# Homework Help: Acceleration of a revolving ball

1. Aug 14, 2009

### bumblebeeliz

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

T = 1/f
v = 2$$\pi$$r / T

3. 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$$\pi$$r / T

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

How did we get the 0.500s?

2. Aug 14, 2009

### RoyalCat

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:
$$f=T^{-1}=\tfrac{1}{T}$$

3. Aug 14, 2009

### tms

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?

4. Aug 14, 2009

### Delphi51

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

5. Aug 14, 2009

### bumblebeeliz

so basically:

f = 2.00s -1 = 0.500s.

I think the extra zero's confused me.

6. Aug 14, 2009

### tms

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.

They shouldn't. You should always keep the right number of significant digits in your calculations.

7. Aug 15, 2009

### RoyalCat

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.