Quiz 1
This is a photograph of an apple and feather free fall in a evacuated chamber. The
apple and feather are released from the top. Suppose the camera opened
flash every 1 s.
(1) Suppose A1 is the distance from t = 0s to t = 1 s, A2 is the distance from t = 1s
to t =2s, A3 is the distance from t = 2s to t = 3s ……
what is the ratio of A1: A2: A3…….
(2) If camera opened flash every 0.06s, is the ratio of the distance between every
0.06 s the same as above or change? Why?
Answer: The ratio is 1:3:5 :7……(2n -1), here n = 1, 2, 3,
• Show it in general case, suppose the time interval is Δt
The distance from 0 –one Δt, A1 = ½*g*(Δt)2
(notice here I consider the distance, so I neglect (-) sign)
The distance from Δt---2 Δt, A2 = ½*g*(2Δt)2 -½*g*(Δt)2
= ½*g*(Δt)2 (22 -1)
Using the same way the distance from 2Δt---3 Δt,
A3 = ½*g*(Δt)2 (32 - 22)
A4 = ½*g*(Δt)2 (42 - 32) …….
An = ½*g*(Δt)2 (n2 – (n-1)2)
= ½*g*(Δt)2 (n2 – n2 +2n-1)
= ½*g*(Δt)2 (2n-1)
Therefore the ratio should be 1:3:5…….(2n-1), the ratio is
independent of Δt
quiz2
(1) A simple method of measuring
the static coefficients of friction:
Suppose a block is placed on a
rough surface inclined relative to the
horizontal as show in the Fig. The
incline angle is increased until the
block stars
to move. Measuring the critical
angle θc at which this slipping just
occurs, you can obtain μs
Proof: μs=tan θc
(2) The 10.2 Kg block is held in place by
massless rope passing over two
massless, frictionless pulleys.
Find the tensions T1, T2, T3, T4, T5
and magnitude of force F
.
Solution: (1)
If the block is not moving, Newton second law applied to the
block. For this balanced situation
 Fx  mg sin   fs  0
 Fy  N  mg cos  0
When the incline angle is increased until the block
Is on the verge of slipping, the force of static
Friction has reached its maximum value
The angle in this situation is the critical angle θc
Solve the above equations, μs N= μs cosθc = mg sinθc
- μs=tan θc
Solution (2)
Quiz 3: Rebounding pendulum ( see the example 10.26, page 286)
1Kg
In this problem, we set mB is 1Kg
(1)What is the speed the ball A just before impact the block ? (2.4m/s)
(2) What is the speed VA and VB after collision?(-1.6m/s, 0.8m/s)
(3) What is rebounding angle θ3 ? (29.7o )
Quiz4:: Figure shows a contest in which a sphere, a cylinder, and
circular hoop all of mass M and radius R. are rolling down from
rest at the same instant of time. ( please read text book P 366367 first)
(1) What is the speed of center
of mass for each of the object
when they reach the bottom
(2) What is acceleration
of the center of mass for each?
(3) What is the time for each of the
object from the top to bottom?
Then you can see which will
win the downhill race?
Race
If we choose the bottom of the ramp as the zero point of
potential energy, from energy conservation
1/ 2 Icm 2  1/ 2MVcm  Mgh,
Icm  cMR 2 c is constant
and Vcm   R
Here C is a constant that depends on the object’s geometry. C = 1 for hoop, =1/2 for cylinder
= 2/5 for sphere From above equation
Vcm 
2 gh
,
1c
Vcm  gh
for hoop,
4 gh
3
for cylinder,
g sin 
Vcm  2aS S  h / sin  , a 
1 c
2
acm  g sin  / 2 for the hoop,
acm  2 g sin  / 3 for the cylinder,
acm  5g sin  / 7 for the sphere
10 gh
for sphere
7
The time from top to bottom:
S  h / sin   1/ 2acmt 2
1
t  2h /(sin   acm) 
sin 
2h(1  c)
g
1
Vcm  acmt  t  Vcm / acm 
sin 
2h(1  c)
g
Since they have the same h and θ, the greater of c, the longer of time.
Hoop’s c =1> cylinder’s c = ½, > sphere’s c = 2/5, that results in
Sphere taking shortest time, cylinder second, hoop is the last
Quiz 5:
Two equal masses are attached to identical ideal strings,
length L, next to one another. One mass is pulled 10o and
the other pulled only 5o , then release. The masses are
released simultaneously. Which mass reaches the vertical
position( θ = 0o) first? Explain your reasoning.
Answer: They reach the equilibrium point at the same time
Reason: For the small angle pendulum, the period
T  2
L
g
If they have the same L, they will have the same period. The
maximum angle does not effect the period.
And the time reaches the equilibrium position should be T/4, since two
pendulums have the same period, two masses should reach the
equilibrium position at same time.
Note: at equilibrium position θ(t) = 0, the function for this oscillation is
 (t )   m cos(t )  0  cos(t )  0

t  cos1 (0) 
2
2

t   / 2, 
t  ,t  T / 4
T
2
( if you pull out mass to the maximum and release, the phase constant
φ=0)