but shouldnt
R(cosα) = R(sqrt(1-(sinα)^2)) ?

anyways. from using the equations you've set up for me, using the cosα = sqrt(1-sin^2α), i arrive at:
α = (Aωcosωt)/(120(sqrt[1-((40sinθ)/(120))^2]))

- α = (4*Pi)/sqrt[11/12]
- α = (4*Pi)(2sqrt(3))/sqrt(11)
- α = (8*Pi)(sqrt(3))/sqrt(11)

but the answer says α = (4*Pi)(sqrt(3))/sqrt(11)?

4. Jul 7, 2009

### rl.bhat

Drop a perpendicular QM on OP. Let QM = y = Asinωt
In triangle QMP, y = Rsinα
α = (Aωcosωt)/(120(sqrt[1-((40sinθ)/(120))^2]))
It should be
α` = (Aωcosθ)/(120(sqrt[1-((40sinθ)/(120))^2]))

Last edited: Jul 7, 2009
5. Jul 7, 2009

### Samuelb88

ahh thanks yo.