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How to find the tangential velocity of an object

Asked Feb 24, 2010, 02:21 PM — 6 Answers
The centripetal force on a 0.82 kg object on the end of a 2.0 massless string being swung in a horizontal circle is 4.0 N. What is the tangential velocity of the object?

The answer I got is 3.1 m/s squared but I don't know how to show the work for it. So please help me.

6 Answers
ebaines's Avatar
ebaines Posts: 10,055, Reputation: 5539
Expert
 
#2

Feb 24, 2010, 02:47 PM
Combine two concepts:

1. F = ma
2. For objects in circular motion with constant tangential velocity:
A = v^2/R

So:

F = ma = mv^2/R for objects in circular motion with constant tangential velocity

You now have everyting you need to deermine the value of v.
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MS_SUMTER2010's Avatar
MS_SUMTER2010 Posts: 55, Reputation: 10
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#3

Feb 24, 2010, 02:56 PM
So it will be this:

4=v sqaured/ 2
Then I get 2.8.

But when I plug it back in the equation v^2/r.. I get 3.9
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Unknown008's Avatar
Unknown008 Posts: 8,147, Reputation: 3745
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#4

Feb 25, 2010, 02:10 AM
You didn't read ebaines post well. You combine F=ma and a = v^2/r.

Then you get:



From there, you 'll get your answer as 3.1 m/s^2 (2 sf)
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MS_SUMTER2010's Avatar
MS_SUMTER2010 Posts: 55, Reputation: 10
Junior Member
 
#5

Feb 25, 2010, 04:51 AM
Ok but what is the v squared part? Kuz it doesn't really say. I know the mass and the radius I just don't know how to get the velocity
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ebaines's Avatar
ebaines Posts: 10,055, Reputation: 5539
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#6

Feb 25, 2010, 06:11 AM
Starting with the formula we gave you:



Raarrange to get by itself:



And take the square root of both sides:



You have valius for and , so you can solve for . Your answer will in units of meters/sec (NOT meters per second squared, as you had in your original post)
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IWishICouldFly's Avatar
IWishICouldFly Posts: 1, Reputation: 1
New Member
 
#7

Aug 23, 2010, 02:59 PM
Quote:
Originally Posted by ebaines View Post
Starting with the formula we gave you:



Raarrange to get by itself:



And take the square root of both sides:



You have valius for and , so you can solve for . Your answer will in units of meters/sec (NOT meters per second squared, as you had in your original post)
Thank you so much I was looking for this
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