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twotaileddemon

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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

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