Download Test 2

Test 2
phy 240
1. a)
b)
c)
d)
physics dealt mostly with
How is the velocity of a particle defined?
bodies.”
What is an inertial reference frame?
Describe friction.
Compare the acceleration of a particle in reference frames related by the
Galilean transformation.
2. Identifying the force (indicate the object, the source and the type of each force),
draw the free body diagram for
a) a van parked on a hill
b) a bucket hanging under a helicopter moving with constant velocity.
(Make sure that the net force is consistent with the indicated motion.)
3. Consider a satellite of mass m moving around a planet of mass M and radius R,
at an altitude h above the planet’s surface, as illustrated in the figure. Relate the
speed of the satellite to the given information (m, M, R, and h). (Hint. Assume
you know the gravitational constant G.)
4. An object (“particle”) has a velocity of [3,0] m/s at one instant. Five seconds
later, its velocity has changed to [8,10] m/s. Assuming that the object was
subject to a constant net force,
a) find the average acceleration of the object (during this time interval)
b) and the displacement of the object (in the same time).
c) Determine the magnitude of the net force.
falling
-1-
a) The rate, at which a particle is changing its position (vector), is
called the velocity (vector) of the particle.
r
r
dr
v( t) ≡
dt
b) If a particle does not interact with other bodies being considered (we
say that all bodies exert a zero (vector) force on the particle), it is
possible to find a reference frame in which that particle has zero
acceleration. This reference frame is called an inertial reference frame.
c) Within certain limits (of the external forces) the static friction
(interaction) prevents the object's surface from moving along the rigid
surface. The maximum magnitude of this force is proportional to the
magnitude of the normal force f s ≤ µs N .
When the external force exerted the limit, the object begins to slide
on the surface, while the kinetic frictional force has a value f k = µ k N
in the direction opposite to the relative velocity of the sliding surface.
d) In a reference frame related by the Galilean transformation the
accelerations are identical at any instant
r
r
a( t ) = a' (t )
-2
W – gravitational force exerted by the earth
on the bucket.
T
T – tension force exerted by the rope on the
bucket.
D
W
D – drag force exerted by the air on the
bucket.
W – gravitational force exerted
by the earth on the van.
N
fs
N – tension force exerted by the
road on the van.
fs – (static) frictional force
exerted by road on the van.
W
-3m
r
FG
h
R
The satellite is subjected only to
gravitational force exerted by the
planet. From the universal law of
gravity, the magnitude of that force
is
1)
Fnet = FG = G
Mm
(R + h )2
The symbols are consistent with the figure.
Because of the large mass of the planet (in comparison with the
satellite) one can assume that the planet’s reference frame is inertial. In
this frame the net force exerted on the object and its acceleration are
related by Newton’s second law of motion. In terms of the magnitudes
2)
Fnet = ma c
where m is the mass of the satellite and ac its centripetal acceleration.
In a uniform circular motion, the angular velocity ω is constant
leading to a constant speed of the satellite
3)
v = ω
ω × r = ωr = ω (R+h)
from which the magnitude of the centripetal acceleration can be related
to the speed of the satellite
4) ac = ω
ω × r’ = ω
ω × (ω
ω × r) = ω2r
The rest is algebra. By substitution
v = ωr = r ⋅
ac
F ⋅r
= net =
r
m
GMm ⋅ r
r2 ⋅ m
=
GM
GM
=
r
R+h
-4-
y
a) From the definition of the average
value of a function
∫ a (t )dt
vf
vf − vi
=
∆t
∆t
[8,10] m − [3,0] m
s
s = [1, 2] m
=
5s
s2
aav =
∆t
=
a
x
vi
b) Since the net force is time independent the particle moves with a
constant acceleration. In such a case the average acceleration is equal
to the acceleration. From the given initial (or final) velocity and the
acceleration function one can determine velocity at an arbitrary instant
t
v(t ) = v i + ∫ adt = v i + at
t i =0
(Note. Using the final velocity is also correct although more complicated
v(t ) = v f +
t
∫ adt = v i + a(t − t f ) )
t f = 8s
With the origin of the coordinate system at the initial location of the
object the position vector can be determined from the velocity function
t
t
1
r (t ) = ri + ∫ vdt = ri + ∫ v i + atdt = ri + v i t + at 2
2
t i =0
ti =0
The displacement (from its definition) is therefore
1
m
1
m
rf − ri = v i t + at 2 = [3,0] ⋅ 5s + ⋅ [1,2] 2 ⋅ (5s )2 = [27.5, 25. 0]m
2
s
2
s
c) The net force and acceleration are related by Newton’s second law,
which requires the mass of the object to be given. There is no
information leading to its value therefore it is not possible to find the net
force.
-1-
-2–
-3-
-4y
vf
x
vi