Download Physics 121 - Salisbury University

Physics 121
5. Kinematics in 2-D and Vectors
5.1 Kinematics of Uniform Circular Motion
5.2 Dynamics of Uniform Circular Motion
5.3 A Car Rounding a Curve
5.6 Newton’s Law of Universal Gravitation
5.7 Gravity near the Earth’s Surface
5.8 Satellites and Weightlessness
Example 5.1 . . . 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
C. The car is not accelerating
D. The car is accelerating but there is
no force acting on it.
Q
Solution 5.1 . . . Round and round
A. The car is accelerating and Q shows the
direction of the force on it.
P
Q
•Centripetal Force changes the direction
•Centripetal Force does not change speed
•Centripetal Force points toward the
center
Example 5.2 . . . Centripetal Force Equation
The correct equation for centripetal force is
A. F = mv
B. F = mvr
C. F = mv / r
D. F = mv2 / r
Solution 5.2 . . . Centripetal Force Equation
The correct equation for centripetal force is
F = mv2 / r
Example 5.3 . . . Centripetal Acceleration
The correct equation for centripetal
acceleration is
A.
B.
C.
D.
a = mvr
a = vr
a=v/r
a = v2 / r
Solution 5.3 . . . Centripetal Acceleration
The correct equation for centripetal
acceleration is
D. a = v2 / r
Example 5.4 . . . 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 5.4 . . . Stoned and Strung
Centripetal Force is F = mv2 / r
90 = (0.25)(v2 / 1.6 )
v = 24 m/s
Example 5.5 . . . 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. 10 m/s
B. 20 m/s
C. 40 m/s
D. 120 m/s
Solution 5.5 . . . Slippery when wet!
Force of Friction = Centripetal force
(0.6)(m)(g) = (m)(v2) / r
v = 11 m/s
Is there gravity on Mars?
Newton's Law of Universal Gravitation
F = GmM/r2
Compare with F = mg so
g = GM/r2
• g depends inversely on the square of the
distance
• g depends on the mass of the planet
• g on the Moon is 1 /6 of g on Earth
Example 5.6 . . . A Neutron Star’s Gravity
An exotic finish to massive stars is that of a
neutron star, which might have as much as
five times the mass of our Sun packed into a
sphere about 10 km in radius! Estimate the
surface gravity on this monster.
Solution 5.6 . . . A Neutron Star’s Gravity
g = GM/r2
g = (6.67x10-11)(5x1.99x1030) / (104)2
g = 6.7x 1012 m/s2
Example 5.7 . . . Satellites
Geosynchronous satellites are used for cable
TV transmission and weather forecasting.
They orbit about 36,000 km above the
Earth’s surface. This is six times the radius
of the Earth which is 6,000 km. What is the
value of “g” at that height?
Solution 5.7 . . . Satellites
The distance of the satellite from the center
of the Earth is 7 RE so the acceleration due
to gravity must be 1 / 49 that on the surface
of the Earth.
(1/49)(9.8) = 0.2 m/s2
That’s all folks!