A LEVEL PROBLEMS
1.A
pilot performs an evasive maneuver by diving vertically at 310 m/s. If
he can withstand an acceleration of 9.0 g’s without blacking out, at what
altitude must he begin to pull out of the dive to avoid crashing into the
sea? [1.1 km.]
SOLUTION:
At
the bottom of the dive, the normal force is upward, which we select as
the positive direction, and the weight is downward. The pilot experiences
the upward centripetal acceleration at the bottom of the dive.We
find the minimum radius of the circle from the maximum acceleration:
amax
= v2/Rmin
(9.0)(9.80
m/s2) = (310 m/s)2/Rmin,
which gives Rmin
= 1.1x103 m.
Because
the pilot is diving vertically, he must begin to pull out at an altitude
equal to the minimum radius: 1.1 km.
2.Determine
the mass of the Sun using the known period of the Earth and the distance
from the Sun. [2.0
´1030 kg]
SOLUTION:
We
relate the speed of the Earth to the period of its orbit from
v
= 2pR/T.
The
gravitational attraction from the Sun must provide the centripetal acceleration
for the circular orbit:
GMEMS/R2
= MEv2/R
= ME(2pR/T)2/R
= ME4p2R/T2,
so we have
GMS
= 4p2R3/T2;
(6.67
´ 10–11 N ·
m2/kg2)MS
= 4p2(1.50
´ 1011 m)3/(3.16
´ 107
s)2, which gives MS
= 2.0
´ 1030
kg.
3.At
what distance between the Earth and Moon should a spacecraft be placed
in order to experience a zero net gravitational force because the Earth
and Moon pull with equal and opposite forces? [3.46
´108 m from Earth’s center]
SOLUTION:
Because
the gravitational force is always attractive, the two forces will be in
opposite directions.If we call the
distance from the Earth to the Moon D and let x be the distance
from the Earth where the magnitudes of the forces are equal, we have
GMMm/(D
– x)2 = GMEm/x2,
which becomes MMx2
= ME(D
– x)2.
(7.35
´ 1022
kg)x2 = (5.98
´ 1024
kg)[(3.84 ´ 108
m) – x]2,
which gives
x
=3.46
´ 108
m from Earth’s center.
4.You
know your mass is 60 kg, but when you stand in an elevator, it says your
mass is 80 kg. What is the acceleration of the elevator and in what direction?
[3.3 m/s2]
SOLUTION:
We
take the positive direction upward.The
spring scale reads the normal force expressed as an effective mass: FN/g.
We write ?F = ma from the force diagram:
FN
– mg = ma, ormeffective
= FN/g
= m (1 + a/g);
80
kg = (60 kg)[1 + a/(9.80 m/s2)],
which gives a = 3.3 m/s2
upward.
The
direction is given by the sign of the result.
5.In
an attempt to distribute the sun’s energy evenly, the astronauts decide
to put themselves in a slow rotation. They started off by accelerating
from no rotation to one revolution per minute during a 10-minute interval.
Think of the spacecraft as if it were a cylinder of radius 8.5 m. Determine
(a) the angular acceleration, and (b) the radial and tangential components
of the acceleration of a point at the edge of the ship 5.0 minutes into
its 10-minute acceleration period. [ 1.8 ´ 10–4
rad/s2; 1.2 ´10–2
m/s2. and 7.7 ´10–4
m/s2]
SOLUTION:
The
final angular speed is
w
= (1 rpm)(2p rad/rev)/(60 s/min) = 0.105 rad/s.
(a)We
find the angular acceleration from
a=
Dw/Dt
=
(0.105 rad/s – 0)/(10.0 min)(60 s/min) =1.8 ´ 10–4
rad/s2.
(b)We
find the angular speed after 5.0 min:
w
= wo + at
= 0 + (1.8 ´ 10-4
rad/s2)(5.0 min)(60
s/min) = 5.45 ´ 10-2
rad/s.
At
this time the radial acceleration of a point on the skin is
aR
= w2r
= (5.45 ´ 10–2
rad/s)2(4.25 m) =1.2 ´ 10–2
m/s2.
The
tangential acceleration is
atan
= ar = (1.8 ´ 10–4
rad/s2)(4.25 m) =7.7 ´ 10–4
m/s2.
