Circular motion & gravitation Flashcards Preview

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Flashcards in Circular motion & gravitation Deck (12)
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1
Q

What is, and how do you work out…
- Angular displacement
- Angular velocity
…of motion along a circular path?

A
  • It is the angle subtended by A and B when an object moves from A to B, calculated by s/r (rad)
  • It is the rate of change of angular displacement,
    ω = Δθ/Δt (rad s^-1)
    ω = 2π/T = 2πf , taken from one complete revolution.
2
Q

What is the frequency and time period?

A
  • The frequency is the number of oscillations per second (Hz) = 1/T
  • The time period is the time it takes for one oscillation = 1/f
3
Q

What is the linear speed of circular motion? How is it derived?

A

The speed with which an object travels along a circular path.
(2πr)/T = v and T = (2π)/ω gives rω = v

4
Q

Describe the speed, velocity, and acceleration of circular motion in the horizontal plane.

A

As an object moves around a circular path at a constant speed, the direction of its velocity constantly changes and, and therefore it accelerates. The magnitude of the velocity remains the same.

5
Q

What is the derivation for centripetal acceleration?

How do you know the direction of centripetal acceleration?

A

As an object moves around a circle from A to B, the initial and final velocities can be combined to find Δv, which is approximately the arc length of a circle radius u or v, angle Δθ. From a = Δv/Δt, a = vΔθ/Δt = ωv.
Substituting ω = v/r, we get a = v²/r
More expressions can be found substituting in things for v.
As Δθ becomes very small, Δv becomes perpendicular to v, which shows that the acceleration is towards the centre of the circle.

6
Q

What is a centripetal force? The formula is in the data book.

A

It is the resultant force acting upon a body travelling in a circular path that is directed towards the centre of the circle, which provides the acceleration. It is a physical force that is different depending upon the situation, either friction, tension in a rope, or gravitational force.

7
Q

How can you solve a problem with an object dangling from a string in circular motion?

A

Resolve the angled tension in the rope into horizontal and vertical components. Ty is equal to the weight, mg of the object, and Tx represents the centripetal force.

8
Q

What is the nature of vertical circular motion?

What are the forces at the top, bottom, and sides of the circle?

A

There is always two forces acting on the object at once, the tension/other and the weight. The resultant of these two forces is the centripetal force, directed towards the centre of the circle. At the top of the circle, mg and T act in the same direction, so F(resultant) = T + mg. At the bottom, F = T - mg.
The tension waves between F + mg and F - mg. At the side, F = T, as mg has no horizontal component.

9
Q

What is the work done by a centripetal force?

A

0 because it is at right angles to the velocity.

10
Q

What is Newton’s universal law of gravitation

A

F = G (Mm)/r² , there is an attractive force between any two point masses (small masses in comparison to their separation). Both masses will experience force F along the line connecting the two masses. Describe in words: force proportional to product of masses, inversely to square of separation.

11
Q

What is…
a gravitational field?
gravitational field strength?

A
  • An imaginary region around a mass where other masses experience a gravitational force, illustrated with radial arrows going into a mass.
  • The gravitational force acting per unit mass on a mass placed in a gravitational field. g (N kg^-1). It is a vector equal to the acceleration in free-fall.
12
Q

How can you work out the forces and velocities involved in orbital motion?

A

Gravity provides the centripetal force, so mv²/r = GMm/r² . Solve for v to get v = root (GM/r).
Substitute in v² = 4π²r²/T² (formula booklet), to show that T² ∝ r³