Centripetal Acceleration while Swinging a Stone on a String

AI Thread Summary
The discussion centers on calculating centripetal acceleration while swinging a stone on a string at a frequency of 5 revolutions per second. The user correctly converts this frequency to angular velocity, resulting in 10π rad/s. There is clarification on the distinction between frequency and angular frequency, emphasizing that frequency is revolutions per unit time while angular frequency is radians per unit time. The user successfully applies the formulas for tangential velocity and centripetal acceleration, ultimately confirming that the calculations yield a centripetal acceleration of 40 m/s². The conversation concludes with the user expressing understanding of their earlier mistakes.
singh101
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Homework Statement
I can't seem to figure out the answer for this question with formulas given.
Relevant Equations
Angle of rot= arc length/ radius
Angular velocity= angle of rotation/ time
v=rw (v=tangential veloctity) (w=angular velocity) r= radius
centripe acceleration= v^2/r
So my initial understanding is that it completes 5 revolutions per second. I converted the 5 rev to radians, so each revolution is 2pi. Now since I got angle of rotation I can plug it into the angular velocity formula which is Angular velocity= angle of rotation/ time. However since I don't have time I can change the formula to angle of rot x frequency. Frequency in this case would be 5. So ang Velocity is 10pi rad/s. From their on I get the tang velocity and plug it into the centrip acceleration formula, however I keep getting the wrong answer. Is there anyway to solve this with the equation given below or am I missing any equations.
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singh101 said:
So ang Velocity is 10pi rad/s.
Right so far.
singh101 said:
From their on I get the tang velocity and plug it into the centrip acceleration formula, however I keep getting the wrong answer.
I cannot tell where you went wrong if you do not show your detailed working.
 
my question is that is 2pi the angle of rotation or is that the angular velocity.
 
Oh wait I think I figured out my mistake. The 5 revolutions per second is the frequency am I correct? How would that be stated as a period
 
singh101 said:
my question is that is 2pi the angle of rotation or is that the angular velocity.
There is no specific angle of rotation in the question, only a rate: 5 revs/s ##=10\pi^c/s##, where the c denotes radians.
 
singh101 said:
Oh wait I think I figured out my mistake. The 5 revolutions per second is the frequency am I correct? How would that be stated as a period
The rotation period is the time to complete one revolution.
 
singh101 said:
so is the 5 revolution per second the frequency or period. I am assuming that is the frequency. If it is the frequency then what would the period be stated as if it were given in the question.
5 revs/s is the rotation rate. "Frequency" needs to be distinguished from "angular frequency ". The frequency is the number of revolutions per unit time, 5 per second, whereas angular frequency is the number of radians per unit time.
In equations like ##v=\omega r## and ##a=\omega^2r## use the angular frequency.
 
oh ok so 5rev/s can be converted into rad by multiplying it by 2pi which gives is 10pi rad/s. This then in added to the formula v=rw. so w is 10pi x 0.4. which gives us 4pi. This is then put into the formula v^2/r (4pi)^2/0.4= 40m/s^2. Am I correct?
 
singh101 said:
oh ok so 5rev/s can be converted into rad by multiplying it by 2pi which gives is 10pi rad/s. This then in added to the formula v=rw. so w is 10pi x 0.4. which gives us 4pi. This is then put into the formula v^2/r (4pi)^2/0.4= 40m/s^2. Am I correct?
Yes. Or you can use ##\omega^2r## for the centripetal acceleration instead, skipping finding v.
 
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Oh ok thank you so much I understand where I made the mistake.
 
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