A ball of mass 0.2 kg is attached to a massless string 1 m long and swung so that it travels in a horizontal circle of radius 0.5 m.
a) Draw a free-body diagram.
b) Find the force of tension in the string as the ball swings in a horizontal circle.
c) Find the period of the ball’s motion.
According to Newton’s Second Law for uniform circular motion, the net force acting on the ball equals mac.
Fnet = mac
The expression Fnet = mac is a vector equation so we can write it as two component equations: Fnet,x = mac and Fnet,y = 0, because the ball is not accelerating vertically.
In the x-direction, there is only FT,x. Thus,
FT,x = mac
FT∙sin(α) = mac (1).
In the y-direction, Fnet,y = 0 becomes
FT,y – mg = 0
FT∙cos(α) = mg (2).
We can determine α by using the dimensions of the string and the circle radius.
sin(α) = r/l
sin(α) = 0.5/1
α = 300
We substitute the values into the equation (2) and solve for FT
FT∙cos(300) = 0.2∙9.8
FT ≈ 2.26 N
We know that
ac = 4π2r/T2 (3)
We substitute the equation (3) into the equation (1) and we get
FT∙sin(α) = m∙4π2r/T2 (4)
We can divide the equation (4) by equation (2) and obtain
tan(α) = 4π2r/(g∙T2)
We can rewrite the last equation as
T = √(4π2r/(g∙tan(α))
and finally, we substitute the values and solve for T
T = √(4π2∙0.5/(9.8∙tan(300))
T ≈ 1.87 s