Horizontal centripetal question?

Question

You rotate a 450 g ball on the end of a string in a horizontal circle of radius 2.5 m. The ball completes eight rotations in 2.0s. What is the centripetal force of the string on the ball?

I know im suppose to use the equation Fc=mv^2/r

but how do i find the v?

Answer 1

if you know the distance the ball traveled in those 2.0 seconds, you can calculate v. The circumference of the circle is 2*pi*r, with r the radius, in meters. The distance traveled is then the number of rotations (let's call it n) =n*2*pi*r. Divide the distance by the time it took to travel the distance (let's call it t), and you get :

v = n*2*pi*r/t

With n=8, r=2.5m and t=2s you get

v = 8*2*pi*2.5/2 = 8*2.5*pi = 20*pi [m/s]

Remember to use m in kg when you fill in the formula for Fc

Answer 2

If the ball completes 8 rotation in two seconds, than it run for a space of 8*(2*pi*r) in 2.0s. You have r and pi (3.141592...)

So v is 8*2*pi*r/2.0

Alternate answer: there is a thing called angular velocity, w, defined as how much angle you run in a given time

the alternate formula is Fc= mw^2r. Try it with the angular velocity you got (2*pi*8)/2.0s) and you'll see they're the same

Cheers,

D

Answer 3

to use the other equation, Fc=mrW *w being angular velocity, you need to convert your 8 rotations in 2 seconds to radians per second.  1 rotation is equal to 2pi radians so just multiply 8 by 2(3.14) to get radians and then divide by 2 so that is per second instead of two seconds.  Now with your angular velocity you can multiply it by the radius and the mass to get the centripetal force.