Simple Pendulum in Uniform Acceleration

1. Physics 101 Learning Object
The Simple Pendulum in
Uniform Acceleration
Jerry, Bingyao Wang
2015.01.31

2. Quick Review
Tangential Axis:
L
𝜽
s
FT = mgcosθ
m
v2
L
1. Equilibrium: 𝜽 = 0˚
2. is centripetal force.
3. When the ball reach its maximum amplitude,
the velocity will be 0, so
FT = mgcosθ + m
v2
L
Radical Axis:
Fnet = −mgsinθ = ma (by Newtons' 2nd Law)
a = −gsinθ & θ =
s
L
(by Arc length formula)
a = −gsin(
s
L
)
sin(
s
L
) =
s
L
(by small-angle approximation)
a = −(
g
L
)s --> ω =
g
L
T = 2π
L
g
Textbook Page358
<— KEY EQUATION

3. The Simple Pendulum in Uniform Acceleration
Key Equation:
A simple case:
The simple Pendulum is still or at uniform motion.
which means a = 0
Q1: A simple pendulum is hung on the ceiling of
a still bus, what is the period of this pendulum?
l

4. Q1: A simple pendulum is hung on the ceiling of a
still bus, what is the period of this pendulum?
l
General Idea:
Consider the period equation of a simple pendulum ;
we know that π is a constant, and the length of the rope l is
also a constant, so the only variable can be changed in the
uniform acceleration is the “gravity acceleration” or more
precisely the “effective gravity acceleration”.
Answer:
In this question, since the bus is not
moving, the only two forces exerted on
the ball are the gravity of the ball and
the tension of the rope. The “g” in the
equation is directly equal to the gravity
acceleration: g, therefore the period is
T
mg

5. Recall: Inertia (Inertial force & Inertial Acceleration)
The ascending or descending elevator in real life:
We know that people will feel heavier (or lighter) in an ascending (or
descending) elevator, which in physics we will say that the apparent
weight is larger (or smaller) than the real weight of a person.
This idea can also be explained by inertia, which basically means that
there is a force (named Inertial Force) of which direction is
always opposite to the direction of acceleration, exerted on an
accelerating or retarded object.
a a
FN
mg
Inertial force: ma
FN = ma + mg
FN = m( a + g ) = mg’
FN is the apparent weight, and g’ can be called effective gravity acceleration.
FN
mg
Inertial force: ma
FN = mg - ma
FN = m( g - a ) = mg’

6. Recall: Inertia (Inertial force & Inertial Acceleration)
The accelerated or retarded bus in real life:
When the bus driver brake hard to stop his bus, the passenger will jerk
forward. Basically, it is very similar to the ascending or descending
elevator case. In general, we can still apply inertia force on this
horizontal uniform acceleration.
a
av
uniform accelerated bus
uniform retarded bus
FN
mg
ma
Inertial force
FN
mg
ma
Inertial force
Vertical Direction
FN = mg
(in equilibrium )
Horizontal Direction
FInertial = ma

7. Q2 (Clicker Question):
A simple pendulum is hung on the ceiling of an ascending elevator
in a uniform acceleration “a”, what is the period of this pendulum?
l
a
A.
B.
C.
D.
E. None of above

8. Q2 (Clicker Question):
A simple pendulum is hung on the ceiling of an ascending elevator
in a uniform acceleration “a”, what is the period of this pendulum?
l
a
B.
TheAnswer is:
mg
T
maInertial force
Explanation:
We know that the ball in a simple
pendulum will reach its equilibrium state
when 𝜽 = 0˚. Therefore, at the equilibrium
state, we have the equation such that FT =
ma + mg = m(a+g). (Since the elevator is
ascending, we have to consider the
inertial force to meet the demand of
Newton’s 2nd Law.) So in general, the
effective gravity acceleration is g’ = (a + g).
Therefore, the period of this pendulum
in an ascending elevator is

9. Q3 (Clicker Question):
A simple pendulum is hung on the ceiling of an descending elevator
in a uniform acceleration “a”, what is the period of this pendulum?
l
a
A.
B.
C.
D.
E. None of above

10. Q3 (Clicker Question):
A simple pendulum is hung on the ceiling of an descending elevator
in a uniform acceleration “a”, what is the period of this pendulum?
l
a
TheAnswer is:
mg
Tma
Inertial force
Explanation:
This one is very similar with the
question Q2. Still we first analyse the
equilibrium state of the ball (in
pendulum) and then find our effective
gravity acceleration. We can easily find
that the equation FT = mg - ma = m(g-a) at
the equilibrium state. Hence the effective
gravity acceleration is g’ = (g - a).
Therefore, the period of this pendulum
in an ascending elevator is
C.

11. Brief Summary
To find the period of a simple pendulum in certain
uniform acceleration, there are few steps in general.
(1) find the equilibrium state of the pendulum (𝜽=0˚).
(2) find the (force) equation of the equilibrium state.
(3) then analyse the acceleration of each direction。
(4) compose the acceleration if necessary. (*)
(5) then find the effective gravity acceleration.
(6) finally use the key formula T=2π√(l/g’)

12. Challenge Problem
Asimple pendulum in an uniform accelerated bus
Q4:
A ball is hung on the ceiling of a motionless bus by
a l meters long rope. The bus suddenly start with
an right acceleration a m/s^2. Note that the
acceleration a is much less than the gravity
acceleration g.
(1) Find the effective gravity acceleration of the
ball in the bus when t = 0.
(2) Assume that tan𝜽 ≒ 𝜽 when 𝜽 is small. Find
the amplitude and period of the ball.

13. We know that at t=0, the bus has a right acceleration a m/s^2, but the
velocity of bus is still 0. And therefore the ball will reach its equilibrium
state in this accelerated bus in an instant.
So we first draw a diagram.
Q4:
A ball is hung on the ceiling of a motionless bus by a l meters long rope.
The bus suddenly start with an right acceleration a m/s^2. Note that the
acceleration a is much less than the gravity acceleration g.
(1) Find the effective gravity acceleration of the ball in the bus when t = 0.
l
a
𝜽 T
mg
ma
Inertial force
𝜽
𝜽
So we have:
FT =
And the effective
gravity acceleration
is

14. We know that the velocity of the ball will become 0 when it reach its maximum amplitude.
And when the bus does not start, the ball’s velocity is 0.
When the bus start, the ball reach its equilibrium state. (𝜽 = 0˚)
Q4:
A ball is hung on the ceiling of a motionless bus by a l meters long rope.
The bus suddenly start with an right acceleration a m/s^2. Note that the
acceleration a is much less than the gravity acceleration g.
(2) Assume that tan𝜽 ≒ 𝜽 when 𝜽 is small. Find the amplitude and period
of the ball.
l
𝜽
the bus not start
equilibrium
state
A
Therefore,
A = l × 𝜽
≒ l × tan𝜽
= l × ma / mg
= la/g
T
mg
ma
Inertial force
𝜽
𝜽
since we already the effective gravity acceleration is
the period of the ball is

15. Simple Pendulum in Uniform Acceleration
SUMMARY
l
a
l
a
l
a
Note that
The real line represents the equilibrium state.
The dotted line represents the maximum
amplitude state. (A & -A)

16. Thanks for watching