Circular Motion; Gravitation

Circular Motion;
Gravitation

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Objectives
Students should understand the uniform circular motion of a particle, so they
can:
O Relate the radius of the circle and the speed or rate of revolution of the
particle to the magnitude of the centripetal acceleration.
O Describe the direction of the particle’s velocity and acceleration at any
instant during the motion.
O Determine the components of the velocity and acceleration vectors at any
instant, and sketch or identify graphs of these quantities.
Students should know Newton’s Law of Universal Gravitation, so they can:
O Determine the force that one spherically symmetrical mass exerts on
another.
O Determine the strength of the gravitational field at a specified point outside
a spherically symmetrical mass.
Students should understand the motion of an object in orbit under the
influence of gravitational forces, so they can:
O Recognize that the motion does not depend on the object’s mass; describe
qualitatively how the velocity, period of revolution, and centripetal
acceleration depend upon the radius of the orbit; and derive expressions
for the velocity and period of revolution in such an orbit.
O Derive Kepler’s Third Law for the case of circular orbits.

5-1: Kinematics of Uniform Circular Motion
O Uniform circular motion describes the movement of an
object in a circular path at a constant speed, v, but at a
constantly changing velocity vector, v, necessary for the
object to stay on the path.
O The velocity vector at a given point in the path is the
tangent to the circular path in the direction of motion.

O The acceleration corresponding to this changing
velocity at a given point is directed toward the center of
the circle, and it is called centripetal or radial
acceleration. Radial acceleration is proportional to the
square of the velocity and inversely proportional to the
radius,
O At a point in the path, the velocity and acceleration of
an object undergoing uniform circular motion are
perpendicular

O The number of revolutions per second by the object is
denoted by the frequency, f. The time in seconds for
each revolution, T, is its reciprocal, T = 1/f, and is
called the period.
O From these definitions, an equation for the speed can
be derived, v = 2πr/T.

Example 5-1: Acceleration of a revolving ball
O A 150-g ball at the end of a string is revolving uniformly
in a horizontal circle of radius 0.600 m. The ball makes
2.00 revolutions in a second. What is its centripetal
acceleration?

Exercise A
O If the string is doubled in length to 1.20 m but all else
stays the same, by what factor will the centripetal
acceleration change?

Example 5-2: Moon’s centripetal acceleration
O The Moon’s nearly circular orbit about the Earth has a
radius of about 384,000 km and a period T of 27.3 days.
Determine the acceleration of the Moon toward the
Earth.

5-2: Dynamics of Uniform Circular Motion
O The force required to keep an object in uniform circular
motion is directed inward toward the center and is the
product of mass and radial acceleration,
O 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.
O If the centripetal force vanishes, the object flies off
tangent to the circle.

O Often the force is provided by the tension of a rope or
string.
O Automobiles require friction between the wheels and
the road to provide the inward force necessary for
moving in a circle, or more often, in an arc of a circle.
O Satellites use gravity to provide the centripetal force for
circular obits.

Example 5-3: Force on revolving ball
(horizontal)
O Estimate the force a person must exert on a string
attached to a 0.150-kg ball to make the ball revolve in a
horizontal circle of radius 0.600 m. The ball makes
2.00 revolutions per second (T=0.500 s), as in Example
5-1.

Example 5-4: Revolving ball (vertical circle)
O A 0.150-kg ball on the end of a 1.10-m-long cord
(negligible mass) is swung in a vertical circle.
O (a) Determine the minimum speed the ball must have at
the top of its arc so that the ball continues moving in a
circle.
O (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)

Exercise B
O In a tumble dryer, the speed of the drum should be just
large enough so that the clothes are carried nearly to
the top of the drum and then fall away, rather than
being pressed against the drum for the whole
revolution. Determine whether this speed will be
different for heavier wet clothes than for lighter dry
clothes.

Exercise C
O A rider on a Ferris wheel moves in a vertical circle of
radius r at constant speed v. Is the normal force that
the seat exerts on the rider at the top of the wheel (a)
less than, (b) more than, or (c) the same as, the force
the seat exerts at the bottom of the wheel?

Conceptual Example 5-5: Tetherball
O The game of tetherball is played with a ball tied to a pole
with a string. After the ball is struck, it revolves around
the pole. In what direction is the acceleration of the ball,
and what force causes the acceleration?

Problem Solving: Uniform Circular Motion
1. Draw a free-body diagram
2. Determine the forces or components of forces that act
radially
3. Choose a convenient coordinate system
4. Apply Newton’s second law to the radial components

5-3: Highway Curves,
Banked and Unbanked
O When a car goes around a curve, there must be a net
force towards the center of the circle of which the curve
is an arc. If the road is flat, that force is supplied by
friction.
O If the frictional force is insufficient, the car will tend to
move more nearly in a straight line, as the skid marks
show.

O As long as the tires do not slip, the friction is static. If
the tires do start to slip, the friction is kinetic, which is
bad in two ways:
O The kinetic frictional force is smaller than the static.
O The static frictional force can point towards the center of
the circle, but the kinetic frictional force opposes the
direction of motion, making it very difficult to regain
control of the car and continue around the curve.

Example 5-6: Skidding on a curve
O A 1,000-kg car rounds a curve on a flat road of radius
50 m at a speed of 50 km/h (14 m/s). Will the car
follow the curve, or will it skid? Assume:
O (a) the pavement is dry and the coefficient of static
friction is µ s = 0.60;
O (b) the pavement is icy and µ s = 0.25

Example 5-7: Banking angle
O (a) For a car traveling with speed v around a curve of
radius r, determine a formula for the angle at which a
road should be banked so that no friction is required.
O (b) What is this angle for an expressway off-ramp curve
of radius 50 m at a design speed of 50 km/h?

Exercise D
O To negotiate an unbanked curve at a faster speed, a
driver puts a couple of sand bags in his van aiming to
increase the force of friction between the tires and the
road. Will the sand bags help?

Exercise E
O Can a heavy truck and a small car travel safely at the
same speed around an icy, banked-curve road?

5-4: Nonuniform Circular Motion
O If an object is moving in a circular
path but at varying speeds, it must
have a tangential component to its
acceleration as well as the radial one.
O This concept can be used for an
object moving along any curved path,
as a small segment of the path will be
approximately circular.

Example 5-8: Two components of acceleration
O A race car starts from rest in the pit area and
accelerates at a uniform rate to a speed of 35 m/s in
11 s, moving on a circular track of radius 500 m.
Assuming constant tangential acceleration, find
O (a) the tangential acceleration, and
O (b) the radial acceleration, at the instant when the speed
is v = 15 m/s.

Exercise F
O When the speed of the race car in Example 5-8 is
30 m/s, how are (a) a tan and (b) a R changed?

5-5: Centrifugation
O A centrifuge works by spinning very fast. This means
there must be a very large centripetal force. The object
at A would go in a straight line but for this force; as it is,
it winds up at B.

Example 5-9: Ultracentrifuge
O The rotor of an ultracentrifuge rotates at 50,000 rpm
(revolutions per minute). The top of a 4.00-cm-long test
tube is 6.00 cm from the rotation axis and is
perpendicular to it. The bottom of the tube is 10.00 cm
form the axis of rotation. Calculate the centripetal
acceleration, in “g’s” at the top and bottom of the tube.

5-6: Newton’s Law of Universal Gravitation
O 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.
O When there is such a disparity in masses, the reaction
force is undetectable, but for bodies more equal in mass
it can be significant.

O Newton’s law of universal gravitation states that for
both masses, the magnitude of the attractive force
between them is proportional to the product of the
masses and inversely proportional to the square of
distance.
O
where
O The magnitude of the gravitational constant G can be
measured in the laboratory. This is the Cavendish
experiment.

Example 5-10: Can you attract another
person gravitationally?
O A 50-kg person and a 75-kg person are sitting on a
bench. Estimate the magnitude of the gravitational
force each exerts on the other.

Example 5-11: Spacecraft at 2r E
O What is the force of gravity acting on a 2,000-kg
spacecraft when it orbits two Earth radii from the
Earth’s center (that is, a distance r E = 6380 km
above the Earth’s surface)? The mass of the Earth
is M E = 5.98 x 10 24 kg.

Example 5-12: Force on the Moon
O Find the net force on the Moon (m M = 7.35 x 10 22 kg)
due to the gravitational attraction of both the Earth
(m E = 5.98 x 10 24 kg) and the Sun (m S = 1.99 x 10 30 kg),
assuming they are at right angles to each other.

5-7: Gravity Near the Earth’s Surface;
Geophysical Applications
O Now we can relate the gravitational constant to the
local acceleration of gravity. We know that, on the
surface of the Earth:
O Solving for g gives:
O Now, knowing g and the radius of the Earth, the mass
of the Earth can be calculated:

O The acceleration due to gravity varies over the Earth’s
surface due to altitude, local geology, and the shape of
the Earth, which is not quite spherical.

Example 5-13: Gravity on Everest
O Estimate the effective value of g on the top of Mt.
Everest, 8,850 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?

5-8: Satellites and “Weightlessness”
O Satellites are routinely put
into orbit around the Earth.
The tangential speed must
be high enough so that the
satellite does not return to
Earth, but not so high that
it escapes Earth’s gravity
altogether.
O The satellite is kept in orbit
by its speed – it is
continually falling, but the
Earth curves from
underneath it.

O Objects in orbit are said to experience weightlessness.
They do have a gravitational force acting on them,
though!
O The satellite and all its contents are in free fall, so there
is no normal force. This is what leads to the experience
of weightlessness.

O More properly, this effect is called apparent
weightlessness, because the gravitational force still
exists. It can be experienced on Earth as well, but only
briefly:

Example 5-14: Geosynchronous satellite
O 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
O (a) the height above the Earth’s surface such a satellite
must orbit, and
O (b) such a satellite’s speed.
O (c) Compare to the speed of a satellite orbiting 200 km
above Earth’s surface.

Exercise G
O Two satellites orbit the Earth in circular orbits of the
same radius. One satellite is twice as massive as the
other. Which of the following statements is true about
the speeds of these satellites? (a) The heavier satellite
moves twice as fast as the lighter one. (b) The two
satellites have the same speed. (c) The lighter satellite
moves twice as fast as the heavier one. (d) The heavier
satellite moves four times as fast as the lighter one.

5-9: Kepler’s Laws and Newton’s Synthesis
O Kepler’s laws describe planetary motion.
O Kepler’s first law states that planetary pathways are
elliptical, orbiting around the Sun, which is at one focus
of the ellipse.

O Kepler’s second law states that an imaginary line from
the sun to the planet sweeps out equal areas in equal
periods of time.
O Kepler’s third law states that the ratio of the cube of a
planet’s mean distance from the Sun to the square of
its period is constant for all planets.

Example 5-15: Where is Mars?
O Mars’ period (its “year”) was noted by Kepler to be about
687 days (Earth days), which is (687 d/365 d) = 1.88 yr.
Determine the distance of Mars form the Sun using the
Earth as a reference.

Example 5-16: The Sun’s mass determined
O Determine the mass of the Sun given the Earth’s
distance from the Sun as r ES = 1.5 x 10 11 m.