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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