Centripetal Force Ball Swing Question

AI Thread Summary
An athlete swings a ball on a chain, achieving different rotation rates at varying chain lengths. The rates of 8.16 rev/s and 6.32 rev/s represent angular frequencies, not periods, which is crucial for calculations. To find linear speed, the angular speed should be multiplied by the radius, and the centripetal acceleration can be calculated using the formula a[c] = ω²r. It's emphasized that converting revolutions per second to radians per second is necessary for accurate results. Understanding these concepts is essential for solving the homework problems correctly.
tcc88
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Homework Statement


An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 8.16 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.32 rev/s.

(a) Which rate of rotation gives the greater speed for the ball? [Which I got is 6.32]
(b) What is the centripetal acceleration of the ball at 8.16 rev/s? [?]
(c) What is the centripetal acceleration at 6.32 rev/s? [?]

Homework Equations


a[c] = v^2/r
T = 2∏r/v

The Attempt at a Solution



I know the I need to find the velocity to solve this question, but that is where I am having trouble. I am assuming 8.16 and 6.32 are the periods. So I get:
v = 2∏(0.300)/(8.16) for B; and then use this to find acceleration which is: a[c] = (0.23)^2/(0.300) = 1.176. But this is not the correct answer! Why?
 
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tcc88 said:

Homework Statement


An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 8.16 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.32 rev/s.

(a) Which rate of rotation gives the greater speed for the ball? [Which I got is 6.32]
(b) What is the centripetal acceleration of the ball at 8.16 rev/s? [?]
(c) What is the centripetal acceleration at 6.32 rev/s? [?]

Homework Equations


a[c] = v^2/r
T = 2∏r/v

The Attempt at a Solution



I know the I need to find the velocity to solve this question, but that is where I am having trouble. I am assuming 8.16 and 6.32 are the periods. So I get:
v = 2∏(0.300)/(8.16) for B; and then use this to find acceleration which is: a[c] = (0.23)^2/(0.300) = 1.176. But this is not the correct answer! Why?

I think your answers for both part a and part b are wrong, and here's why:

The "rev/s: in 6.32 rev/s and 8.16 rev/s stands for "revolutions PER second." So it is most emphatically NOT a period. It tells you how many revolutions occur in one second. So, you might say, it tells you how often a revolution occurs. In fact, if you were inclined to phrase it differently, you might say that this number tells you how frequently a revolution occurs.

:wink: :wink: nudge nudge
 
cepheid said:
I think your answers for both part a and part b are wrong, and here's why:

The "rev/s: in 6.32 rev/s and 8.16 rev/s stands for "revolutions PER second." So it is most emphatically NOT a period. It tells you how many revolutions occur in one second. So, you might say, it tells you how often a revolution occurs. In fact, if you were inclined to phrase it differently, you might say that this number tells you how frequently a revolution occurs.

:wink: :wink: nudge nudge

So would the frequency be 1/8.16 [Which I am leaning somewhat more towards] or just 1 [Which if it is, can you tell me why]? And if not either, please just tell me the answer... :(
 
I am using 1/6.32 and 1/8.16 and I am still getting the wrong answer?? o.O WTH is going on?!?
 
tcc88 said:
I am using 1/6.32 and 1/8.16 and I am still getting the wrong answer?? o.O WTH is going on?!?

The thing I was trying to hint at very strongly was that the 6.32 and 8.16 ARE the angular frequencies. I even said that they tell you how OFTEN or FREQUENTLY a revolution occurs.

The angular frequency is also the angular speed in this case. Do you know how to find the linear speed given the angular speed?
 
cepheid said:
The thing I was trying to hint at very strongly was that the 6.32 and 8.16 ARE the angular frequencies. I even said that they tell you how OFTEN or FREQUENTLY a revolution occurs.

The angular frequency is also the angular speed in this case. Do you know how to find the linear speed given the angular speed?

Would the Linear speed be rps[angular speed or frequency] * pi * d? Also will I need both to find the circular acceleration or just one? I legit don't remember my professor teaching this, but I am willing to learn it if it means getting my h.w in on time! :/
 
No, the linear speed is the angular speed multiplied by r (the radius). The reason for this comes from the definition of an angle (in radians). The angle is the arc length you travel around the circle divided by the radius. Linear speed would be arc length/time, and if you divided this by the radius you'd get angle/time, which is angular speed. Make sense?

If you went a full revolution, the distance (arc length) would be 2*pi*r (a full circumference) and so the angle would be this divided by r, which would be 2pi radians.
 
You have to be careful to convert the angular speed from rev/sec to radians/sec before working with these formulas. As I said above, one revolution is 2*pi radians.
 
Just to follow up on what Cephid has said, it is easier and less confusing to use ac2r rather than converting to tangential speed. ω=2πf where f is the frequency of rotation = 1/period of rotation (f=1/T).

AM
 
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