Download Tue Aug 31 - LSU Physics

Phys 2101
Gabriela González
Physics having predictive value!
Newton’s laws: two qualitative, one quantitative:
1.  If no net force, no acceleration (either at rest, or
moving with constant velocity)
2. 


F =ma
3.  Action and reaction forces are equal and opposite.
2
1
Motion in two dimensions, with constant acceleration:
g


F = −mg ˆj = m a

a(t) = −g ˆj


v (t) = v 0− gt ˆj

  1
x (t) = x 0 + v 0 t − gt 2 ˆj
2
v x = v 0 cosθ ; v y = v 0 cosθ − gt
x = x 0 + v 0 cosθ t ; y = y 0 + v 0 sinθ t − 1 gt 2
€
2
3
€

r = x iˆ + y ˆj
= R(cosθ iˆ + sin θ ˆj )

r (t) = R(cosωt iˆ + sin ωt ˆj )

v (t) = Rω (−sinωt iˆ + cosωt

a(t) = −Rω 2 (cosωt iˆ + sin ωt


F = ma
ˆj )
ˆj )
Uniform: constant speed (angular speed = ω, linear speed=v=ωR)
Circular: constant radius R, velocity€vector is not constant!
a = Rω2=v2/R
4
2
Circular: velocity vector is not constant, so there’s always acceleration.
A force is always needed to produce the acceleration.
If the force (and acceleration) points to the center, speed is constant.
If there is a tangential component of the force, the speed will change. 5
A roller-coaster car has a mass of 1200 kg when fully loaded with passengers. As
the car passes over the top of a circular hill of radius 18 m, its speed is not
changing. At the top of the hill, what are the (a) magnitude FN and (b) of the
normal force on the car from the track if the car’s speed is 14m/s? What are (c) FN
and (d) the direction if v=11 m/s?
6
3
A roller-coaster car has a mass of 1200 kg when fully loaded with passengers. As the car passes over the top
of a circular hill of radius 18 m, its speed is not changing. At the top of the hill, what are the (a) magnitude FN
and (b) direction of the normal force on the car from the track if the car’s speed is 14m/s? What are (c) FN and
(d) the direction if v=11 m/s?
Mg + FN = Ma = Mv 2 /R
v
FN
Mg
FN = M(v 2 /R − g)
€
If v=14m/s, FN=+1,306 N : it points down (tracks push on the car)
If v=11m/s, FN=-3,693 N : it points up (tracks need to hold the slow car!)
8
F = ma is all you need!
9
4
Work done by a force to move an object from point a to point b:
 
W = ∫ F • dx
b
Units: [W]=Nm=J
a
If the force is constant,
   b    
 
F
•
d
x
=
F
•
d
x
=
F
•
r
−
r
=
F
( b a ) • d = Fd cosθ
∫
∫
b
W=
a
a
€

F
a

d
θ
b
Work can be negative (!?)
whenever the force “opposes”
motion (ex: friction).
10
Whenever an object is moving with speed v,
it has kinetic energy:
K=
1 2
mv
2
Kinetic energy is never
negative!
Units:
[Work]=[Force][displacement]=Nm=(kgm/s2) m
=kgm2/s2=Joule (J)
[K]=[mass][velocity2]=kg (m/s)2=J: same as work!
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5
Work-energy theorem: The change in kinetic energy
of a particle is equal to the net work done on the
particle by all external forces.
 
1
1
F
∫ • dx = ΔKE = 2 mv a2 − 2 mv b2
a
b
W =
Consider a constant 1-D force producing constant acceleration a, starting from
rest. In time t, the work done by the force is Fd, where d is the distance
traveled, ½ a t2. At time t, the velocity is v=at, so the kinetic energy is
€
K = ½ mv2 = ½ m(at)2 = (ma) (½ a t2) = Fd
12
A horizontal force Fa of magnitude 20.0 N is applied to a 3.00 kg psychology
book as the book slides a distance d=0.5m up a frictionless ramp at angle
θ=30o.
(a) During the displacement, what is the net work done on the book by Fa, by
the gravitational force on the book, and by the normal force on the book?
(b) If the book has zero kinetic energy at the start of the displacement, what is
its speed at the end of the displacement?
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6
A horizontal force Fa of magnitude 20.0 N is applied to a 3.00 kg psychology book as the book slides a distance d=0.5m up a
frictionless ramp at angle θ=30o.
(a) During the displacement, what is the net work done on the book by Fa, by the gravitational force on the book, and by the
normal force on the book?
(b) If the book has zero kinetic energy at the start of the displacement, what is its speed at the end of the displacement?
 
W = ∫ F • dx
b
d
a
Work done by Fs is Ws=+Fs d cosθ = +8.66 J
Work done by gravity force is Wg=-Mgdsinθ=-7.35J
Work of normal force is zero.
ΔK = Kf = Wnet = 1.31 J
½ mv2 =Wnet
v=(2Wnet/m)½ = 0.93 m/s
15
A variable force F acts along the x-axis on a 10kg mass. The particle
starts moving to the right with speed 2 m/s at x=0.
a)  Describe the motion of the mass.
b)  At what position does the particle has maximum and minimum kinetic
energy?
c)  Plot the position versus time for the particle.
d)  What is the expression for the force as a function of position?
F(N)
20
10
-2 -1
0
-10
1
2
x(m)
-20
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7
Hooke’s law:


F = −k d
Key concepts:
•  d: displacement from free, relaxed end
•  -ve sign: “restoring” force
•  large k: “stiff”, small k= “soft”
In one dimension, with the origin at the free,
relaxed end:
F=-kx
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•  Gravitational force: magnitude = mg, always vertical, pointing down.
•  sometimes does positive work, sometimes negative work.
•  Normal force: magnitude =whatever is needed for the objects not to
penetrate surface; always perpendicular to surface.
•  never does any work !!
•  Friction force: magnitude = µκN (if moving), ≤ µsN (if not); always
opposing motion and parallel to the surface.
•  always negative work ! (or no work, if there is no motion).
•  Tension: magnitude determined by acceleration; always parallel to
string or cord.
•  does positive work when lifting or pulling, negative work when
lowering or resisting pull.
•  Spring forces: magnitude = - kx; always parallel to spring, trying to
“restore” its relaxed state.
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•  sometimes does positive work, sometimes negative work.
8

xf
  xf
W = ∫ F • dx = ∫ F dx

xi
xi
•  If force is constant (gravity, tension, friction),
xf
W = ∫ F dx = F (x f − xi )
x
i
•  For a spring force,
xf
xf
1
W = ∫ F dx = ∫ − k x dx = − k x 2f − xi2
2
xi
xi
(
)
Work-energy theorem valid for all forces:
The change in kinetic energy of a particle is equal to the net work
done on the particle by all external forces.
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•  Power = energy delivered by a force F per unit time
•  Average power: P=ΔW/Δt
•  Instantaneous power: P=dW/dt
•  Constant force : W=Fx, P=dW/dt = F dx/dt = Fv
•  Spring force F=-kx, W= - ½ k x2
P = dW/dt = - ½ k (2 x) dx/dt = - k x v = Fv
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