UNIT I: CONSERVATION LAWS
Circular Motion
Horizontal Circular Motion
▪ One complete rotation is called a cycle or revolution
▪ The time for one cycle is the period (T), measured in seconds
▪ The frequency (f) measures the number of cycles per second (Hz)
▪ Each is the inverse of the other:
T = 1 f = 1
f T
▪ Rotational frequency can also be measured in rpm (revolutions per minute)
▪ The speed of an object in circular motion is given by:
[pic]
where R = radius of circle (m)
T = period (s)
▪ The centripetal force is the force keeping the object in a circular path
[pic]
▪ Objects experience centripetal acceleration towards the centre of the circle (due to the centripetal force), causing travel in a circular path
v
ac [pic]
▪ For a vehicle travelling in a circular path, friction is the force keeping the vehicle in a circle, so friction is the centripetal force
Fc = Ff
Vertical Circular Motion
▪ The centripetal force varies for an object in vertical circular motion
Eg1) A 1.7 kg object is swung from the end of a 0.60 m string in a vertical circle. If the time of one revolution is 1.1 s, find the tension in the string
a) when it is at the top
T Fg
Fc = T + Fg
4π2Rm = T + mg
T2
4π2(0.60 m)(1.7 kg) = T + (1.7 kg)(9.81 m/s2)
(1.1s)2
T = 17 N down
b) when it is at the bottom
T
-Fg
Fc = T + (-Fg)
4π2Rm = T + (-mg)
T2
4π2(0.60 m)(1.7 kg) = T - (1.7 kg)(9.81 m/s2)
(1.1 s)2
T = 50 N up
Eg2) A 10 kg object is swung in a vertical circle with a radius of 0.75 m. Find the minimum speed of the object at the top of its motion for it to remain in circular motion.
T Fg
Fc = T + Fg
Minimum speed will occur when T = 0
Fc = Fg
[pic] = mg
(10 kg)v2 = (10 kg)(9.81 m/s2)
0.75 m
v = 2.7 m/s
Eg3) A person with a weight of 588.6 N is riding on a roller coaster that is moving at 25.0 m/s through a dip with a radius of 20.0 m. Find the person’s apparent weight.
FN
-Fg
Fc = FN + (– Fg)
[pic] = FN + (–mg)
(60 kg)(25 m/s)2 = FN – (60 kg)(9.81 m/s2)
20 m
FN = 2.46 x 103 N
If travelling over a hill:
FN
-Fg
Fc = FN + (-Fg)
Eg4) A string requires 135 N of force in order to break. A 2.00 kg mass is tied to the string and whirled in a vertical circle with a radius of 1.10 m. Find the maximum speed the mass can be whirled without breaking the string.
The greatest tension will occur at the bottom of the circle:
T
-Fg
Fc = T + (– Fg)
[pic] = T – Fg
Find v when T = 135 N
(2 kg)(v)2 = 135 N – (2 kg)(9.81 m/s2)
1.10 m
v = 7.97 m/s
Kepler’s Laws
▪ Kepler’s 1st Law:
The motion of an orbiting body follows an elliptical path, with the orbited body at one of the foci.
[pic]
▪ Kepler’s 2nd Law:
An orbiting body sweeps out equal areas in equal times.
▪ Kepler’s 3rd Law:
All objects which orbit the SAME body have the same KEPLER’S CONSTANT …
K = T2
R3
where T = period of the object’s orbit
R = radius of object’s orbit
Since two objects orbiting the same body have the same K then:
K1 = K2
T12 = T22
R13 R23 (units have to be the same on both sides)
Eg1) Calculate Kepler’s Constant for Earth. The radius of Earth’s orbit around the Sun is 1.490 x 1011 m.
K = T2
R3
K = (1 year)2
(1.49 x 1011)3
K = 3.023 x 10-34 year2/m3
Eg2) Neptune has an orbit of 4.50 x 1012 m from the Sun. How long does it take Neptune to complete one orbit?
- Since Earth and Neptune orbit the same body,
KEarth = KNeptune
TE2 = TN2
RE3 RN3
(1 year)2 = TN2 _
(1.49 x 1011)3 (4.50 x 1012)3
T = 166 years
Satellites in Orbit
▪ The centripetal force acting on satellite is the gravitational force attracting the satellite to the orbited body
Fc = Fg
[pic]
[pic]
where v = speed of satellite
R = radius of satellite’s orbit
mE = mass of ORBITED BODY
Eg1) Calculate the speed of a 985 kg artificial satellite around Earth that has an orbital radius of 6.90 x 106 m.
[pic]
[pic]
v = 7.60 x 103 m/s
Eg2) LandStat is an Earth-imaging satellite that takes pictures of Earth’s ozone layer and geological features. It orbits Earth at a height of 912 km. How many minutes does it take LandStat to orbit the Earth?
[pic]
[pic]
v = 7.402 x 103 m/s
[pic]
[pic]
T = 6.1813 x 103 s
T = 103 minutes
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