Question

A tetherball set has a ball with mass 0.411 kg and a nylon string with diameter 2.50 mm, Young's modulus 4.00 GPa, and density 1150 kg/m3. The nylon string has a length of 2.200 m when the ball is at rest (hanging straight down). While playing tetherball, Monty hits the ball around the pole so it moves in a horizontal circle with the string at an angle of 65.0° to the pole. 

(a) How much does the string stretch compared with when the ball is at rest?

(b) What is the ball's kinetic energy? 

(c) How long would it take a transverse wave pulse to travel the length of the string from the ball to the top of the pole?

 

Answer 1

(a)

The free-body diagram is shown below:

 

The amount of length of the nylon string stretch when the ball is at rest is

ΔL = mgL/πr2Y

 

Here, m is mass of the weight at the other end of the string, g is acceleration due to gravity, L is length of the string, r is radius of the string, and Y is Young’s modulus of the material of the string.

 

At point C, the net force acting on the tetherball is,

F = mgcos(65.0°) + mv2/L

 

Now, the change in the potential energy from B to C is equal to the kinetic energy of the tetherball.

ΔU = K

mgL[1 – cos(65°)] – 1/2 mv2

v2 = 2gL[1 – cos(65°)]

 

Substitute the value of v2 in the equation F = mgcos(65.0°) + mv2/L.

F = mg cos(65.0°) + 2mgL[1 – cos(65.0°)]/L

   = mg [3cos(65.0°) + 2]

   = 3.2679mg

 

The amount of length of the nylon string stretch when the ball is at C is,

ΔLC = FL/πr2y

 

Substitute the value 3.2679 mg for F in equation.

ΔLC = FL/πr2Y

        = (3.2679 mg)L/πr2Y

 

Now consider the difference of ΔLC and ΔL.

ΔLC – ΔL = (3.2679mg)L/πr2Y – (mg)L/πr2Y

                = (2.2679)(mg)L/πr2Y

 

Substitute 0.411 kg for m, 1.25 mm for r, 9.8 m/s2 for g, 2.200 m for L, and 4.00 × 109 Pa for Y in the equation.

ΔLC – ΔL = (3.2679)(mg)L/πr2Y

             = [(2.2679) (0.411 kg)(9.8 m/s2)(2.200 m)]/[π(1.25 mm)2(10-3 m/1 mm)2(4.00 × 109 Pa)(1 kg/m∙s2/1 Pa)]

             = 1.02 × 10-3 m

 

Therefore, the string is stretched by an amount more than 1.02 × 10-3 m when it is at rest.

 

(b)

The kinetic energy of the ball is,

K = mgL[1 – cos(65.0°)]

 

Substitute 0.411 kg for m, 9.8 m/s2 for g, and 2.200 m for L in the equation.

 

K = mgL[1 – cos(65.0°)]

    = (0.411 kg)(9.8 m/s2)(2.200 m) [1 – cos(65.0°)]

    = (5.1163 kg∙m2/s2)(1 J/1 kg∙m2/s2)

    = 5.116 J

 

Round off to three significant figures, the kinetic energy of the ball, when it is in motion is 5.12 J.

 

(c)

The speed of the transverse wave in the string is,

v = √Y/ρ

 

Here, ρ is density of the material of the string.

 

Substitute 4.00 × 109 Pa for Y and 1150 kg/m3 for ρ in the equation v = √Y/ρ.

v = √Y/ρ

   = √(4.00 × 109 Pa)(1kg/m∙s2/1 Pa)/1150 kg/m3

   = 1.865 × 103 m/s

 

The time taken by the wave n the transverse wave to travel the length of the string from the ball to the top of the pole is,

t = L/v

 

Substitute 2.200 m for L and 1.865 × 103 m/s for v in the above equation.

t = L/V

  = 2.200 m/1.865 × 103 m/s

  = 1.179 × 10-3 s

 

Round off to three significant figures, the time taken the transverse waves in the string from the ball to the top of the pole is 1.18 × 10-3 s.