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
 Fy  N  mg cos  0
 Fx  mg sin   fs  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, in that
situation, fsmax = μs N= μs mgcosθc = mg sinθc
Solve the above equations, μs N= μs mgcosθc = mg sinθc
- μs=tan θc
(2)
Quiz 3
You have a new job designing rides for an amusement park. In one wide, the
rider’s chair is attached by a 9.0-m-long chain to the top of a tall rotation
tower. The tower spins the chair at rate of 1.0rev every 4.0 s. You’ve assumed
that the maximum possible combined weight of the chair and rider is 150 Kg.
You’ve found a great price for chain at the local discount store, but your
supervisor wonders if the chain is strong enough. You contact the
manufacturer and learn that the chain is rated to withstand a tension of
3000N. Will this chain be strong enough for the ride? Why?
Solution:
Newton’s second law along the r-axis is
T sinθ = Fr = mω2r
Notice here, r is the radius of the circle, it is not the
length of chain. r=L sinθ, also tension T is along the
chain, it is not along the r-axis.
From above equation: T sinθ = mω2r = mω2Lsinθ
→ T = mω2L = 3330N
Thus, the 3000 N chain is not strong enough for the ride.
Quiz4: Figure shows a contest in which a sphere, a cylinder, and
circular hoop all of mass M and radius R( notice particle is not
taking the race). are rolling down from rest at the same instant
of time. ( please read text book P 366-367 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) Which will win the downhill
race? why
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,
Vcm 2  2aS S  h / sin  , a 
acm  g sin  / 2 for the hoop,
acm  2 g sin  / 3 for the cylinder,
acm  5 g sin  / 7 for the sphere
4 gh
3
for cylinder,
g sin 
1 c
10 gh
for sphere
7
Continue:
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 springs
next to one another. One mass is pulled so its spring stretches 20
cm and the other pulled so its spring stretches only 10 cm. The
masses are released simultaneously. Which mass reaches the
equilibrium point first? Explain your reasoning
Quiz 5: Answer: They reach the equilibrium point at the same
time
T  2
m
k
If they have the same m and k, they will have the same period. The
maximum displacement does not effect the period.
The time reaches the equilibrium position should be T/4, since two
spring-mass systems have the same period, two masses should
reach the equilibrium position at same time.
Note: at equilibrium position x(t) = 0, the function for this oscillation is
x(t )  A cos(t )  0  cos(t )  0
t  cos1 (0) 

2
2

t   / 2, 
t  ,t  T / 4
T
2
Quiz 6:
(1)In Isothermal process, the work on the ideal gas as the volume
changes from Vi to Vf is:
Vi
W  nRT ln( )
Vf
Derive it
(2) A mole of gas at temperature 20oC compressing at constant
temperature until the gas pressure double. What is the work done
on the gas?
What is the thermal energy change in this process? How much heat
released?
(1) start from the definition:
W  
vf
vi
nRT
pdV   
dV
vi V
vf
Vi
 nRT (ln Vi  ln Vf )  nRT ln
Vf
2. A mole of gas at temperature 20oC compressing at constant
temperature until the gas pressure double. What is the work done
on the gas?
What is the thermal energy change in this process? How much
heat released?
2. Answer:
Notice when temperature is a constant, gas pressure double, the
volume should be half, Vi/Vf = 2. From above equation:
W = nRT ln(2) = 1* 8.31 * 293* ln(2) = 1687 J ( Work should be
positive
That means external force does the work on the gas)
In isothermal process, ΔT = 0 so the thermal energy change ΔE =
0 too.
ΔE = W + Q, so Q = -1687 J, negative sign means heat is
released from gas