Centripetal force of an object swinging

[JWest]

I have to find the centripetal force and velocity of an object being swung around on a string that is attached to a weight (starting at 200 grams up to 1100 grams). The length of the string (radius) that is swinging the object is .5 meters. I have the times for each mass (200g-1100g). How would I approach this problem? To find the velocity, would I use the equation V=d/t, where "d" is the .5 meters and "t" is the time it takes for the object to be swung around once, or should I use the equation Fc=m(v^2/r), where "Fc" is the weight (Fg), mass is the weight, and radius is the string distance of .5 meters? Plus, I need to make 2 graphs the data where "Fc" is the y-axis and "v" is the x-axis and it has to be a parabola and "Fc" is the y-axis and "v^2" is the x-axis and it has to be a straight line like this - /. Can anyone help me?

 

[apchemstudent]

I have to find the centripetal force and velocity of an object being swung around on a string that is attached to a weight (starting at 200 grams up to 1100 grams). The length of the string (radius) that is swinging the object is .5 meters. I have the times for each mass (200g-1100g). How would I approach this problem? To find the velocity, would I use the equation V=d/t, where "d" is the .5 meters and "t" is the time it takes for the object to be swung around once, or should I use the equation Fc=m(v^2/r), where "Fc" is the weight (Fg), mass is the weight, and radius is the string distance of .5 meters? Plus, I need to make 2 graphs the data where "Fc" is the y-axis and "v" is the x-axis and it has to be a parabola and "Fc" is the y-axis and "v^2" is the x-axis and it has to be a straight line like this - /. Can anyone help me?

You can figure out V from the circumference of the circle divided by the amount of time it took the object to go around once. From there it's pretty much straightforward.

The graph being a parabola also makes perfect sense since the F(v) of Fc is v^2/r where 1/r is a constant meaning the vertical stretch.

The graph is a line when F(v^2) of Fc since the slope will be 1/r and again is a constant, which implies that the graph is a line.

 

[JWest]

And v^2 is just the velocity squared? I didn't get the parabola for the Fc vs. V graph.

 

[apchemstudent]

And v^2 is just the velocity squared? I didn't get the parabola for the Fc vs. V graph.

You should... the graph is suppose to be F(v) = v^2/.5 where F(v) = Fc

Say v = 1 then F(c) = 2
...2 ......8
...3......18
...-1......2
...-2.....8
...-3......18

If you continue in this fashion and plot the points, it will be a parabolic. Even graph this on a graphing calculator with x^2/.5...