This is what the figure looks like. http://i.imgur.com/9dwYpcT.jpg
A piston-to-wheel linkage is common in mechanical systems. In this question we study how the two rates - rat at which piston moves back and forth and the rate at which the wheel turns are related to each other.
a) The first step always is to relate variables themselves. Looking at the figure, show tha he following equation holds at all times
L^2 = (x - r cosθ)^2 + r^2 sin^2(θ)
b) Now differentiate the equation in previous part with respect to time t to establish the following relation between the rates:
2(x-r cos(θ))(dx/dt + r sin(θ) dθ/dt) + 2r^2 sin(θ) cos(θ) dθ/dt = 0
c) calculate the speed of the piston when θ = pi/2, assuming that r=10cm, L = 30cm and the wheel rotates at 4 revolutions per minute.
For part a, I know I am suppose to use the law of cosines to somehow prove it. But I can't get anywhere. Any Help is appreciated I really can't figure this out.
A piston-to-wheel linkage is common in mechanical systems. In this question we study how the two rates - rat at which piston moves back and forth and the rate at which the wheel turns are related to each other.
a) The first step always is to relate variables themselves. Looking at the figure, show tha he following equation holds at all times
L^2 = (x - r cosθ)^2 + r^2 sin^2(θ)
b) Now differentiate the equation in previous part with respect to time t to establish the following relation between the rates:
2(x-r cos(θ))(dx/dt + r sin(θ) dθ/dt) + 2r^2 sin(θ) cos(θ) dθ/dt = 0
c) calculate the speed of the piston when θ = pi/2, assuming that r=10cm, L = 30cm and the wheel rotates at 4 revolutions per minute.
For part a, I know I am suppose to use the law of cosines to somehow prove it. But I can't get anywhere. Any Help is appreciated I really can't figure this out.
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Yeah, you have to apply cosine rule.
Define θ first, it is an angle subtended by radius and length of x (it is unclear in the diagram -pardon me)
Therefore, by cosine rule , L^2 = x^2 + r^2 - 2(x)(r)cosθ
=x^2 -2xrcosθ + r^2 *1 (recall your trig identity)
= x^2 - 2xrcosθ + r^2 (sin^2θ + cos^2θ )
=x^2 - 2xrcosθ + r^2cos^2θ + r^2sin^2θ
=(x-rcosθ )^2 + r^2sin^2θ (shown)
dL^2/dt = dL^2/dθ * dθ / dt
dL^2/dθ = (2)(x-rcosθ)(rsinθ ) + 2(rsinθ)(rcosθ )
= 2(x-rcosθ)________+2r^2sinθ cosθ______
There is something between the spaces... Im sorry , im just a high school student , i can't seem to get the rate of change of x :(
This is all i can help. Good Luck
EDIT: NEVERMIND I GOT IT
dL^2/dx = 2(x-rcosθ )
dL^2/dt = dL^2/dx * dx/dt
(2)(x-rcosθ)(rsinθ ) + 2(rsinθ)(rcosθ ))*dθ/dt = 2(x-rcosθ)*dx/dt (WHY NOT NEGATIVE)
Rearrange and factorize to get
2(x-r cos(θ))(-dx/dt + r sin(θ) dθ/dt) + 2r^2 sin(θ) cos(θ) dθ/dt = 0 (shown)
For part c, this time, I really don't know how to.
Define θ first, it is an angle subtended by radius and length of x (it is unclear in the diagram -pardon me)
Therefore, by cosine rule , L^2 = x^2 + r^2 - 2(x)(r)cosθ
=x^2 -2xrcosθ + r^2 *1 (recall your trig identity)
= x^2 - 2xrcosθ + r^2 (sin^2θ + cos^2θ )
=x^2 - 2xrcosθ + r^2cos^2θ + r^2sin^2θ
=(x-rcosθ )^2 + r^2sin^2θ (shown)
dL^2/dt = dL^2/dθ * dθ / dt
dL^2/dθ = (2)(x-rcosθ)(rsinθ ) + 2(rsinθ)(rcosθ )
= 2(x-rcosθ)________+2r^2sinθ cosθ______
There is something between the spaces... Im sorry , im just a high school student , i can't seem to get the rate of change of x :(
This is all i can help. Good Luck
EDIT: NEVERMIND I GOT IT
dL^2/dx = 2(x-rcosθ )
dL^2/dt = dL^2/dx * dx/dt
(2)(x-rcosθ)(rsinθ ) + 2(rsinθ)(rcosθ ))*dθ/dt = 2(x-rcosθ)*dx/dt (WHY NOT NEGATIVE)
Rearrange and factorize to get
2(x-r cos(θ))(-dx/dt + r sin(θ) dθ/dt) + 2r^2 sin(θ) cos(θ) dθ/dt = 0 (shown)
For part c, this time, I really don't know how to.