LESSON 21 – CENTRIPETAL
FORCE: TENSION PROBLEMS
Fernando Morales
August 28, 2013
PEER INSTRUCTION

A beginning physics student, confused by a
seeming contradiction in Newton’s laws, asks her
teacher the following question: “If, for every force
there is an equal and opposite reaction force,
then all forces in nature come in equal and
opposite pairs, and are therefore balanced. Thus,
since there can never be such a thing as an
unbalanced force, how can any object ever
accelerate?” Explain the fault in this common
misconception.
PEER INSTRUCTION
A rodeo performer spins a lasso above his head.
a) Explain the purpose of twirling the rope before
throwing it.
b) Describe how he could maximize the distance
the rope can be thrown.
c) Describe the path the rope will take once he
releases it.

TWO BIRDS WITH ONE STONE

A brave warrior places a 1.0 kg rock in a sling
and swings it in a horizontal circle around his
head on a 1.0 m long vine. If the vine breaks at a
tension of 200 N, what is the maximum speed,
with which he can swing the rock? How would
the answer change, if instead he used a vertical
circle?
PEER INSTRUCTION

Rank in order, from largest to smallest, the
centripetal accelerations (ar)a to (ar)e of particles
a to e.
TARZAN

Tarzan swings on a vine that can only support
1,200N. The vine is 25m long and Tarzan has a
mass of 90kg. What is the maximum speed that
Tarzan can swing from one tree to the next and
not have his vine break?
SIR ISAAC NEWTON CONTEST QUESTION
Jane, of mass 60 kg, stands on a 15 m high cliff
holding a 12 m horizontal vine tied to a branch
above the level plain below. She starts from rest
and swings freely along the circular arc shown.
When she reaches position 30 degrees below
horizontal, she releases the vine. Calculate the
tension in the vine at position II just before Jane
let go.
 Ans. T= 882 N

A ball moving in a circle
a=
v²
r
According to Newton’s 2nd Law
net force  acceleration
Fnet = ma = m
v²
r
NB: this force often denoted Fc (for Centripetal Force) is a NET
force – the actual force(s) involved might be tension in a string,
gravity, electromagnetism or friction
AN OUTWARD FORCE? NO
A common misconception: many people think an
object moving in a circle must have an outward
force on it (after all, why do we feel as though we
are crushed in our car seat when going around a
corner if there is no outward force?)
THIS IS INCORRECT – what we feel is the
result of the conflict between our own inertia (we
would continue moving in a straight line) and
the force of the door pushing us around the
corner
The force on the object
moving in a circle
MUST be inward
for it to go
around the circle
The force on the
hand is outward
but the force on
the ball is inward

4. Tarzan plans to cross a gorge by swinging in
an arc from a hanging vine. If his arms are
capable of exerting a force of 1500 N on the rope,
what is the maximum speed he can tolerate at
the lowest point of his swing? His mass is 85 kg;
the vine is 4.0 ml long. v = 5.6m/s
7. Tarzan, at 96 kg, is swinging on a 9.3 m vine
to reach Jane! At the bottom of the swing the
tension in the vine is 1580 N.
 a) What is Tarzan’s speed at the bottom of the
swing? 7.87 m/s
 b) What is Tarzan’s speed as he reaches Jane at
an angle of 42º? 3.9 m/s
 c) What is the tension in the vine just before he
lets go and grabs onto Jane? 855 N

X5
-3
EXAMPLE
Force on a revolving ball A 0.250 kg ball on a 0.400 m string
revolves horizontally at 3 revolutions per second. What is the centripetal force – the
tension in the string, FT?
Solution
The cord sags at angle θ. Why? Because some force must
balance mg down  a component of FT must be upward
N2L: F = FT = ma = mv ²
rg
tan θ =
r
2r
v²
v= T
v
²
X: FTcosθ = m r Y: FTsinθ = mg
g T²
mv ²
tan θ =
mg
FT =
FT =
T = 0.333 s
4² r
rcos θ
sin θ
(3 revs / s)
= 0.0690
mv ²
mg
θ = 3.95º ~ 4° (sb rcos θ, not just
=
rcos θ
sin θ
r)
mg
sin θ
FT =
rg
= 35 N
=
sin 4
cos θ
v²
EXAMPLE
X5 - 4
Tetherball When the ball is struck, it whirls around the pole.
What is the direction of the acceleration, and what supplies that acceleration?
Solution
The circle is horizontal  the
acceleration is also horizontal
(centripetal)
The horizontal component of the
tension in the string supplies
the centripetal force
EXAMPLE
X5 - 5
Vertical Circle A 0.150 kg ball is swung in a vertical circle on a
1.10 m cord. (a) Find the minimum speed of the ball at the top so that it makes it around
the circle. (b) Find the tension in the cord at the bottom of the circle if the ball is moving
twice as fast as in (a)
Solution
(a) The FBD at A shows mg and the
tension in the same direction
v²
F = Fnet = mg + FT = m
r
The speed will be minimum when
FT goes to 0
v²
mg = m
r
v = (gr) = 3.28
m/s
EXAMPLE
X5 - 5
Vertical Circle A 0.150 kg ball is swung in a vertical circle on a
1.10 m cord. (a) Find the minimum speed of the ball at the top so that it makes it around
the circle. (b) Find the tension in the cord at the bottom of the circle if the ball is moving
twice as fast as in (a)
Solution
(b) The FBD at B shows mg and the
tension in opposite directions
The speed is to be 2(3.28) = 6.56 m/s
v²
F = Fnet = FT – mg = m
r
v²
FT = m
+ mg
r
= 7.34 N
REQUIRED BEFORE NEXT CLASS

Chapter Review from Nelson Textbook # 31,
34, 35, 37, 38, 44, 48, 54