Register to reply

Centripetal Force, Tension, Circular Motion

by twotaileddemon
Tags: centripetal, circular, force, motion, tension
Share this thread:
twotaileddemon
#1
Nov17-06, 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 hand-held 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
\
Phys.Org News Partner Science news on Phys.org
Bees able to spot which flowers offer best rewards before landing
Classic Lewis Carroll character inspires new ecological model
When cooperation counts: Researchers find sperm benefit from grouping together in mice
PRodQuanta
#2
Nov17-06, 07:02 PM
P: 354
Quote Quote by twotaileddemon
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?
The ball is not accelerating/moving up or down in the example. Thus, the y-hat component of force must = 0. Thus, the y-hat component of tension must equal mg. Thus, the tension vector must be at some angle up and to the left.

Quote Quote by twotaileddemon
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.
You are a little off here. Think about why the ball is not horizontal to start off with. Look at your force diagram. There will always downward force due to gravity. Thus, there must always be a downward force on the ball that you can't get rid of.

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
Doc Al
#3
Nov17-06, 07:04 PM
Mentor
Doc Al's Avatar
P: 41,325
Quote Quote by twotaileddemon
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.
OK.

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
Realize that gravity acts downward, but the centripetal acceleration is horizontal. If the cord is horizontal, its tension is the centripetal force. Furthermore, since forces are vectors you can't just add a horizontal force (T) to a vertical force (-mg); you must add components in each direction separately.

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).
Redo after correcting 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?
The tension is along the cord; if the cord isn't horizontal, neither is the tension.

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.
No. The tension does two things: its horizontal component provides the centripetal force; its vertical component supports the object's weight. Since there must be a vertical component, the tension--and thus the cord--cannot be purely horizontal.

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
Redo, correcting the earlier errors, and considering horizontal and vertical components separately. Apply Newton's 2nd law to each direction, then combine those two equations to solve for theta.

twotaileddemon
#4
Nov17-06, 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!
twotaileddemon
#5
Nov18-06, 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..
OlderDan
#6
Nov18-06, 12:56 AM
Sci Advisor
HW Helper
P: 3,033
Quote Quote by twotaileddemon
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
For part c) you were told to assume the motion is horizontal. The horizontal motion is circular motion and g is not a factor in this motion. The tension is the force that is making the mass move in a circle with the stated speed of 3.7m/s.
twotaileddemon
#7
Nov18-06, 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.
Doc Al
#8
Nov18-06, 06:54 AM
Mentor
Doc Al's Avatar
P: 41,325
Quote Quote by twotaileddemon
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?
That's all there is to it.

And I would calculate for d accordingly?
Right.
Doc Al
#9
Nov18-06, 07:13 AM
Mentor
Doc Al's Avatar
P: 41,325
Quote Quote by twotaileddemon
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...
If "south" represents down, then it seems that you have the cord angling down from the mass. Remember that the vertical component of the cord tension must balance the weight of the mass. Since weight acts down, the vertical component of tension must act up. Therefore the cord--and tension--must angle up and towards the axis of rotation from the mass. (It's not clear what you are calling east and west.)
twotaileddemon
#10
Nov18-06, 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.
Doc Al
#11
Nov18-06, 03:48 PM
Mentor
Doc Al's Avatar
P: 41,325
Quote Quote by twotaileddemon
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.
The tension--and thus cord--must be at a northwest angle to provide the centripetal force and balance the weight.


And judging by the picture and my answer for part eiii, the angle looks to be about 85 degrees or so.
You might want to redo that calculation. (Check your value for ma_c.) If the angle were 85 degrees, realize that the cord would be practically vertical.
twotaileddemon
#12
Nov18-06, 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*
Doc Al
#13
Nov18-06, 08:27 PM
Mentor
Doc Al's Avatar
P: 41,325
That's more like it.
twotaileddemon
#14
Nov18-06, 08:55 PM
P: 263
Thanks ^_^! I feel like I'm really getting it :)


Register to reply

Related Discussions
Questions about circular motion/ centripetal acceleration/force Introductory Physics Homework 5
Always centripetal force when object in circular motion? Introductory Physics Homework 10
Uniform circular motion - centripetal force and banked curves Introductory Physics Homework 3
Challenging physics problems.....uniform circular motion, centripetal force Introductory Physics Homework 1
Uniform Circular Motion-Centripetal Force Introductory Physics Homework 1