16m

A family decides to create a tire swing in their backyard for their son Ryan. They tie a nylon rope to a branch that is located 16 m above the earth, and adjust it so that the tire swings 1 meter above the ground. To make the swing more exciting, and so they don't have to push Ryan all the time, they construct a launch point that is 13 m above

the ground. You are their neighbor, and you are concerned that the swing might not be safe, so you calculate the maximum tension (Tmax) in the rope to see if it will hold.

Calculate the maximum tension in the rope, assuming that Ryan starts from rest from his launch pad. Is it greater than the rated value of the rope (1500 N)?

y

x

16m

1m

h0=13m

Ryan’s mass:

mR=30kg

Mass of tire:

mT=5kg

Neglect the mass of the rope.

Ryan’s mass:

mR=30kg

Mass of tire:

mT=5kg

Neglect the mass of the rope.

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Reflection

1. At what point of the swinging motion will the tension in the rope be at it’s

maximum.

A) On the way down

B) At the lowest point

C) On the way back up

At the bottom of the trajectory, all forces acting on Ryan are vertical. The sum of the tension (up) and weight (down) produce a centripetal force (up).

The more the force of weight pulls down on the rope, the greater the tension will be. Only a component of the force of weight pulls on the rope on the way up or down.

Choice AIncorrect

Choice BCorrect

€

T

€

θ

yx

€

T

€

W

At the bottom of the trajectory, all forces acting on Ryan are

vertical. The sum of the tension (up) and weight (down)

produce a centripetal force (up).

The more the force of weight pulls down on the rope, the greater the tension will be.

Only a component of the force of weight pulls on the rope on

the way up or down.

The tension is at a maximum at the bottom of the trajectory.

At the bottom of the trajectory, all forces acting on Ryan are

vertical. The sum of the tension (up) and weight (down)

produce a centripetal force (up).

The more the force of weight pulls down on the rope, the greater the tension will be.

Only a component of the force of weight pulls on the rope on

the way up or down.

The tension is at a maximum at the bottom of the trajectory.

€

W

Choice CIncorrect

At the bottom of the trajectory, all forces acting on Ryan are vertical. The sum of the tension (up) and weight (down) produce a centripetal force (up).

The more the force of weight pulls down on the rope, the greater the tension will be. Only a component of the force of weight pulls on the rope on the way up or down.

2. Which physics principle should we use to solve this problem.

A) Conservation of Mechanical Energy

B) Newton’s 2nd Law

C) Both A and B

Choice A

It’s a good choice, but only part of the story.

Since we know that the maximum tension is when the swing reaches the bottom of its trajectory, we can use the conservation of energy law to

find the speed of the tire at this point.

But, once we find the speed we will need to use Newton’s 2nd Law to find the solution.

€

Initial Energy = Final Energy

We need to use Newton’s 2nd Law, but we must also apply the law of conservation of energy to find important information first.

Choice B

It’s a good choice, but only part of the story.

€

Fx =0 and ∑ Fy = ma ∑

There are no forces acting on the tire in the x-direction at the

instant when the tire is at the bottom of its trajectory.

There are no forces acting on the tire in the x-direction at the

instant when the tire is at the bottom of its trajectory.

Choice CCorrect

We can use the conservation of energy law to find the speed of the tire at the point of

maximum tension:

Once we find the speed we will need to use the vertical component of Newton’s 2nd Law to

find the solution:

€

Initial Energy = Final Energy

€

Fy =ma ∑

3. Applying the law of conservation of energy, which of the following simplified expressions

do we find for Ryan’s speed when the swing is at its lowest point.

A)

B)

C)

€

VB = 2ghf

VB= Ryan’s speed at the bottom

Subscripts 0 and f stand for initial and final.

VB= Ryan’s speed at the bottom

Subscripts 0 and f stand for initial and final.

€

VB = 2g h0 −hf( )

€

VB = 2gh0

Choice AIncorrect

You must take the initial gravitational energy into consideration. Ryan swings

from rest atop the launch pad.

Please try again.

Choice BCorrect

Setting the initial energy equal to the final energy we find:

€

mR +mT( )gh0 =12

mR +mT( )vB2 + mR +mT( )ghf

gh0 =12

vB2 +ghf

2 gh0 −ghf( ) =vB2

2g h0 −hf( ) =vB

At the bottom point, the swing is moving with a velocity vB, so there is kinetic energy.

There is also gravitational potential energy, because the swing is still above the ground in our coordinate system.

There is only gravitational potential energy at the initial point, because the swing is not moving yet.

Since the tire swings 1 meter above the ground at its lowest point, the system

has gravitational potential energy at this point. Please try again.

Choice CIncorrect

4. What value do we get for Ryan’s speed at the point of maximum tension?

A) vB=17.1m/s

B) vB=15.3 m/s

C) vB=7.7m/s

h0=13m This is the height from which Ryan is released from rest.

hf=1m This is the height that Ryan will still have when he reaches the lowest

point in the swing’s motion.

g=9.8m/s2

Choice AIncorrect

Please try again, make sure you are using the correct values:

Choice BCorrect

€

vB = 2g h0 −hf( )

vB = 2 9.8m /s 2( ) 13m −1m( )

vB ≈15.3m /s

Choice CIncorrect

h0=13m This is the height from which Ryan is released from rest.

hf=1m This is the height that Ryan will still have when he reaches the lowest

point in the swing’s motion.

g=9.8m/s2

Please try again, make sure you are using the correct values:

5. Which of the following freebody diagrams correctly depicts all of the forces acting on Ryan and the tire at the lowest

point?

A)

B)

C)

(mR+mT)g

(mR+mT)g

(mR+mT)g

Tmax

Tmax

Tmax

Fc

Fc

Tmax=Maximum Tension

mr=Ryan’s mass

mT=Tire’s mass

g=9.8m/s2

Fc= Centripetal Force

Tmax=Maximum Tension

mr=Ryan’s mass

mT=Tire’s mass

g=9.8m/s2

Fc= Centripetal Force

Choice ACorrect

This diagram correctly depicts the only forces acting on the tire and rider.

Choice BIncorrect

The centripetal force is not a new force, but only the name of the component of the net force acting

towards the center of curvature when the path is not straight.

When we use Newton’s 2nd Law we will set the right hand side equal to the centripetal force (mac),

because this expresses the response of the system to the forces acting upon it.

Choice CIncorrect

The centripetal force is not a new force, but only the name of the component of the net force acting

towards the center of curvature when the path is not straight.

When we use Newton’s 2nd Law we will set the right hand side equal to the centripetal force (mac),

because this expresses the response of the system to the forces acting upon it.

Note:

Centripetal force is generally expressed as:

This is a statement of the “radial” or ”centripetal” component of the net force.

Note:

Centripetal force is generally expressed as:

This is a statement of the “radial” or ”centripetal” component of the net force.

6. Apply Newton’s 2nd Law to find an expression for Tmax.

Which one of the following expressions is correct?

A)

B)

C)

€

Fc =mac

€

Fy =∑ T−mR +mT( )g= mR +mT( )ac

€

Fy =∑ T−mR +mT( )g=0

€

Fy =∑ T + mR +mT( )g= mR +mT( )ac

Choice ACorrect

€


Since this is a non-equilibrium problem, our net force is not zero.

Choice BIncorrect

Since this is a non-equilibrium problem, our net force is not zero.

€


Correct expression:

Choice CIncorrect

€

Fy =∑ T + mR +mT( )g= mR +mT( )ac

This sign should be negative, because the weight of Ryan and the tire acts in the opposite direction of

the tension force.

7. Find an expression for Tmax in terms of vB and other given quantities.

Which of the following expressions is correct?

A)

B)

C)

Hint

Centripetal acceleration:

R is the radius of the curve that is being traveled.

Hint

Centripetal acceleration:

R is the radius of the curve that is being traveled.

€

ac =v2

R

€

Tmax = mR + mT( )vB

2

R− g

⎛

⎝ ⎜

⎞

⎠ ⎟

€

Tmax = mR + mT( )vB

2

R+ g

⎛

⎝ ⎜

⎞

⎠ ⎟

€

Tmax = mR −mT( )vB

2

R− g

⎛

⎝ ⎜

⎞

⎠ ⎟

Choice AIncorrect

Be careful with your signs.

Correct reasoning:

€

Fy =∑ Tmax − mR + mT( )g = mR + mT( )a c

Tmax − mR + mT( )g = mR + mT( )vB

2

R

⎛

⎝ ⎜

⎞

⎠ ⎟

Tmax = mR + mT( )vB

2

R

⎛

⎝ ⎜

⎞

⎠ ⎟+ mR + mT( )g

Tmax = mR + mT( )vB

2

R+ g

⎛

⎝ ⎜

⎞

⎠ ⎟

Choice BCorrect

€



2

R

⎛

⎝ ⎜

⎞

⎠ ⎟

Tmax = mR + mT( )vB

2

R

⎛

⎝ ⎜

⎞

⎠ ⎟+ mR + mT( )g

Tmax = mR + mT( )vB

2

R+ g

⎛

⎝ ⎜

⎞

⎠ ⎟

Choice CIncorrect Correct reasoning:

Be careful with your signs.

€



2

R

⎛

⎝ ⎜

⎞

⎠ ⎟

Tmax = mR + mT( )vB

2

R

⎛

⎝ ⎜

⎞

⎠ ⎟+ mR + mT( )g

Tmax = mR + mT( )vB

2

R+ g

⎛

⎝ ⎜

⎞

⎠ ⎟

8. What is the value of Tmax?

Is the swing safe for Ryan if the rope will break at a tension of 1500N?

Answer

Reflection

Problems

€

Tmax = mR + mT( )vB

2

R+ g

⎛

⎝ ⎜

⎞

⎠ ⎟

Tmax = 30kg + 5kg( )15 .32

15m+ 9.8m /s 2

⎛

⎝ ⎜

⎞

⎠ ⎟

Tmax = 889N

16m

1m

15m

R=15m

The swing is safe for Ryan!

Reflection Questions

• What is the maximum mass that a rider can have and swing safely?

• If the launch pad was lowered to 10m, would Tmax become higher or lower for Ryan?

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