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Centripetal Force, Tension, Circular Motion 
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#1
Nov1706, 06:12 PM

P: 263

Hi again ^^. I worked to the best of my ability to solve my homework problem, but it's very hard and confusing me. I showed all of the work to the best of my knowledge, and I was hoping someone could check my work. I'm not necessarily looking for the answer as much as an explanation so that I can understand what I'm doing. Thanks for any time ^^. I appreciate it!
To study circular motion, two students use a handheld device, which consists of a rod on which a spring scale is attached. A polished glass tube attached at the top serves as a guide for a light cord attached to the spring scale. A ball of mass .200 kg is attached to the other end of the cord. One student swings the ball around at a constant speed in a horizontal circle with a radius of .5 m. Assume friction and air resistance are negligible. a) Explain how the students, by using a timer and the information given above, can determine the speed of the ball as it is revolving. v = (2πr) / T We are given the radius (.5 m) and if we have a timer, we can determine the period, or time required for the ball to travel once around the circle. By using the equation above and plugging in the values, we can determine the speed of the ball as it is revolving. b) How much work is done by the cord in one revolution? Explain how you arrived at your answer. Since the centripetal force on an object is always perpendicular to its velocity, no work is done on the object because no energy is gained or lost. c) The speed of the ball is determined to be 3.7 m/s. Assuming that the cord is horizontal as it swings, calculate the expected tension in the cord. Fc = mac = T – mg T = mac + mg T = m(v2/r) + mg T = .2 kg ((3.7m/s)2/.5 m) + .2 kg (9.8 m/s2) = 7.44 N d) The actual tension in the cord as measured by the spring scale is 5.8 N. What is the percent difference between this measured value and the tension and the value calculated in part (c)? (5.8 N / 7.44 N) x 100 = 78% 100% – 78% = 22% There is 22% difference between the actual tension and the value calculated in part (c). e) The students find that, despite their best efforts, they cannot swing the ball so that the cord remains exactly horizontal. i) On the picture of the ball below, draw vectors to represent the forces acting on the ball and identify the force that each vector represents. (The image is just a single dot). I drew tension to the left, and weight (mg) to the south. I'm wondering though, could tension be north west? ii) Explain why it is not possible for the ball to swing so that the cord remains exactly horizontal. It is not possible to swing the ball so that the cord remains exactly horizontal because the centripetal force always draws the ball to the center of the circle and will not allow it to go straight. iii) Calculate the angle that the cord makes with the horizontal. Fc = mac = 5.8 N  mgsinθ (mv2/r – 5.8 N)/(mg) = sinθ ((.2kg ((3.7 m/s)2)/.5 m) – 5.8 N)/((.2kg)(9.8m/s2)) = .165 sin θ = .165 θ = 9.5 \ 


#2
Nov1706, 07:02 PM

P: 354

Other than that, your problems look pretty good. Good job. *I didn't check any of your math. But the concepts look right. Paden Roder 


#3
Nov1706, 07:04 PM

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#4
Nov1706, 08:51 PM

P: 263

Centripetal Force, Tension, Circular Motion
Okay thanks :).
I will try them later/tomorrow and repost my work. I appreciate it! 


#5
Nov1806, 12:40 AM

P: 263

Okay, for part C I broke up the components (x and y) and added the separately.
(x) F_c = ma_c = Fncos(theta) (y) F_c = 0 = Fn  mg Fn = mg And by plugging mg that I got from the y coordinates into the x, I got ma_c = mgcos(theta) And if tension equals ma_c (because if the cord is horizontal, the tension is the centripetal force  as you said)  then (.2)(9.8)cos0 = 1.96 N So the expected tension is 1.96 N Then for part d, 1.96/5.8 x 100 = 33.8% 100  33.8% = 66.2% Therefore, the percent difference is 66.2% for e i) I have mg going down, and the tension going southeast. I'm still a little confused on this.. I know it's along the cord, but the cord is above the radius... It's like a straight pole that extends above the radius, then a line goes SE until it meets the end of the radius... for e ii) It is not possible for the ball to swing so that the cord remains exactly horizontal because the tension acts in both horizontal and verticle components. The horizontal acounts for F_c and the verticle supports the objects weight, so the cord cannot be just horizontal (from what you said) and for e iii), I applied the directions and newtons second law and got... (x) F_c = ma_c = Tcos(theta) (same as before, but the line isn't horizontal and we dont know theta) ma_c/T = cos (theta) 2.794/5.8 = cos (theta) .096 = cos(theta) theta = 84.5* I worked on this for quite some time.. I hope it makes at least more sense or is a little bit closer to the answer.. 


#6
Nov1806, 12:56 AM

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#7
Nov1806, 01:08 AM

P: 263

So for part c, can the answer be as simple as
F_c = T = ma_c = mv^2/r = (.2)(3.7^2)/.5 = 5.476 N? And I would calculate for d accordingly? I don't know though.. seems too easy. I'm probably missing some great detail. 


#9
Nov1806, 07:13 AM

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#10
Nov1806, 03:01 PM

P: 263

I have a diagram of the mass on the cord if you want to see it .. but I dont know how to upload images on here..
From what you've said, tension must angle up and towards the axis of rotation from the mass.. so if the mass is to the right originally (though hanging from a diagonal cord that is going up northwest), then the tension might be northwest in order to keep the ball from rotating further away from the rod. And judging by the picture and my answer for part eiii, the angle looks to be about 85 degrees or so. 


#11
Nov1806, 03:48 PM

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#12
Nov1806, 08:21 PM

P: 263

e iii)
Well, if the 'actual' tension is 5.8 N, I should use that in the calculation for part iii... ma_c should equal (.2)(3.7^2)/(.5) = 5.476 N And I want the angle the cord makes with the horizontal... and if the angle is that between the cord and radius.. the ball being to the right of the angle.. then. it should definitely be much less than 85.. F_c = ma_c = Tcos(theta) 5.476/5.8 = cos(theta) .944 = cos (theta) (theta) = 19.5* 


#14
Nov1806, 08:55 PM

P: 263

Thanks ^_^! I feel like I'm really getting it :)



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