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.

###### Solution.

a)

b)

According to Newton’s Second Law for uniform circular motion, the net force acting on the ball equals ma_{c}.

F_{net} = ma_{c}

The expression F_{net} = ma_{c} is a vector equation so we can write it as two component equations: F_{net,x} = ma_{c} and F_{net,y} = 0, because the ball is not accelerating vertically.

In the x-direction, there is only F_{T,x}. Thus,

F_{T,x} = ma_{c}

or

F_{T}∙sin(α) = ma_{c} (1).

In the y-direction, F_{net,y} = 0 becomes

F_{T,y} – mg = 0

or

F_{T}∙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 = 30^{0}

We substitute the values into the equation (2) and solve for F_{T}

F_{T}∙cos(30^{0}) = 0.2∙9.8

F_{T} ≈ 2.26 N

c)

We know that

a_{c} = 4π^{2}r/T^{2} (3)

We substitute the equation (3) into the equation (1) and we get

F_{T}∙sin(α) = m∙4π^{2}r/T^{2} (4)

We can divide the equation (4) by equation (2) and obtain

tan(α) = 4π^{2}r/(g∙T^{2})

We can rewrite the last equation as

T = √(4π^{2}r/(g∙tan(α))

and finally, we substitute the values and solve for T

T = √(4π^{2}∙0.5/(9.8∙tan(30^{0}))

T ≈ 1.87 s