Chapter 6 and 7. Circular Motion + Gravitation

Question 1

What is

1. Angular Velocity (ω)?
2. Instantaneous Angular Velocity?
3. Units of ω?

Answer 1

1. Rate of Change of Angular Displacement
   ∆θ/∆t
2. Angular velocity that exists at any given instant
   dθ/dt
3. rad/s

Question 2

Formula of Tangential Speed (v)?

Answer 2

v = rω

Question 3

Formula of Angular Velocity (ω)?

Answer 3

ω = 2π/T = 2πf

Question 4

Formula of

1. Centripetal Acceleration and
2. Centripetal Force and where is it acting?

Answer 4

1. ac = v^2/r = rω^2 = vω

2. Fac = m * ac , always pointed to centre of circle

Question 5

What is the tension in the string connected to a ball swirling (nearly) horizontally?

Answer 5

The centripetal force acting on the ball

Question 6

What force is centripetal force considered? Do you label it in FBD?

Answer 6

Resultant force, therefore it is not a new force, and is caused by normal contact+-weight or tension+-weight.
No, you never label resultant forces in FBD.

Question 7

Does magnitude of centripetal force change at different positions in a vertical circle? How?

Answer 7

Yes, largest at bottom, smallest at the top

Question 8

Solving Circular Motion Qns Framework

Answer 8

1. Identify Object undergoing circular motion
2. Draw FBD
3. Resolve forces into perpendicular components
4. Apply Newton’s 2nd Law and solve

Question 9

Define Newton’s Law of Universal Gravitation.

Answer 9

Every particle attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them

F = (GMm)/(r^2)

Question 10

State the key features of Geostationary Satellites and its advantages and disadvantages

Answer 10

Key Features :

1. Period is 24h
2. Direction of rotation is same direction as the rotation of the Earth. (West to East)
3. Plane of rotation contains the equator

Advantages :

1. Remain stationary above the same point on Earth, there require no tracking and is ideal for use as communication satellites since there is no need to adjust the satellite dish to receive TV signals
2. Positioned at a high altitude so they can view almost the whole Earth disk below them. Ideal for remote imaging.

Disadvantages :

1. High altitude so spatial resolution of their images tend to be not as good
2. Since they are positioned above the equator, they cannot see the north or south poles and are of limited use for latitudes greater than 60-70 degrees north or south.

Question 11

Give 3 reasons why value of g varies at various regions on Earth.

Answer 11

Earth is not a perfect sphere. Flattened at poles and bulging at the equator

2. Density of Earth is not uniform. (Eg. Northern hemisphere has more land area than the southern hemisphere)
3. Earth is rotating. There gravitational pull on a body has to also provide the body with a centripetal acceleration. Therefore measured gravitational acceleration is smaller. (N=Fg-mac)

Question 12

Compare the differences between True Weightlessness and Apparent Weightlessness

Answer 12

1. True weightlessness - no net gravitational force (g=0)
   Eg. infinite distance away from any other body or b/w Earth and Moon where gravitational fields cancel each other
2. Apparent weightlessness - exerts no contact force on its support. W=N=0 . Object in freefall

Question 13

Define Kepler’s Third Law

Answer 13

Square of the time of revolution of the planet T about the Sun is proportional to the cube of its mean distances r from it

Question 14

Name and define and give formula for :

1. Fg
2. g
3. U
4. ø

Answer 14

1. F = (GMm)/(r^2)
2. Gravitational field strength : The gravitational field strength g at a point is defined as the gravitational force per unit mass action on a mass placed at the point. g=(Fg)/m = (GM)/(r^2)
3. Gravitational potential energy : The gravitational potential energy of a mass at a point in a gravitational field is the work done by an external force in bringing that mass from infinity to the point, without any change in KE. U = -(GMm)/(r)
4. Gravitational potential : The gravitational potential at a point in a gravitational field is the work done per unit mass, by an external force, in bringing the mass from infinity to a point, without any change in KE.
   ø = -(GM)/r

Question 15

Explain why U is negative.

Answer 15

Gravitational force is attractive in nature. Therefore, POSITIVE work needs to be done by an external force to INCREASE the separation between 2 masses. The work done by the external force causes GPE of the system to increase as the distances between the masses increases. By definition, GPE is 0 at infinity. Hence the masses have a GPE lower than zero when they are a finite distance apart. Thus, GPE is negative

Question 16

Derive KE, PE and TE

Answer 16

1. Fac = (GMm)/(r^2)
   m(v^2/r) = (GMm)/(r^2)
2. change h is a lot smaller than Radius, so R(R+H) = r^2
   change in U = GMm(h/R^2) = mgh

TE = KE + PE = -(GMm)/(2r)

Question 17

What is the value of Gravitational Constant G?

Answer 17

6.67x10-11 Nm^2 kg^-2

Question 18

Finding centripetal acceleration and what to note?

Answer 18

Fg - N = Fac

Wrong to believe Fac = Fg

Question 19

Finding total GPE of a system and work done in bringing another mass to the centre

Answer 19

1. -Gm1m2/r - Gm2m3/r - Gm1m3/r

2. -Gm1m4/r - Gm2m4/r - Gm3m4/r, take angle/2 to locate centre