Acceleration of a revolving ball

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Homework Help Overview

The problem involves a 150-g ball revolving uniformly in a horizontal circle with a radius of 0.600 m, completing 2.00 revolutions per second. The focus is on determining the centripetal acceleration and understanding the relationship between period and frequency.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the period (T) from the frequency (f) and how it relates to the time taken for one revolution. There is confusion regarding the conversion between frequency and period, particularly in the context of significant figures.

Discussion Status

Participants are exploring the relationship between frequency and period, with some providing insights into the correct expressions for these concepts. There is acknowledgment of confusion regarding numerical values and significant digits, but no consensus has been reached on the interpretation of the calculations.

Contextual Notes

Some participants note the importance of maintaining significant digits in calculations and question the clarity of the frequency and period definitions. The discussion reflects a mix of understanding and confusion regarding these fundamental concepts.

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Homework Statement



A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600m. The ball makes 2.00 revolutions in a second. What is the centripetal acceleration?

Homework Equations



T = 1/f
v = 2\pir / T


The Attempt at a Solution



I know the answer to this question because its an exercise but I am confused with the "T" equation (Period equation).

v = 2\pir / T

v = 2(3.14)(0.600m) / (0.500s)

How did we get the 0.500s?
 
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bumblebeeliz said:

Homework Statement



A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600m. The ball makes 2.00 revolutions in a second. What is the centripetal acceleration?

Homework Equations



T = 1/f
v = 2\pir / T


The Attempt at a Solution



I know the answer to this question because its an exercise but I am confused with the "T" equation (Period equation).

v = 2\pir / T

v = 2(3.14)(0.600m) / (0.500s)

How did we get the 0.500s?
If in one second, the mass completes two revolutions, how long does it take for it to complete one revolution?

The relationship between the frequency (How many occurrences per unit of time) and the period (How long it is between one occurrence and the next) is:
f=T^{-1}=\tfrac{1}{T}
 
From the statement of the problem. If the ball makes 2 revolutions in one second, how much time does it take to make one revolution?
 
F = ma = m*v^2/R
cancel the m's to get a formula for acceleration.
There is an alternative one where v is replaced by 2*pi*R/T
 
so basically:

f = 2.00s -1 = 0.500s.

I think the extra zero's confused me.
 
bumblebeeliz said:
so basically:

f = 2.00s -1 = 0.500s.

You got the right answer numerically, but you got it by confusing the frequency. The correct expression for the frequency is

f = 2.00 \: \frac{1}{\text{sec}},

while you seem to have used

f = \frac{1}{2.00 \:\text{sec}}

and then fudged the units in the result.

I think the extra zero's confused me.

They shouldn't. You should always keep the right number of significant digits in your calculations.
 
bumblebeeliz said:
so basically:

f = 2.00s -1 = 0.500s.

I think the extra zero's confused me.

Like tms said, the dimensions for frequency are [T]^{-1}, and its units in the MKS system are \tfrac{1}{sec}\equiv 1 Hz
1 Hz (1 Hertz) is defined as a frequency of 1 per second.
 

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