A ball is attached to a string and swung so that it travels in a horizontal circle

ballstringswung.jpgA 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


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