Physics 221
Chapter 6
Problem 1 . . . Round and round
• A red car goes counterclockwise around a
track at a constant speed (see figure)
P
• A. The car is accelerating and Q shows the
direction of the force on it
• B. The car is accelerating and P shows the
direction of the force on it
Q
• C. The car is not accelerating
• D. The car is accelerating but there is no force
acting on it.
Solution 1 . . . Round and round
• A red car goes counterclockwise around a
track at a constant speed (see figure)
P
• A. The car is accelerating and Q shows the
direction of the force on it.
• Centripetal Force changes the direction
• Centripetal Force does not change speed
• Centripetal Force points toward the center
Q
Problem 2 . . . Equation for C.F.
• The correct equation for centripetal force is
• A. F = mv
• B. F = mvr
• C. F = mv / r
• D. F = mv2 / r
Solution 2 . . . Equation for C.F.
• The correct equation for centripetal force is
F = mv2 / r
Problem 3 . . . Stoned and strung
• Dennis the menace ties a 250 g rock to a 160
cm. string and whirls it above his head. The
string will break if the tension exceeds 90 N.
What minimum speed will endanger the
windshield on his neighbor’s car?
Solution 3 . . . Dennis the Menace
• Centripetal Force is F = mv2 / r
• 90 = (0.25)(v2 / 1.6 )
• v = 24 m/s
Problem 4 . . . Slippery when wet!
• A car exits on a ramp (unbanked) of radius 20
m. The coefficient of friction is 0.6. The
maximum speed before slipping starts is most
nearly
•
•
•
•
A.
B.
C.
D.
10 m/s
20 m/s
40 m/s
120 m/s
Solution 4 . . . Slippery when wet!
• Force of Friction = Centripetal force
• (0.6)(m)(g) = (m)(v2) / r
• v = 11 m/s
Problem 5 . . . Spaced out!
• Residents in a colony in outer space live in a
cylindrical spaceship that has a diameter of 90
m and its speed at the rim is 15 m/s. The
simulated “g” felt by the occupants is most
nearly
•
•
•
•
A.
B.
C.
D.
0.5 g
1.0 g
2.0 g
5g
Solution 5 . . . Spaced out!
• (m)(v2) / r = m “g”
•
•
•
•
•
“g” = v2 / r
“g” = (15)(15)/45
“g” = 5 m/s2
“g” = 0.5 g
Correct answer is A.
Satellites
• Many LEO (Low Earth Orbit) satellites are
fairly close to the Earth. There is substantial
“g” out there. Yet people “feel” weightless!
This has to do with “free-fall” like when you
are in an elevator and it accelerates down.
• Problem 8: How much would a 120 lb woman
“feel” she weighs in an elevator decelerating
at 2 m/s2 ?
Satellites
•
•
•
•
•
•
Solution8: F = ma
mg - T = ma
T = mg - ma
T = 12 x 10 - 12 x 2
T = 120 - 24
T = 96 lbs.
T
a
mg
• Note: Satellites are in “free-fall” so the
occupants feel weightless!
Air Drag
• Free-fall acceleration is g= 9.8 m/s2. However,
if air resistance is taken into account then the
acceleration keeps decreasing as the object’s
speed keeps increasing. This is because the
resistive force of air drag increases as the
speed increases. When the resistive force
becomes equal to the force of gravity (weight)
the object attains a constant speed. This final
constant speed is called Terminal Velocity.
Terminal Velocity
What does “drag” depend on?
Air Resistance (drag) is generally proportional to
the square of the speed
R = Cv2
R = ½ D  A v2
R = resistive force
D = Drag coefficient (0.5 to 2)
 = density of air
A = cross-sectional area
Generally in a liquid, the resistance is proportional to the speed
R = - bv
That’s all Folks!