Download Circular Motion

Circular Motion
AP PHYSICS 2
Uniform Circular Motion
In uniform circular motion a particle is not changing
speed, but is accelerating since it is constantly changing
direction.
The velocity vector is always tangent to the direction of
motion, the acceleration vector points toward the center.
𝑎=
𝑣2
𝑟
𝑇=
2𝜋𝑟
𝑣
(centripetal acceleration)
(period of revolution)
Example 1
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.
𝑟 = 3.84 × 108 𝑚
𝑎=
𝑣2
,
𝑟
𝑣=
2𝜋𝑟
𝑇
𝑇 = 27.3 𝑑
∴ 𝑎=
(2𝜋𝑟)2
𝑇2𝑟
=
ℎ
24.0 𝑑
𝑠
3600 ℎ = 2.36 × 106 𝑠
[2 3.14 3.84 × 108 𝑚 ]2
2.36 × 106
𝑚
2 (3.84 × 108 ) = .00272 𝑠 2
Example 2
A 0.150 kg ball on the end of a 1.10 m-long cord is swung in a vertical
circle. A) Determine the minimum speed the ball must have at the top
of its arc so that it 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.
Remember: 𝑎 =
𝑣2
𝑟
and Σ𝐹 = 𝑚𝑎, so at the top of
the swing: 𝐹𝑇1 + 𝑚𝑔 =
𝑣2
𝑚𝑟
What is the minimum tension on the cord at the top of
the swing if it is to continue swinging?
𝐹𝑇1 = 0 𝑚𝑔 =
𝑣2
𝑚𝑟
, 𝑣 = 𝑔𝑟 =
𝑚
(9.80 𝑠2)(1.10 𝑚) = 3.28
𝑚
𝑠
Example 2 cont’d
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.
𝑣2
𝑟
Again: 𝑎 = and Σ𝐹 = 𝑚𝑎. What forces are acting on the
ball at the bottom of the swing?
𝑣2
𝐹𝑇2 − 𝑚𝑔 = 𝑚
𝑟
𝑣2
𝐹𝑇2 = 𝑚 𝑟 + 𝑚𝑔
= 0.150 𝑘𝑔
= 7.34 𝑁
𝑚 2
6.56 𝑠
+ 0.150 𝑘𝑔
1.10 𝑚
𝑚
9.80 2
𝑠
Newton’s Law of Universal
Gravitation
Every particle in the universe attracts every other particle with a force
that is proportional to the product of their masses and inversely
proportional to the square of the distance between them. This force
acts along the line joining the two particles.
𝐹=
𝑚 𝑚
𝐺 1 2 2,
𝑟
𝐺 = 6.67 ×
2
𝑁∙𝑚
10−11 2
𝑘𝑔
Example 3
What is the force of gravity acting on a 2000 kg spacecraft when it orbits
two Earth radii from Earth’s center (that is, a distance 𝑟𝑒 = 6380 𝑘𝑚
above the earth’s surface)?
The mass of the Earth is 𝑀𝐸 = 5.98 × 1024 𝑘𝑔.
(5.98 × 1024 𝑘𝑔)(2000 𝑘𝑔)
𝐹=𝐺
= 4900 𝑁
(2 ∗ 6.380 × 106 𝑚)2
Example 4
A geosynchronous satellite is one that stays above the same point on
the equator of the Earth. Such satellites are used for such purposes as
cable TV transmission, for weather forecasting, and as communications
relays. Determine a) the height above the Earth’s surface such a
satellite must orbit and b) such a satellite’s speed.
What forces are acting on the satellite?
𝑚𝑠𝑎𝑡 𝑚𝐸
𝑣2
𝐹=𝐺
= 𝑚𝑠𝑎𝑡
𝑟2
𝑟
Since it is geosynchronous it orbits once every 24 hours, thus:
𝑣=
2𝜋𝑟
𝑇
where 𝑇 = 1 𝑑𝑎𝑦 = 86400 𝑠. Substituting this equation gives us:
𝑚𝑠𝑎𝑡 𝑚𝐸
(2𝜋𝑟)2
𝐺
= 𝑚𝑠𝑎𝑡
𝑟2
𝑟𝑇 2
Example 4 cont’d
𝑚 𝑚
𝐺 𝑠𝑎𝑡2 𝐸
𝑟
𝑟=
3
=
(2𝜋𝑟)2
𝑚𝑠𝑎𝑡
,
𝑟𝑇 2
𝐺𝑚𝐸 𝑇 2
4𝜋2
solve for r:
= 4.23 × 107 or 42,300 km from the Earth’s center.
Subtract 6380 km (the diameter of the Earth) to give (about) 36,000 km
b)
𝑚 𝑚
𝐺 𝑟1 2 2
=
𝑣2
𝑚 𝑟,
solve for v: 𝑣 =
𝐺𝑚𝐸
𝑟
= 3070
𝑚
𝑠
Homework!
Chapter 5: Questions 1, 9 Problems 7, 25, 29, 39