Analyzing Reciprocating Motion: Shaft Attached to a Rotating Wheel

Analyzing reciprocating motion: Shaft attached to a rotating wheel

Najeha Rashid* and Dr. Sabieh Anwar May 25, 2019

1 Abstract

In a steam engine, the expansion of compressed gasses in a cylinder is responsible for the to and fro motion of the piston. The linear motion of the piston is converted into the rotating movement of a ﬂy-wheel by a connecting rod or crankshaft. This mechanism, nowadays, is most commonly used in internal combustion engines, in motor vehicles running on petrol or diesel. In this report we will study the reverse mechanism: Conversion of rotary motion of a wheel into the reciprocating motion of a shaft. The dependence of the motion on the lengths of various components of the setup and angle (θ) traversed by the wheel will be investigated.

2 Theory

As the wheel rotates in Figure (1b), the connecting rod pulls on or pushes the shaft horizontally, thereby changing the distance, x of the screen from the sonic detector[1]. Using the diagram we can write: R sin θ = L sin β and d = R cos θ + L cos β. Solving the equations simultaneously gives us the expression for d as follows: r d R R = cos θ + 1 − ( sin θ)2. L L L Now, d + L0 + x = C =⇒ x = K − d.

Here, C is the horizontal distance between the center of the wheel and the sensor, which remains the same throughout the experiment. We have grouped it with L0, another constant, in the later equation as K. Therefore we have: r R x = −R cos θ − L 1 − ( sin θ)2 + K. (1) L By taking single and double time derivatives respectively we get: !−1/2! v 1 dx R R 2 = = sin θ 1 + cos θ 1 − sin θ and (2) Rω Rω dt L L

! !−1/2 a 1 dv R cos2 θ R 2 = = sin2 θ − 1 − sin θ − cos θ, (3) Rω2 Rω2 dt L R 2 L 1 − L sin θ

*[email protected] [email protected] 1 (a) The pendulum setup used in this experiment. However, our experiment only involves the to and fro motion of the black disc (screen) attached to the right end of the shaft. (b)

Figure 1

Voltage (v) Angular Frequency, ω (rad/s) Period, T (s) 30 8.266 ± 0.002 0.760 ± 0.001 18 4.744 ± 0.002 1.324 ± 0.004

Table 1: Values of the angular frequency and their corresponding time periods obtained from our curve ﬁttings where, v is the velocity of the shaft, a is the acceleration, and ω is the angular frequency of the wheel, which is constant and depends on the electrical power supplied.

3 The Experiment 3.1 Motion sensor The Vernier Motion Detector used, is an ultra-sonic device that determines the position of the object in front of it by the echo time of ultra-sonic waves. It has a resolution of 1 mm.

3.2 Procedure Figure (1a) shows the complete setup of the experiment. The wheel is powered by a DC power supply (HY5003). The Vernier Motion Detector was placed with its center coinciding with that of the black screen for maximum echo. It was interfaced with the computer through the Vernier LabPro Interface. Before taking readings for a particular value of voltage, the sensor’s distance from the screen had to be adjusted for an accurate detection of motion and then maintained for the rest of the experiment. This is necessary because if the distance is too short then the time between successive echos is greatly diminished and the detector is unable to diﬀerentiate between incoming signals. In the opposite case, the resolution of the detector is diminished and minor changes in the disc motion go undetected. This required observing the signal for various distances before deciding upon the most suitable one. In our case, suitable distances were found to be around 27 cm and 25 − 26 cm for 30 V and 18 V respectively. The lengths, L and R were measured using vernier calipers to be: R = (15.00 ± 0.5) mm and L = (36.00 ± 0.5) mm. (4)

3.3 Obtaining Results The signals were recorded on Logger Pro software and plots for the position, velocity and acceleration of the shaft, w.r.t time were obtained at 30 V and 18 V electric supply. Table (1) lists the results. The equations, derived in the ﬁrst section, were used along with Eq. (4) for this purpose. and are stated as follows. 2 Angular Frequency ω (rad/s) xmax (cm) xmin (cm) ∆x(cm) A 4.744 ± 0.002 25.5 22.8 2.70 0.900 8.266 ± 0.002 26.6 24.2 2.40 0.900

Table 2: The maximum and minimum displacement of the screen from the sensor. ∆x should ideally be equals to 2R.

Angular Frequency, ω (rad/s) vmax (cm/s) 4.744 ± 0.002 6.70 8.266 ± 0.002 10.0

Table 3: The maximum magnitudes of velocity obtained from the best-ﬁt curve.

3.3.1 Position The best-ﬁt curve equation plotted w.r.t time in Figure (3a) and Figure (4a) is: r R 2 x = A − R cos tω − L 1 − sin tω + K (5) L

The fitting parameters, ω, A (unitless), and K (m) were adjusted manually to fit our experimental results in both the cases. If we compare Eq. (5) to Eq. (1), A comes out to be unity and K represents a constant distance. Also, θ = t × ω, (6) where, t is the time elapsed in seconds, after the wheel starts rotating, and ω is the angular frequency in rad s−1. The maximum distance of the screen from the sensor occurs at θ = π rad and the minimum distance at 0 rad. Applying this to Eq. (5) and using the values of fitting parameters, we get the following equations. The values are given in Table (2).

xmin = A(−R − L) + K and xmax = A(R − L) + K

and ∆ x = A(2R).

3.3.2 Velocity The best ﬁt equation for velocity plotted in Figure (3b) and Figure (4b): ! R R 2−1/2 v = B sin tω 1 + cos tω 1 − sin tω (7) L L

The equation uses the same value of ω, as Eq. (5). The ﬁtting parameter, B, was adjusted manually.

3.3.3 Acceleration The best-ﬁt curve equation used in Figure (3c) and Figure (4c):

! !−1/2 ! R cos2 tω R 2 a = D sin2 tω − 1 − sin tω − cos tω (8) L R 2 L 1 − L sin tω

Units of D are: cm/s2 and the value of ω same as deduced in Eq. (5).

3 Coordinate axis Position at π/2 rad Mid-point displacement Exp. value x 1.57 1.36 1.46 y 0.245 0.242 0.244

Table 4: The intersection points for half cycle from θ in the range 0 to π rad.

Pin joint A, as seen at di erent angles (θ) of rotation

a. A L θ β B C A’ B’

d r Figure 2: Figure (a) and Figure (b) shows the Pin joints A at equal height above the horizon- b. tal (θ = 2π − θ) in the right and left semicircle A of the wheel respectively. If the connecting rod at AB, were to swing around a ﬁxed point B towards the horizontal axis, it will cut the axis at 2π-θ β B C. A0B0 in Figure (a) and Figure (b), are the A’ C B’ positions of the connecting rod corresponding to the minimum and maximum displacement of d l the shaft from the sensor respectively. dr and dl give the linear displacement of the shaft when c. the rotation of the wheel brings A to A0. Fig- A ure (c) gives the angular displacement of A corresponding half the linear displacement of the θ=77.8 shaft between 0 rad to π rad. It can be seen that this doesn’t occur at π/2 rad as one would have A’’ C B’’ A’ B B’ expected, but occurs at an acute angle. The linear displacement of the shaft will be equal, if A dl = dr dl 00 0 d were to move either towards A or A because r C lies on exactly the center of the wheel.

4 Discussion

As Pin-joint A traverses one full rotation around the wheel, the distance it covers in the left semicircle of rotation is less than the distance it covers in the right semicircle. This is evident in in Figure (5). Figure (2) gives an intuitive understanding of why this is so. We will analyze the behaviour of the shaft rotating at ω = (4.744 ± 0.002)s over one complete period. We can see in Figure (2) that dl will always be less than dr meaning that although A spends equal amount of time in both halves (ω is constant), the to and fro displacement the shaft covers as A spans the right side is greater than when it is moving across the left. Therefore the value of θ corresponding to the expectation value of position is an acute angle, as stated in Table (4). Finding the expectation value of Pin-joint A using Figure (5) :

Area under the plot Exp. value of Position = = 0.2438 rad 2π

4 5 3 s s d d 4.5 n n . o o s 2.5 c c 4 e e s s 5 3 004) . r r 0 3.5 o o 2 f f . ± t t s 3 o o l l ) ) p p s s 324 ( ( . e e e e 001) . 2.5 1.5 m m m m 0 i i i i T T T T (c) (c) 2 ± s s v v 1 n n 760 o o . 1.5 i i t t a a ) equals to (1 r r 1 e e = (0 l l T e e 0.5 T c c c c 0.5 A A 0 0 0 1 0

0.4 0.3 0.2 0.1 1.2 0.8 0.6 0.4 0.2

-0.1 -0.2 -0.3 -0.2 -0.4 -0.6 -0.8

A A c c c c e e l l e e r r a a t t i i o o n n ( ( m m = = s s ) )

2 2 ! ! 5 3 4.5 s s d d 2.5 n n 4 o o c c e e s s 3.5 5 3 2 r r o o f f 3 t t ) ) s s o o l l ( ( e e p p 2.5 1.5 m m e e i i m m T T i i (b) (b) 2 T T s s 1 v v 1.5 y y t t i i c c 1 o o l l e e 0.5 V V 0.5 0 0 0 0

0.1

-0.1

0.08 0.06 0.04 0.02 0.15 0.05

-0.02 -0.04 -0.06 -0.08 -0.05 -0.15

V V e e l l o o c c i i t t y y ( ( m m = = s s ) )

1 1 ! ! 5 3 4.5 s s d d 2.5 n n 4 o o c c e e s s 3.5 5 3 2 r r o o f f 3 t t ) ) o o s s l l ( ( e e p p 2.5 1.5 m m e e i i m m T T i i (a) (a) 2 T T s s 1 v v 1.5 n n o o i i Figure 4: The results obtained at 30 V. In this case, a single period of the rotation is for a time period, t t i i 1 s s o o 0.5 P P 0.5 Figure 3: The results obtained at 18 V from our curve ﬁttings. Enclosed in between dashed lines is a single period ( 0 0

0.26 0.25 0.24 0.23 0.27 0.26 0.25 0.24

0.255 0.245 0.235 0.225 0.275 0.265 0.255 0.245 P o s i t P i o o n s i ( t m i o n ) ( m )

5 Plot of Position between 0 to 2:

0.26 Mid-point displacement 3=:/2 & 3:/4 Expectation value 0.255

0.25

) X: 1.57

m X: 1.46

( Y: 0.2454

n Y: 0.2439

o 0.245

s o

P Figure 5: A compar- 0.24 X: 1.363 ison of the expecta- Y: 0.2425 tion value, the mid- point of displacement 0.235 between θ = 0 rad and π rad and the position at π/2 rad. 0.23 The plot is symmetric about θ = π and the remaining inter- 0 1 2 3 4 5 6 section points can be 3(rad) easily calculated.

Plot of Velocity between 0 to 2: Plot of Acceleration between 0 to 2: 0.08 0.5 Peak = (1.217, 0.067) 3=:/2 & 3:/4 3=:/2 & 3:/4 0.06 0.4 X: 2.142

Y: 0.03937 )

0.04 L 2 0.3

)

! =

s 0.02 0.2

(

m n ( M

0 o 0.1

t t

i X: 3.14

a c

Y: -0.0002064 r e

o -0.02 0

c V N c

-0.04 A -0.1 X: 4.153 L N Y: -0.04036 -0.06 -0.2 M Peak = (5.063, -0.067) -0.08 -0.3 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 3(rad) 3(rad) (a) (b)

Figure 6: As θ increases from 0 rad, the velocity of the shaft increases in the positive direction (decreasing acceleration) until in reaches it’s maximum value at 1.217 rad (69.70◦) (acceleration =0 rad/s2). After this, the shaft starts to slow down (deceleration) while still moving in the positive direction. The deceleration is maximum at point L, after which it becomes increasingly lower till M is reached. After this, the shaft accelerates in the negative direction with increasing magnitude till N. From here the acceleration reduces to zero at 5.063 rad (290.2◦). This is the maximum velocity in the negative direction). Beyond this point, the shaft decelerates till the velocity is 0 rad/s, before the start of the next cycle.

6 4.1 Understanding the extreme values of velocity It can be seen in Figure (6a) that the extreme values of the velocity occur in the right semi circle of rotation. In Figure (1b) if θ = θ0 is such that the connecting rod is tangential to the wheel, then the centripetal velocity (R × ω) will be the same as the velocity of the rod being pulled/pushed at a corresponding angle β0. In that case, the speed of the shaft in the forward/backward direction is simply: |v| = Rω cos β where β = π/2 − θ0. (9) We can say that at θ0, the rotary motion of the wheel is being converted most eﬀectively into the linear motion of the shaft and the velocity of the shaft in either direction is maximum. We use values of θ0 from Figure (6a) and inserting them into Eq. (9). We then have: |v| = Rω cos (0.353) = 0.668 for θ = 1.217 rad (10) |v| = Rω cos (−0.349) = 0.668 for θ = 5.063 rad (11) Our answers match the values in Table (3) with just an error of 0.3%.

4.2 Dependence of motion on length segments It can be shown that the disparity in the maximum displacement of the shaft toward the left and right

depends upon the ratio α = R/L. Using Eq. (1) we Dependence of dl=2R on , = R=L calculate the ratio of the maximum displacement to 0.5

the left, dl to the total peak to peak displacement, 0.45 2R, of the shaft: 0.4 d x(θ = π) − x(θ = π/2) l = 0.35 2R 2R √ 0.3 2 ! α − 1 + 1 − α 1 R 2 0.25

= , = l

α 2R d 0.2

The relation is mapped in Figure (7). For our setup: 0.15

α = 0.405 ± 0.020 and 0.1 d 0.05 l = 0.394 ± 0.005 2R 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 The ratio R/L must be greater than unity, oth- R=L erwise as the Pin joint A moves from 0 rad to π/2 rad the shaft will be pulled to the left. Once A has crossed π/2 rad, the connecting rod will simply Figure 7 swing about point B in all consecutive rotations, concentric to the wheel and no reciprocating motion occurs (consult Figure(1b). At very large value of L, the displacement on both sides of the wheel is the same where as for L = R it is zero.

5 Conclusion

We investigated the motion of a shaft undergoing reciprocating motion and concluded that the motion is not symmetric about the vertical axis through the center of the wheel. This is proven by the fact that the expectation value of position of the shaft is such that Pin joint A occurs at an acute angle. The ratio, R/L, was found to be crucial to this phenomenon and for connecting rods having lengths much greater than the radius of the wheel (L >> R), the maximum displacement on either side of the vertical was found to be the same.