Uniform centripetal acceleration

[Panix]

Homework Statement

A small ball with a mass of 3.0kg revolves uniformly in a circle on a frictionless horizontal surface. The ball is attached to a string of 1.40m to a pin set at center of circle. The object makes 2rev/sec.

a)find the centripetal accel
b)find the tension exerted on the string

Homework Equations

ac=v^2/r
Fc= ma=m(4pi^2r/T^2)
T= .5

The Attempt at a Solution

if its uniform wouldn't the Fc=FT(force of tension)
T would =.5 because 2 rev/1 sec?
I am not sure how to find centripetal accel because i do not know how to find the velocity unless i am missing something..

 

[kuruman]

If it makes one revolution and the string is 1.40 m, how much distance does it cover?

 

[kerimek]

Firstly, to find the centripetal acceleration, we can use the equation ac = v^2/r, where v is the velocity and r is the radius of the circle. In this case, v can be found by dividing the number of revolutions (2 rev/sec) by the time period of one revolution (T= 0.5 sec), giving us a velocity of 4 m/s. Plugging this into the equation, we get ac = (4 m/s)^2 / 1.4m = 11.43 m/s^2.

To find the tension exerted on the string, we can use the equation Fc = ma, where m is the mass of the ball and a is the centripetal acceleration we just calculated. Plugging in the values, we get Fc = (3.0 kg)(11.43 m/s^2) = 34.3 N.

In summary, the centripetal acceleration of the ball is 11.43 m/s^2 and the tension exerted on the string is 34.3 N. It is important to note that these values are only accurate if the ball is truly revolving uniformly and there is no friction present.