WORKSHEET 8: ROTATIONAL
Problem 1: A
conical pendulum. A mass M attached to a string of length L hangs from
the ceiling. If I rotate the mass such that it makes a horizontal circle
of radius R, describe the mechanics of the system.
The force which provides the centripetal force
is the horizontal component of the Tension. The system is in equilibrium
in the y-direction but accelerating in the x-direction.
-mac = -mv2/r
SFy = 0
mg = 0
From the equations above you could solve
for T, v or any quantity that may be asked for.
Problem 2: A
"screaming drum". An amusement park ride in which a person is pinned to
the wall of a huge drum that rotates.
The force that provides the centripetal force is the Normal reaction force
of the drum on the person. Again, there is acceleration in the x-direction,
and equilibrium in the y-direction. There is a force of static friction
between the body and the drum.
SFx = mac = mv2/r
SFy = 0
FN = mv2/r
fs- mg = 0
Problem 3: A
car of mass M is rounding an unbanked or flat curve of radius R. Obviously,
there is friction between the tires of the car and the road. If there is
NO skidding, then static friction, if there is skidding, kinetic friction.
Describe the system mechanics.
The force that provides the centripetal force is the static friction between
the car and the road.
SFx = mac = mv2/r
SFy = 0
fs = mv2/r
FN- mg = 0
Problem 4: In 1901, Allo "Dare Devil"
Diavolo introduces the stunt of riding a bicycle in a loop-the-loop. If
the loop is a circle (R = 2.7 m), what is the minimum speed Diavolo must
have at the top of the loop if he was to remain in contact with the loop?
At the top of the loop, the forces acting on the daredevil are the normal
force and gravity both directed down. Since he is moving in a circular
path, the net force acting on him is equal to the centripetal force also
directed down. Therefore:
-N – mg = -mv2/R
To find his minimum speed he must have to
keep contact with the loop, we set N = 0, therefore
v2 = gR = (9.80 m/s2 )(2.7 m) = 26.46 m2 /s2
v > 5.14 m/s. (answer)
If his speed is less than or equal
to 5.14 m/s, he falls.
Problem 5: Planet
X has a mass MX = 1.50 x 1027 kg, and a radius RX
= 8.5 x 107 m.
(a) Find the acceleration due to gravity
(gX ) on Planet X.
gx = [GMx/Rx2]
= [(6.67 x 10-11)(1.50 x 1027)/(8.5 x 107)2]
= 13.8 m/s2
(b) What is the speed of a 2000 kg satellite
orbiting Planet X if it is orbiting at a height h = 2000 km above the surface
of the planet?
v = [GMx/(Rx + h)]1/2
= [(6.67 x 10-11)(1.50 x 1027)/(2.00 x 106
+ 8.5 x 107]1/2
= 33.9 km/s
(c) Find the period of the satellite.
T = (2pr)/v
= [2p(8.7 x
107)/33.9 x 103 m/s]
= 16,100 s = 4.48 hours.
Problem 6: An Earth satellite, in
circular orbit at an altitude h = 230 km above the Earth’s surface, has
a period T = 89 min. What mass of the Earth follows from this data?
We apply Kepler’s Law of periods to the Earth-satellite
The radius of the satellite orbit is r = RE
+ h = 6.37 x 106 m + 230 x 103 m = 6.60 x 106
Can you attract a person gravitationally? A 50-kg person and a 75-kg person
are sitting on a bench so that their center of masses are 50-cm apart.
Estimate the force of gravitational attraction between the two.
This value is too small to be observed.
Geosynchronous Satellite. A geosynchronous satellite is one that stays
above the same point of the Earth at all times in its orbit (like cable
TV satellites, communications satellites). Determine the height above the
surface of the Earth and the orbital speed for such a satellite.
The satellite in circular motion experiences
a centripetal force provided by the force of gravitational attraction between
the satellite and the Earth. Therefore:
FG = mv2 /r where m
= mass of the satellite and r = RE + h.
A box of empty film canisters is dumped from a rocket traveling outward
from the Earth at a speed of 1800 m/s from a height of 1600 km above the
surface of the Earth. The package eventually falls to the Earth. Ignoring
air resistance, estimate its speed just before impact.
We use conservation of mechanical energy.
An astronaut puts a bowling ball of mass m = 7.20 kg into a circular orbit
about the Earth at a height h = 350 km.
a) What is the kinetic energy of the
The orbital radius r = RE + h = 6370 km + 350 km = 6.72 x 106
K = ½ mv2 = (GMm)/(2r)
= [(6.67 x 10-11)(5.98 x 1024 kg)(7.20 kg)]/2(6.72
x 106 m)
= 2.14 x 108 J
b) What is the potential energy of the
U = -(GMm)/r = -2K
= -2(2.14 x 108 J)
= -4.28 x 108 J
c) What is the mechanical energy of
E = K+ U
= ½ mv2 - (GMm)/r
= K – 2K
= -2.14 x 108 J