Transcript Tutorial_swing

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)?
Ryan’s mass:
mR=30kg
Mass of tire:
mT=5kg
Neglect the mass
of the rope.
16m
h0=13m
y
x
1m
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
Choice A
Incorrect
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 B
Correct

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).

T
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.
T

W
The tension is at a maximum at the
bottom of the trajectory.

W
y
x

Choice C
Incorrect
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.
Initial Energy  Final Energy
But, once we find the speed we will need to use
Newton’s 2nd Law to find the solution.
Choice B
It’s a good choice, but only part of the story.
We need to use Newton’s 2nd Law, but we
must also apply the law of conservation of
energy to find important information first.
F
x
0
and
F
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.

y
 ma
Choice C
Correct
We can use the conservation of energy law to
find the speed of the tire at the point of
maximum tension:
Initial Energy  Final Energy
Once we find the speed we will need to use the
vertical component of Newton’s 2nd Law to
find the solution:
F
y
 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.
VB= Ryan’s speed at
the bottom


A)
VB  2ghf
B)
VB  2gh0 hf 
C)
VB  2gh0
Subscripts 0 and f
stand for initial and
final.
Choice A
Incorrect
You must take the initial gravitational
energy into consideration. Ryan swings
from rest atop the launch pad.
Please try again.
Choice B
Correct
There is only
gravitational
potential
energy at the
initial point,
because the
swing is not
moving yet.
Setting the initial energy equal to
the final energy we find:
1
2
mR  mT gh0  mR  mT vB  mR  mT ghf
2
1 2
gh0  vB  ghf
2
At the bottom point, the swing
is moving with a velocity vB, so
there is kinetic energy.
2gh0  ghf   v
There is also gravitational
potential energy, because the
swing is still above the ground
in our coordinate system.
2
B
2gh0  hf   vB
Choice C
Incorrect
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.
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
Choice A
Incorrect
Please try again, make sure you are
using the correct values:
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 B
Correct
vB  2gh0  hf 
vB  29.8m/s2 13m 1m
vB  15.3m/s

Choice C
Incorrect
Please try again, make sure you are
using the correct values:
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
5. Which of the following freebody diagrams correctly depicts
all of the forces acting on Ryan and the tire at the lowest
point?
A)
Tmax
(mR+mT)g
Tmax
B)
Fc
C)
Fc
(mR+mT)g
Tmax
Tmax=Maximum Tension
mr=Ryan’s mass
mT=Tire’s mass
(mR+mT)g
g=9.8m/s2
Fc= Centripetal Force
Choice A
Correct
This diagram correctly depicts the only
forces acting on the tire and rider.
Choice B
Incorrect
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 C
Incorrect
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.


6. Apply Newton’s 2nd Law to find an
expression for Tmax.
Which one of the following expressions is
correct?
Note:
A)
Fy T  mR  mT g  mR  mT ac
B)
F
C)
Fc  mac
T  mR  mT g  0
y
F
y
Centripetal force is
generally
expressed as:
This is a statement
of the “radial” or
mR  mT ac ”centripetal”
component of the
net force.
T  mR  mT g  



Choice A
Correct
F
y
Since this is a non-equilibrium
problem, our net force is not zero.
T  mR  mT g  mR  mT ac

Choice B
Incorrect
Since this is a non-equilibrium
problem, our net force is not zero.
Correct expression:
F
y
T  mR  mT g  mR  mT ac
Choice C
Incorrect
F
y

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)


Tmax
Tmax
Tmax
v 2

B
 mR  mT   g
 R

Hint
Centripetal
acceleration:
v 2

B
 mR  mT   g
 R

v 2

B
 mR  mT   g
 R

v2
ac 
R

R is the radius
of the curve that
is being
traveled.
Choice A
Incorrect
Correct reasoning:
F
y
Be careful with
your signs.
Tmax  mR  mT g  mR  mT ac
Tmax
v 2 
 mR  mT g  mR  mT  B 
 R 
Tmax
v 2 
 mR  mT  B  mR  mT g
 R 
Tmax
v 2

B
 mR  mT   g
 R

Choice B
Correct
F
y
Tmax  mR  mT g  mR  mT ac
Tmax
v 2 
 mR  mT g  mR  mT  B 
 R 
Tmax
v 2 
 mR  mT  B  mR  mT g
 R 
Tmax
v 2

B
 mR  mT   g
 R

Choice C
Incorrect
Correct reasoning:
F
y
Be careful with
your signs.
Tmax  mR  mT g  mR  mT ac
Tmax
v 2 
 mR  mT g  mR  mT  B 
 R 
Tmax
v 2 
 mR  mT  B  mR  mT g
 R 
Tmax
v 2

B
 mR  mT   g
 R

8. What is the value of Tmax?
Is the swing safe for Ryan if the rope will break
at a tension of 1500N?
Answer
15m
16m
Tmax
v 2

B
 mR  mT   g
 R

Tmax
15.32

2
 30kg  5kg
 9.8m/s 
15m

1m
R=15m
Tmax  889N
Reflection

The swing is safe for Ryan!
Problems
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?