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Lecture - 12
More examples of Newton’s laws
EXAMPLE: Merry-go-round
Little Jacob (15 kg) sits on the edge of a merrygo-round of radius 1.0 m while his big sister
makes it turn… faster and faster. How fast can
the system go before Jacob takes off if the
coefficient of static friction between Jacob’s
pants and the merry-go-round is 0.50?
Little Jacob (15 kg) sits on the edge of a merry-go-round of radius
1.0 m while big sister makes it turn… faster and faster. How fast
can the system go before Jacob takes off if the coefficient of
static friction between Jacob’s pants and the merry-go-round is
0.50?
Centrifugal force
N
(fictitious force),
it is the direction Jacob
tends to be pushed out
fs
mg
To oppose the tendency, the force towards the center comes
from the contact between Jacob and the merry go round:
force of static friction
2
Static friction provides the needed radial acceleration: fS = mrω
Maximum speed → Maximum static friction:
ωMAX =
µS g
r
=
(0.50)(9.8 m/s2 )
1.0 m
2
µS mg = mr ωMAX
= 2.2 rad/s ≈
1
3
turn/s
Example: Flat Curve
A car of mass m with constant speed v drives through a curve of
radius R. What is the minimum value of the coefficient of static
friction between the tires and the road for the car not to slip?
Fnet = ma
a
fs
v2
fs = m
R
fs ≤ µS N = µS mg
v
v2
m ≤ µS mg
R
Or, given ms, there is a maximum v: v2
µS ≥
gR
v ≤ µ s gR
Example: Bucket
A stone of mass m sits at the bottom of a bucket. A string is
attached to the bucket and the bucket is made to move in
circles. What is the minimum speed that the bucket needs to
have at the highest point of the trajectory in order to keep the
stone inside the bucket?
N
R
mg + N = ma
mg
v2
a needs to be a =
R
v2
mg + N = m
R
N
R
mg
v2
mg + N = m
R
N= m(
v2/R)
- mg
DEMO: Bucket
with water
• If v increases, N needs to be larger (if v becomes too large, since
N is also the force on the bucket by the stone, the bottom of the
bucket might end up broken…)
• If v decreases, N needs to be smaller. But at some point, N will
become zero! This is the condition for the minimum speed:
2
v min
mg = m
R
⇒
v min = gR
The speed cannot get any smaller or the trajectory will
not be a circle anymore (because the remaining forces –
mg-- will produce an acceleration that is too strong for a
circle of radius R ―at that speed)
R
If v < vmin, the stone will do
something like this...
ACT: Tension and mass
Example Test Question
Example Test Ques.on Example: Banked Curve
A car of mass m with constant speed v drives through a
banked curve of radius R. If there is no friction, what
speed does the car need to maintain for it not to slide
up or down the roadway?
y
x
x: N sinθ = max = mv2/R Y: N cosθ – mg = may = 0
N
θ
a
tanθ =
v2
Rg
⇒
v = Rg tanθ
If R = 400 m, v = 49 m/s = 109 mph
mg
v = velocity
out of page
θ= 31o
R
DEMO:
Conical
pendulum
Example: Box on truck
A box with mass m = 50 kg sits on a truck. The
coefficients of friction between the box and the truck
are μK = 0.2 and μS = 0.4.
What is the maximum acceleration that the truck
can have without the box slipping?
a
m
a
NB,T Direction of motion
fS B,T
Not slipping: static
fS = mBa
relative to the truck in
the absence of friction
WB,E
fS MAX = mBa MAX
N-W = 0
amax
µSN µS mB g
fS ,MAX
= µS g = 0.4 g = 3.9 m/s2
=
=
=
mB
mB
mB
Drag forces
For solid-fluid relative motion, friction force (called “drag
force” or “resistance”) depends on the relative speed:
fD = kv
for low speeds
fD = Dv 2
for high speeds
k and D depend on the
geometry and the
materials.
Terminal speed
Acceleration of a suitcase that falls from a plane:
fD
mg − fD = ma
Increases with v
Eventually, fD = mg , so a = 0!
When this happens, the system has reached its
terminal speed:
2
mg − Dv terminal
=0
v terminal
mg
=
D
mg
This is how
parachutes work!
DEMO:
Parachute
EXAMPLE: Pulling yourself up
A kid with mass m = 30 kg has
designed a rough elevator to
get to his tree-house. It’s
made of a seat of mass M = 5
kg, a rope and a pulley. To use
the elevator, you sit on the seat
and pull on the rope as shown
below.
How strong is the kid pulling if
the elevator is moving at
constant speed?
2T − (m + M ) g = (m + M )a = 0
T =
m +M
2
g
T
T
35 kg
T =
(9.8 m/s2 ) = 172 N
2
(M+m)g
EXAMPLE: Box on another box
A box of mass m1 = 1.5 kg is being pulled by a horizontal
string with tension T = 45 N. It slides with friction (μK =
0.50, μS = 0.70) on top of a second box of mass m2 = 3.0
kg, which in turn sits on a frictionless floor. Find the
acceleration of box 2.
a1
T
a2 ?
μK = 0.5
m1
m2
frictionless
T
m1
N
fK
m2
fK
m1g
For box 2: fK = μKN = m2a2 (only force in x- direction)
From box 1, we know that N - m1g = 0
a2 =
µk N
m2
= µk
m1
m2
g = 0.50
1.5 kg
3.0 kg
(9.8 m/s2 ) = 2.5 m/s2
The magnitude of the tension did not play any role!
The tension just needs to be large enough so the boxes cannot move
together.
The magnitude of the tension did not play any role!
The tension just needs to be large enough so the boxes cannot move
together.
If they moved together, the acceleration of both blocks would be:
T
45 N
m1
a=
=
= 10 m/s2
T
m 1 + m2 4.5 kg
fS
The static friction would be the
only horizontal force on m2:
m2
m2a = fS
But static friction has a maximum value:
This imposes a lower limit to the
coefficient of static friction:
µS ≥
m2a
m1 g
=
m2T
=
fS
f S ≤ µSN = µSm 1 g
m2a ≤ µSm1 g
(3.0 kg)(45 N)
= 2.0 > 0.7
kg)(9.8 m/s )
(m + m ) m gBlocks(4.5willkg)(1.5
not move together for this tension
1
2
1
2
Example Test Question