Periodic Motion: Circular Motion

Question 1

What is uniform circular motion?

Answer 1

An object will follow a circular path at a constant speed if a constant perpendicular resultant force is applied to the body.

Question 2

Angular displacement is measured in?

Answer 2

radians

Question 3

Centripetal acceleration is?

Answer 3

Acceleration towards the centre of the circle.

Question 4

What’s centripetal force?

Answer 4

The resultant force required to produce uniform circular motion. Acts towards centre of circle (same direction as circular acceleration).

Question 5

Why is no work done by the centripetal force on the object?

Answer 5

Centripetal force acts at right angles to the direction of the object’s velocity, ∴ no work done by centripetal force on the object bc there’s no displacement in the direction of the force.

Question 6

Vertical circles: write out force eq for ball at top, bottom, and side, relating weight to tension in string.

Answer 6

At top of circle: F = T + W
At bottom of circle: F = T - W
Halfway between: F = T bc weight acting perpendicular ∴ has no effect on force.

Question 7

Over the top of a hill, force eq?

Answer 7

mg - S = mv^2/r

(S = normal contact force)

Question 8

Leaving the ground - condition, and eq for speed?

Answer 8

Vehicle will lose contact w road if speed equal to or greater than a certain speed, v0.
If this happens, S = 0.

∴ mg = mv0^2/r
v0^2 = gr

Question 9

Centripetal force is provided by sideways force of friction between tyres and road.
There’s a max friction that can exist ∴ max velocity for uniform circular motion. What is this limiting force of friction?
What’s the condition for no skidding?

Answer 9

F0 = mv0^2/r

For no skidding to occur v < v0

Question 10

(textbook)
On a banked track: w/o banking, centripetal force = friction. On banked, track, speed can be higher. For no sideways friction…

Answer 10

For no sideways friction, horizontal component of support forces must act as centripetal force.
(N1 + N2) sinθ = mv^2/r

(N1 + N2) cosθ - mg (vertical balances the weight)

∴ tanθ = (N1 + N2) sinθ/(N1 + N2) cosθ = mv^2/mgr
Condition for no sideways friction = tanθ = v^2/gr

In other words, no sideways friction if v^2 = grtanθ

Question 11

Sv balances weight. Sh provides centripetal force.
F=?

Answer 11

F= mv^2/r
F = max friction + (N1 + N2) sinθ
Can have larger resultant force ∴ larger velocity
v = sqrt Fr/m

Question 12

Q Explain why a circular athletics track is banked for sprinters but not for marathon runners. (6)

Answer 12

-Centripetal force is resultant force required for uniform circular motion.
-W/o banking the centripetal force is provided by friction.
-For a given radius and mass of runners, centripetal force is proportional to v^2
-Typical speed of sprinters is greater than marathon runners so sprinters need a greater resultant force towards centre of circle.
-Adding banking to tracks increases resultant force (towards centre of circle)
-Bc reaction force will have component towards centre of circle.

Question 13

Q: If a car took a bend too fast?

Answer 13

increased v = increased centripetal force. if centripetal > resultant, radius of path increase until centripetal = resultant. car may leave track.

Question 14

What is ‘g force’?

Answer 14

Used to compare the reaction force we experience during a fairground ride vs reaction force we experience when on the ground.

Question 15

What’s the experience ‘weightlessness’?

Answer 15

Occurs when reaction force on us is zero (∴0g force)

mv^2/r = mg
v^2/r = g

Question 16

Big dipper?

Answer 16

S - mg = mv^2/r

Will experience more than 1g.

Question 17

Very long swing?

Answer 17

1/2mv^2 = mgh
S - mg = mv^2/r
S - mg = 2mgh/r

Question 18

Big wheel?

R = reaction force

Answer 18

mg + R = mv^2/r
Weight of person contributes to centripetal force ∴ reaction force at a minimum. Less than 1g.
At v0, where v0^2 = gr, R=0 ∴ no force on person due to the wheel.