PHYS 1020 Lecture 16 Circular Motion€¦ · G == gravitational constant . C+J 5.35 A satellite is in a circular orbit about the earth (M E = ... radius about the Sun to that of the

PHYS 1020 Lecture 16 Circular Motion

Black Hole in the centre of our Milky Way Galaxy.

Satellites in circular orbits c == centripetal g == gravity

Fc = Fg

Fc = m~a = mv2

r

Fg =G m M

r2

Therefore in general : mv2

r=

G m M

r2

v2 =G M

r

m == mass of 1 object, v == velocity of this object r == radius of orbit

M == mass of other object G == gravitational constant

C+J 5.35 A satellite is in a circular orbit about the earth (ME = 5.98 x 1024 kg). The period of the satellite is 1.85 x 104 s. What is the speed at which the satellite travels?

r =3

rGMT 2

4⇡2

To make comparisons, form ratios:

r1r2

=3

sT 21

T 22

Enroute to the answer we found out that:

It is the year 2094; and people are designing a new space station that will be placed in a circular orbit around the Sun. The orbital period of the station will be 6.0 years. Determine the ratio of the station’s orbital radius about the Sun to that of the Earth’s orbital radius about the Sun. Assume that the Earth’s orbit about the Sun is circular. a) 2.4 d) 5.2 b) 3.3 e) 6.0

c) 4.0

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Extra Practice: C&J 5.53 Two satellites, A and B, are in different circular orbits about the earth. The orbital speed of satellite A is three times that of satellite B. Find the ratio (TA/TB) of the periods of the satellites.

Mass of Black Hole

What is the mass of the black hole in the centre of our galaxy? (See calculation shown in class.)

Free fall and artificial gravity

•  Apparent weight = 0 because person, elevator and scale all accelerate together.

•  Apparent weight = 0 because person, satellite and scale all accelerate towards centre of earth:

€

a = v2 r

Image reprinted with permission of John Wiley and Sons, Inc.

To provide a sense of weight, spin the satellite.

•  Person in satellite is moving in a circular orbit about the centre of the satellite. There must be a force to provide the centripetal acceleration: FN.

•  Want FN to equal weight on earth, mg.

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mvspin

2

r= mg

vspin = rg

5.40 To create artificial gravity, the space station shown in the drawing is rotating at a rate of 1.00 rpm. The radii of the cylindrically shaped chambers have the ratio rA/rB = 4 . Each chamber A simulates an acceleration due to gravity of 10.0 m/s2. Find values for (a) rA, (b) rB, and (c) the acceleration due to gravity that is simulated in chamber B.

Vertical circular motion

•  On earth’s surface, don’t forget the weight of an object moving in a vertical circle. Example: ball on a string.

top

bottom

•  To keep the ball in circular motion, T must be different at the top and bottom.

•  If T = 0 at the top, then all the centripetal acceleration at that point is provided by the weight. Then the orbital speed is minimum.

•  Max. orb speed when T at bottom is maximum.

•  See text for same discussion with FN instead of T.

top bottom

C&J 5.55 A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing upward on her with a force equal to twice her weight as she goes through the dip. If r = 20.0 m, how fast is the roller coaster traveling at the bottom of the dip?

Work and energy Energy: a property of matter or radiation that manifests itself as the ability to do work.

•  A scalar quantity associated with a system.

•  S.I. units: Joules (J) 1 J = 1 kg m2/s2

See Table 6.1 for other units of energy.

Different forms of energy can be converted from one to another.

•  work: energy added to or removed from the system by an external force.

•  kinetic energy: energy of motion.

•  potential energy: energy of the configuration of the system that is available to do work.

•  internal energy: energy internal to an object (ex. heat).

Work done by a constant force

Since -1 ≤ cosθ ≤ 1, W can be positive or negative.

The maximum amount of work that can be done by a force on a object occurs when the force points in the same direction as the displacement:

Here θ = 0, cos θ = 1 so W = Fs

Only the component of the force parallel to the displacement (Fcosθ) does work.

Work is zero when:

•  displacement is zero

•  force is zero

•  force is perpendicular to displacement

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W < 0W > 0

Energy removed

Energy added

Image reprinted with permission of John Wiley and Sons, Inc.

C&J Conceptual question 6.1 The same force F pushes in four different ways on a box moving with a velocity v, as the drawings show. Rank the work done by the force F in ascending order (smallest first):

A d, c, b, a B c, b, a and d tied C a, b, c, d

D b, c, a and d tied E d, c, a, b

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C&J 6.76 A cable lifts a 1200-kg elevator at a constant velocity for a distance of 35 m. What is the work done by (a) the tension in the cable and (b) the elevator's weight?

Homework •  Read 6.1, 6.2, 6.3, •  C&J 6.20

Midterm: Preparation session with Dr. Basnet Tuesday, October 25, 12:00 – 2:00 in room 224 Education Building (next page)

Midterm: Preparation with OPUS:

The Organization of Physics Undergraduate Students is offering some study sessions for the Phys 1020 students. They’ve had really great feedback about them being useful for students in the past. This year: •  Saturday Oct 22nd … Armes 200, 1pm to 6pm •  Tuesday Oct 25th… Buller 207, 3:30pm to 8:30pm. Charge $20 for a single 5 hour session, the proceeds of which go entirely to the organization. (next page)

Midterm: •  Please continually check the course website

http://www.physics.umanitoba.ca/undergraduate/phys1020/exams.htm

•  There will be a formula sheet http://www.physics.umanitoba.ca/undergraduate/phys1020/wa_files/Phys1020_20formula_20sheet_20midterm_202016.pdf

•  Prohibited calculators are described at http://www.act.org/content/act/en/products-and-services/the-act/help.html