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AP Physics B – Lesson 11
Section 5-1 Uniform Circular Motion
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The motion of a mass in a circle at a constant speed.
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The Magnitude (size) of the velocity vector v is constant, but the DIRECTION of v changes
continually!
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At any point the instantaneous velocity is tangent to the circular path. This means velocity is
perpendicular to the radius.
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A mass moving in circle at a constant speed is accelerating, because it is changing direction.
Acceleration = Rate of change of velocity or a = (Δv/Δt)
In any time interval (Δt) the change in velocity (Δv) is the difference in the vectors.
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If we look at the change in velocity Δv in the limit that the time interval Δt becomes
infinitesimally small then v1 and v2 become essentially perpendicular. The vector difference
then points to the center of the circle.
Since the acceleration vector must be in the same direction as the velocity vector, the
acceleration points toward the center of the circle.
Centripetal  “Toward the center”
Centripetal acceleration = acceleration toward the center. Another term used synonymously is
radial acceleration.
The frequency, f, of a revolving object is the number of revolutions it makes each second.
The period, T, of an object is the time for one complete revolution.
Example: A 150 g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600m.
The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?
Example: The moon’s nearly circular orbit about the Earth has a radius of about 384,000km and a period
T of 27.3 days. Determine the acceleration of the Moon toward the Earth.
5-2 Dynamics of UCM
For an object to be in uniform circular motion, there must be a net force acting on it.
We already know the acceleration, so can immediately write the force:
There is no centrifugal force pointing outward; what happens is that the natural tendency of the object
to move in a straight line must be overcome.
If the centripetal force vanishes, the object flies off tangent to the circle.
Example: Estimate the force a person must exert on a string attached to a 0.150kg ball to make the ball
revolve in a horizontal circle of radius 0.600m. The ball makes 2.00 revolutions per second.
Example: A 0.150-kg ball on the end of a 1.10-m-long cord (negligible mass) is swung in a vertical circle.
(a) Determine the minimum speed the ball must have at the top of its arc so that the ball continues
moving in a circle.
(b) Calculate the tension in the cord at the bottom of the arc, assuming the ball is moving at twice the
speed of part (a).
Section 5-3 Highway Curves
Case 1 - Unbanked Curve: When a car rounds a curve, there MUST be a net force toward the circle
center (a centripetal force) of which the curve is an arc. If there weren’t such a force, the car couldn’t
follow the curve, but would (by Newton’s 2nd Law) go in a straight line. On a flat road (without tire
slipping or sliding), this centripetal force is the static friction force.
Example: A car, mass m = 1,000 kg car rounds a curve on a flat road of radius r = 50 m at a constant
speed v = 14 m/s (50 km/h). Will the car follow the curve, or will it skid?
Assume:
a.
Dry pavement with the coefficient of static friction μs = 0.6
b.
Icy pavement with μs = 0.25.
If the friction force isn’t sufficient, the car will tend to move more nearly in a straight line (Newton’s 1st
Law). As long as the tires don’t slip, the friction is static. If the tires start to slip, the friction is kinetic,
which is bad in 2 ways!!
1. The kinetic friction force is smaller than the static friction force.
2. The static friction force points toward the circle center, but the kinetic friction force opposes the
direction of motion, making it difficult to regain control of the car & continue around the curve.
Section 5-6
If the force of gravity is being exerted on objects on Earth, what is the origin of that force? Newton’s
realization was that the force must come from the Earth. He further realized that this force must be
what keeps the Moon in its orbit.
The gravitational force on you is one-half of a Third Law pair: the Earth exerts a downward force on you,
and you exert an upward force on the Earth. When there is such a disparity in masses, the reaction
force is undetectable, but for bodies more equal in mass it can be significant.
Therefore, the gravitational force must be proportional to both masses.
By observing planetary orbits, Newton also concluded that the gravitational force must decrease as the
inverse of the square of the distance between the masses.
In its final form, the Law of Universal Gravitation reads:
𝑚1 𝑚2
𝐹=𝐺
𝑟2
Where F is the gravitational force, G is a constant 6.67 X 10-11 N · m2/ kg2 , m are the masses of the
objects, and r is the distance between the objects.
Example: Calculate the gravitational force acting on you as a result of the person sitting next to you.
Assume that one person has a mass of 50 kg and the other is 75 kg.
What is the force of gravity acting on a 2000 kg spacecraft when it orbits two Earth radii from the earth’s
center (that is, a distance rE=6380 km above the earth’s surface)? The mass of the earth is
ME=5.98x1024 kg.
Estimate the effective value of g on the top of Mt. Everest, 8850 m (29,035 ft) above sea level. That is,
what is the acceleration due to gravity of objects allowed to fall freely at this altitude?
A geosynchronous satellite is one that stays above the same point on the earth, which is possible only if
it is above a point on the equator. Such satellites are used for TV and radio transmission, for weather
forecasting, and as communication relays. Determine (a) the height above the earth’s surface such a
satellite must orbit, and (b) such a satellite’s speed. (c) Compare to the speed of a satellite orbiting 200
km above Earth’s surface.
Section 5-9
Kepler’s laws describe planetary motion.
1st Law: The orbit of each planet is an ellipse, with the Sun at one focus.
2nd Law: An imaginary line drawn from each planet to the Sun sweeps out equal areas in equal times.
3rd Law: The ratio of the square of a planet’s orbital period is proportional to the cube of its mean
distance from the Sun
Derive Kepler’s 3rd Law from Newton’s Laws