0153 Script - Using a Pendulum to find the Radius of ...

0153 Script - Using a Pendulum to find the Radius of Curvature of a Roundabout.docx of 14

Whiteboard: 1/60, f5.6, iso 640, WB Fluorescent 4, Manual Audio at level 11, focal length ½ way to ∞. H4n INPUT @ 93% gain. BBB 1/60, f5.6, iso 320, WB Fluorescent 4, Manual Audio at level 11, focal length ½ way to ∞. H4n INPUT @ 93% gain. Title: Using a Pendulum to find the Radius of Curvature of a Roundabout Mr.p: [Standing stage right] Good morning. Today we are

going to determine the radius of curvature of a roundabout using a pendulum. Sean Vander Meulen and Stuary Arey in Istanbul, Turkey are going to show us the example. Sean, please take it away now… [Look at Sean and pause 10 seconds for transition]

Sean: [Standing stage left. Look at mr.p and pause 10 seconds for transition. Please speak clearly and at a moderate pace.] Thanks mr.p. We started by hanging a pendulum in the middle of a car. With the car at rest, the pendulum hangs straight down and creates a vertical reference line. … [Video with 3 parts: Top view of car using drone. Speedometer.


Pendulum] (We then drove the car in a circle at a constant speed of 19 kilometers per hour. … With the car moving at a constant speed and describing a constant radius, the inertia of the pendulum causes it to swing out away from the vertical. … By overlaying the video of the pendulum with the car in motion over the vertical pendulum, we can measure the angle. … We can then predict, using physics, the radius through which the car is driving.) [Again, stage Left.] Now it’s your turn to do the physics. Good luck. [pause 10 seconds for transition]

Mr.p: [standing stage right, 10 second pause] Thanks Sean. Let’s summarize this as a problem. Billy, please?

Billy: A pendulum hangs at an 8.0° angle with the vertical in a car being driven at 19 kilometers per hour around a


roundabout. What is the radius of the circle the car describes?

Mr.p: Bobby, what do we know and what are we trying to find? Bobby: We know the angle, theta, is 8.0 degrees. We also

know the tangential speed of the car is 19 kilometers per hour. We need to convert that to meters per second so multiply by 1000 meters over 1 kilometer and 1 hour over 3600 seconds to get {calc} 5.27 repeating meters per second. And we are solving for the radius, so radius equals question mark?

[show same 3 simultaneous videos again with all the numbers] mr.p: [

θ = 8.0°; v

t=19

kmhr

×1000m1km

× 1hr3600s

= 5.27ms

; r = ? ] The angle of the pendulum was not quite constant, (so we actually measured the maximum angle of 12 degrees and the minimum angle


of 4 degrees to get an average angle of 8 degrees.) Bo, what do we do next? • Bo: Draw a free body diagram. • Mr.p: (Please, go ahead.) • Bo: The force of gravity is straight down and the force of

tension is in the direction of the string. … I think that is it, right?

• Bobby: There is no surface, so … no Force Normal. • Billy: That means there is no Force of Friction either. • Bo: I guess that’s it, just the Force of Gravity and the Force

of Tension. Mr.p: [FBD Stand where FBD isn’t] Billy, in what direction do we need to

sum the forces in? • Billy: Uh. The x and y direction?


• Bobby: Actually no, remember, when an object is moving in a circle, you need to sum the forces in the in direction and a direction perpendicular to the in direction. [top view w/ circle?]

• Bo: That’s why mr.p said, “in what direction do we need to sum the forces in”, get it?

• Billy: Oh right. So we need to sum the forces in the in direction and … the y direction.

Mr.p: [Stand where FBD isn’t] But before we do so, we need to break forces into components. Bobby, please? • Bobby: The force of gravity is in the y direction already, so

we don’t need to break that into components. The force of tension … we start by drawing the components and … sine of theta equals opposite over hypotenuse which equals


Tension Force in the in direction over the Tension Force, therefore the Tension Force in the in direction equals tension force times sine theta. And cosine of theta equals adjacent over hypotenuse, which equals tension force in the y direction over tension force. Therefore the tension force in the y direction equals the tension force times the cosine of theta. But we don’t know the tension force, do we?

• Bo: Let go of your numbers dependency. • Billy: and utilize your equation holster!!

[transition to small live screen over paper background to keep track of all equations in equation holster.]

Mr.p: [ sinθ = O

H=

FTin

FT

⇒ FTin

= FT

sinθ& cosθ = A

H=

FTy

FT

⇒ FTy= F

Tcosθ with Holsters

Stand where FBD isn’t] Keep going Bobby.


Bobby: We need to … redraw the free body diagram. Force of Gravity is still down, but now, instead of the force of tension we have its components, the force of tension in the y direction and the force of tension in the “in” direction.

Mr.p: [redraw FBD Stand where FBD isn’t] And now Bo? • Bo: Let’s sum the forces in the y direction. The net force in

the y direction equals the tension force in the y direction minus the force of gravity and the net force in the y direction also equals mass times acceleration in the y direction. Can we assume the acceleration of the pendulum in the y direction is zero?

• Mr.p: (Yes, we need to assume a constant angle to solve the problem. So yes, the acceleration in the y direction is approximately zero.) [maybe not Voice Over]


• Bo: Good. Then the tension force in the y direction equals the force of gravity and … we have an equation for the force of tension in the y direction in our equation holster. … We substitute in to get the tension force times cosine theta equals the force of gravity which equals mass times the acceleration due to gravity. Which we should put into our equation holster.

Mr.p: [

Fy= F

Ty− F

g= ma

y= m 0( ) = 0∑ ⇒ F

Ty= F

g⇒ F

Tcosθ = mg ⇒ F

T= mg

cosθ with holster] Actually, for reasons which will be clear very soon, I solved for the Tension Force which is equal to mass times acceleration due to gravity all divided by the cosine of the angle and put that in our equation holster. Billy, what next?


• Billy: We need to sum the force in the in direction. So Net force in the in direction equals, negative Force of Tension in the in-direction, negative because it is to the left and … the net force in the in direction also equals mass times acceleration in the in-direction which is the centripetal acceleration and we can substitute into that equation … from our equation holster for Tension Force in the in direction we can substitute Tension Force times sine theta and for centripetal acceleration we know that equals the tangential velocity squared divided by the radius. … Oh, and also from our equation holster we know the Tension Force equals mass times acceleration due to gravity divided by the cosine of theta.

• Bobby: [write it out, get exited, jump in with] Everybody


brought mass to the party! [YouTube card] • BBB: [general reactions]

Mr.p: [

Fin∑ = −F

Tin= ma

c⇒−F

Tsinθ = m

vt2

r

⎛

⎝⎜

⎞

⎠⎟ ⇒ − mg

cosθ⎛⎝⎜

⎞⎠⎟

sinθ = mv

t2

r

⎛

⎝⎜

⎞

⎠⎟ ⇒ −g tanθ =

vt2

r ] (chuckle) Billy, please? • Billy: And we can rearrange the formula to solve for the

radius the car describes. … radius equals the negative of the tangential velocity squared divided by the quantity acceleration due to gravity times the tangent of theta. … That doesn’t seem right, the radius can’t be negative?

• BBB: [pause, with entire Fin∑ equation on screen with

FBD] • Bobby: The Tension Force in the in direction is in the in direction.


• Bo: And the in direction is positive. • Bobby: And the out direction is negative. • Billy: So when we sum the forces in the in direction,

because the Force of Tension in the in direction is in toward the center of the circle, the force of tension in the in direction is positive and not negative. That makes sense. That follows through the whole equation and the radius ends up being positive, not negative. Thanks guys. [top view with circle?]

• Bobby: You’re welcome. • Bo: Yup.

Mr.p: [Fix negative & ⇒ r =

vt2

g tanθ Top view with circle?] Absolutely. That is a very common mistake, remember


forces in toward the center of the circle are positive and forces out away from the center of the circle are negative. … Now we can substitute in numbers.

[

⇒ r =5.27( )2

9.81( )tan 8( ) = 20.2037m ≈ 2.0×101m ] We get the tangential speed of 5.27 repeating squared divided by the quantity 9.81 times the tangent of 8 degrees which works out to be 20.2037 or, rounded to two significant figures, 2.0 x 10 to the first meters, which is 20 meters with 2 significant digits. … That’s what we calculated. Now, let’s compare that to the measured radius. (We can measure the radius the car describes using the top view video of the car. Let’s pause the video for a moment. The length of the car is 4.45 meters which we can use as our measuring tool. You can see we


can fit just over 4 cars in the radius {crop out one car and manipulate to show on screen}, which goes just beyond the center of the car where the pendulum was. In fact, we can measure the last bit of car to be very nearly one fourth of a car. The radius then is 4.25 cars. Which we can multiply by 4.45 meters per car, to get 18.9125 or approximately 19 meters.) [

r = 4.25cars × 4.45m

car=18.9125m ≈19m &

E

r= O − A

A×100 = 20.2037−18.9125

18.9125×100 = 6.8272 ≈ 6.8%] And using the equation

for relative error to determine the percent difference between our measured and observed value, we get a difference of approximately 6.8 percent. Which means, I think you would all agree that, {smiling and hearing the first four beats of “The Physics Works by Emre}


BBB: The physics works, the physics works, uh huh, uh hu, the physics works. {singing to Emre’s song} [Play song on phone on repeat for video/audio sync. Turn down volume and sing over song.] Turkey: [cut to Turkey. Everybody singing and dancing to Emre’s song. Hopefully with Emre up singing his song. Mr.p: [at about 35 seconds into the song] Thank you very much for learning with me today, I enjoyed learning with you. [Various outtakes from Turkey video for the remainder of Emre’s song.]

0153 Script - Using a Pendulum to find the Radius of ...

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