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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a) I cannot decipher your diagram but the equation looks like Pythagoras.
b) L and r are constants so differentiating with respect to t gives
0 = 2(x-rcosΘ)(dx/dt - rsinΘdΘ/dt)+r^2.2sinΘcosΘ.dΘ/dt
hence 2(x-rcosΘ)(dx/dt - rsinΘdΘ/dt)+2r^2sinΘcosΘ.dΘ/dt =0
c) Sub Θ=π/2, r=10, L=30 into a) to give x=20√2
Also dΘ/dt = 8π rad per min.
Sub into the answers into b) and you have an equation for dx/dt which is
the value of the speed of the piston in cm per min.
I make it 80π≈251.3 cm per min
b) L and r are constants so differentiating with respect to t gives
0 = 2(x-rcosΘ)(dx/dt - rsinΘdΘ/dt)+r^2.2sinΘcosΘ.dΘ/dt
hence 2(x-rcosΘ)(dx/dt - rsinΘdΘ/dt)+2r^2sinΘcosΘ.dΘ/dt =0
c) Sub Θ=π/2, r=10, L=30 into a) to give x=20√2
Also dΘ/dt = 8π rad per min.
Sub into the answers into b) and you have an equation for dx/dt which is
the value of the speed of the piston in cm per min.
I make it 80π≈251.3 cm per min