Calculating Revolutions Per Minute of a Ball

[Kajayacht]

Homework Statement

A boy whirls a ball on a string in a horizontal circle of radius 1.0 m. How many revolutions per minute must the ball make if its acceleration towards the center of the circle is to have the same magnitude as the acceleration due to gravity?

Homework Equations

a=v^2/R
V={omega}R
{omega}=2pi*f

The Attempt at a Solution

9.8/1= v^2
v= 3.13

{omega}=v/R=3.13
f=3.13/2pi
f= 4.92

 

[alphysicist]

Hi Kajayacht,

Homework Statement

A boy whirls a ball on a string in a horizontal circle of radius 1.0 m. How many revolutions per minute must the ball make if its acceleration towards the center of the circle is to have the same magnitude as the acceleration due to gravity?

Homework Equations

a=v^2/R
V={omega}R
{omega}=2pi*f

The Attempt at a Solution

9.8/1= v^2

This gives the right answer for v (since R=1), but of course from your equation above it needs to be v^2 = a R ( not divided by R).

v= 3.13

{omega}=v/R=3.13
f=3.13/2pi
f= 4.92

I think you probably have some calculator problems here. The numerator is about 3, and the denominator is about 6, so your answer should be around 1/2.

It looks like you calculated 3.13 / 2 * pi. If you type it in like that without parenthesis around the 2pi it multiplies by pi instead of dividing by pi. (So you might try entering 3.13/(2pi), for example.)

Also, you forgot to list your units for f. What are they?

 

[Kajayacht]

Ahh wow yes of course
well now I got .498 rev/min but it's still saying that I'm wrong

it also says "Consider the acceleration due to circular motion."

 

[alphysicist]

Ahh wow yes of course
well now I got .498 rev/min but it's still saying that I'm wrong

it also says "Consider the acceleration due to circular motion."

They are asking for the frequency in rev/min, but I don't think you yet have those units. What units is the answer 0.498 in? (Look back at the quantities you put into the problem, and see which one has a time unit with it.)