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Conical Pendulum Problem -Right Way of Solving?

by webren
Tags: conical, pendulum, solving
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webren
#1
Jul15-06, 05:07 PM
P: 34
Hello,
I wanted to make sure that I am solving this problem in the correct manner, because one of my answers seems a little off from the book's.

"Consider a conical pendulum with an 80.0-kg bob on a 10.0-m wire making an angle of 5.00(degrees) with the veritcal. Determine (a) the horizontal and vertical components of the force exerted by the wire on the pendulum and (b) the radial acceleration on the bob."

Here are my steps in solving this problem:

I drew a free-body diagram of the bob and the string and noticed immediately that above the bob (the string) is tension, and that tension is causing the centripetal force, and the bottom of the bob is mg (weight). I realized that a right triangle could be drawn in this scenario. The hypotenuse is equal to 10 m, because that's the length of the wire.

I broke down the scenario into x (or r) and y force components.

Fx(or Fr) = T(sin5) = m(v^2/r)
Fy = T(cos5) - mg = 0

From here, I went ahead and solved for T. So T(cos5) = mg, which equals 787 N. After that, I plugged the T value into the Fx(or Fr) equation and got 68.6 N. The book agrees with me that the x component of the force is 68.6 N, but it says for 784 N for the y component. My answer is three Newtons off. That seems too much?

In solving for (b), I divided the two force equations together, which cancelled out m and converted the sin and cos into a single tan. I knew I would need to know the values for velocity, and the radius to continue. First, I solved for the radius using my right triangle and basic geometry to realize that the radius = the hypotenuse multiplied by sin5. Because the hypotenuse = 10 m, my radius = 0.872 m. After that, I solved for velocity, which = 0.864 m/s. To find the radial acceleration, I divided the two (with velocity being squared) and got 0.856 m/s^2. The book agrees with me there as well. To check my radial acceleration answer, I made sure that it was equal to Tsin5, and it is.

Is that the correct way in solving this kind of problem? Any opinions or suggestions would be great. Also, is it acceptable to be off by three Newtons in the answer above? Thank you.
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nrqed
#2
Jul15-06, 05:52 PM
Sci Advisor
HW Helper
P: 2,957
Quote Quote by webren
Hello,
I wanted to make sure that I am solving this problem in the correct manner, because one of my answers seems a little off from the book's.

"Consider a conical pendulum with an 80.0-kg bob on a 10.0-m wire making an angle of 5.00(degrees) with the veritcal. Determine (a) the horizontal and vertical components of the force exerted by the wire on the pendulum and (b) the radial acceleration on the bob."

Here are my steps in solving this problem:

I drew a free-body diagram of the bob and the string and noticed immediately that above the bob (the string) is tension, and that tension is causing the centripetal force, and the bottom of the bob is mg (weight). I realized that a right triangle could be drawn in this scenario. The hypotenuse is equal to 10 m, because that's the length of the wire.

I broke down the scenario into x (or r) and y force components.

Fx(or Fr) = T(sin5) = m(v^2/r)
Fy = T(cos5) - mg = 0

From here, I went ahead and solved for T. So T(cos5) = mg, which equals 787 N.
Just a comment: the way you wrote it is confusing because it sounds as if you are saying that T cos(5) is 787 N, when you mean that T is 787 N.
After that, I plugged the T value into the Fx(or Fr) equation and got 68.6 N. The book agrees with me that the x component of the force is 68.6 N, but it says for 784 N for the y component. My answer is three Newtons off. That seems too much?
what your 787 N represents is the tension which is NOT vertical. They want the vertical component of the force exterted by the wire, so they want the y component of the tension, which is T cos(5).

In solving for (b), I divided the two force equations together, which cancelled out m and converted the sin and cos into a single tan. I knew I would need to know the values for velocity, and the radius to continue. First, I solved for the radius using my right triangle and basic geometry to realize that the radius = the hypotenuse multiplied by sin5. Because the hypotenuse = 10 m, my radius = 0.872 m. After that, I solved for velocity, which = 0.864 m/s. To find the radial acceleration, I divided the two (with velocity being squared) and got 0.856 m/s^2. The book agrees with me there as well. To check my radial acceleration answer, I made sure that it was equal to Tsin5, and it is.
In the last line, you mean that the mass times the radial acceleration gives T sin (5), right?

You did it right except that you could have done it much much faster! They did not ask about the speed or the radius, just the radial acceleration, so you could have used [itex] T_x = T sin(5)= m a_r [/itex] and just divide Tsin(5) by the mass and be done!!
webren
#3
Jul15-06, 06:14 PM
P: 34
Ahh, okay. I thought that the tension was the y component the problem was talking about. Knowing that, the force of the y component is 784, which agrees with the book. Thanks for clearing that up, and for showing me a quicker way of solving.

nrqed
#4
Jul15-06, 06:19 PM
Sci Advisor
HW Helper
P: 2,957
Conical Pendulum Problem -Right Way of Solving?

Quote Quote by webren
Ahh, okay. I thought that the tension was the y component the problem was talking about. Knowing that, the force of the y component is 784, which agrees with the book. Thanks for clearing that up, and for showing me a quicker way of solving.
You are very welcome.
And good for you, you did all the work by yourself!

Patrick
fabbo
#5
Sep7-06, 01:44 PM
P: 31
looking at a similar problem but a bit confused on how webren gets
Tsinθ = m (v^2/r)
?


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