Circular Motion. The situation below is called uniform circular motion. 20ms -1.

Circular Motion

The situation below is called uniform circular motion.

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WHY is this called uniform circular motion?

Because the car is travelling in a circle at a Constant Speed

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Is a car accelerating when it is undertaking uniform circular motion?

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Because:The velocity is

changing. The speed may be constant but the direction is continually changing and so the car is continually accelerating as it moves around the circle.

YES?

What are two important characteristics about the acceleration on the car below?

The acceleration:• is always towards

the centre of the circle

• constant in magnitude

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Why must the acceleration be towards the centre of the circle in uniform circular motion?

When the acceleration is towards the centre of the circle it is perpendicular to the motion.Being perpendicular to the motion there is no component of the acceleration along the direction of motion so the there is no increase in the speed, BUT THERE IS A CHANGE IN DIRECTION.

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Why must the acceleration be constant in uniform circular motion?

Because the acceleration is only changing the direction of the velocity and the direction is constantly changing by the same amount to form a circle, the acceleration must be constant.

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Is there a net force on an object undergoing uniform circular motion?

Since the car is accelerating towards the centre of the circle there must be net force towards the centre of the circle.

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How can we work out the magnitude of the acceleration?

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How can we work out the magnitude of the net force?

𝑭𝒏𝒆𝒕=𝒎𝒗𝟐

𝒓20ms-1

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Possible Questions

Question 1

The 1000kg car below is travelling around the curve at a constant speed of 10.0ms-1.

(a) On the diagram below draw an arrow showing the net force on the car at X.

Fnet

Question 1The 1000kg car below is travelling around the curve at a constant speed of 10.0ms-1.

(b) On the diagram below draw arrows to show all the separate forces acting on the car at point X, ignoring air resistance. The arrows must show both the direction of the forces represented and the point of application of each force

W

N

Fr

Question 1


(c) Draw an arrow below showing the force of the car on the road at X.

Fcar on road

Fnet

Question 1


(d) At Y the car starts slowing down, draw an arrow showing the net force on the car.

Ftan

Fcent

Fnet

Question 1


(e) Draw an arrow showing the net force at Y if the car was actually speeding up.

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Question 1


(f) Draw an arrow showing the force of the car on the road in question e).

FtanFcent

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Fcar on road

Question 1


(g) What is the net force on the 1000kg car at X?

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Fnet = ? m = 1000kg v = 10ms-1 r = 10m

Fnet = 10000NFnet = 10kN

Question 1


(h) What is the acceleration on the car at X?

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a = ? m = 1000kg v = 10ms-1 r = 10m

a

a = 10ms-2

Question 1The 1000kg car below is travelling around the curve at a constant speed of 10.0ms-1.

(i) If the maximum safe speed for the car on this curve is 20ms-1

, what is the minimum safe curve radius for the car on this sort of surface if it was travelling at 40ms-1?

This question will be dealt with next Physics lesson

Question 2(a) The diagram opposite shows a car

that has been able to safely travel around a corner in icy conditions where there is negligible friction. Draw in the forces on the car.


Question 2

(b) Explain why banking will allow the car to be able to safely negotiate the curve without skidding off the road?


Question 2

(c) What is the normal reaction of the road on the car if bank is 10o to the horizontal, the curve has a radius of 100m and the car is travelling at 25ms-1 ?


Question 3

Derive the formula for working out the banking angle would be required for a car to go around a curve of radius r at a speed of v?


Question 4

What speed limit should be put on the 80m banked curve opposite if the road is subject to ice?


Question 5In the sketch below Jim is sitting in a carriage frictionlessly moving through a valley in a roller coaster ride. Jim has a mass of m kg, and the valley is close to circular, with a radius of r m. The carriage he is sitting in is travelling at v ms-1. The gravitational field is represented by g.

(a) Draw in the forces acting on Jim with appropriate relative sizes

W

N Give a qualitative reason to explain why must N be greater than WBecause Jim is undergoing uniform circular motion the net force has to be the towards the circle so N has to be greater than W.

Question 5In the sketch below Jim is sitting in a carriage frictionlessly moving through a valley in a roller coaster ride. Jim has a mass of m kg, and the valley is close to circular, with a radius of r m. The carriage he is sitting in is travelling at v ms-1. The gravitational field is represented by g.

(b) Why does Jim feel heavier?

W

NJim’s sensation of weight comes from N and the equation below shows that N will be greater than W at the bottom of the valley so Jim will feel heavier. (note: the calculation takes towards the centre of the circle as the positive direction).

= N – W + W = N

+

Question 5(c) If Jim’s mass is 65kg, r =

20m and v = 10ms-1 what is the magnitude of the acceleration of Jim at the bottom?

a = ? m = 65kg r = 20m v = 10ms-1

a = 5 ms-2

Question 5(d) If Jim’s mass is 65kg, r =

20m and v = 10ms-1 what is the Normal Reaction on Jim?

N = ? m = 65kg r = 20m v = 10ms-1

N = + W N + 65 10

N 975N

Question 5(e) If Jim’s mass is 65kg, r =

20m and v = 10ms-1 what is the G Force on Jim?

G Force = ? N = 975N W = 650NG Force = G Force =

G Force =1.5 g’s

Question 5(f) If Jim’s mass is 65kg, r =

20m and v = 10ms-1 how much heavier will Jim feel (as a percentage)?

% AW = ? N = 975N W = 650N%AW = 100%AW = 100%AW = 150%

So the Jim will feel 50% heavier% Apparent Weight

Question 5extra If Jim’s mass is

65kg, r = 20m and v = 10ms-1 what mass would register on a set of spring scales?

AM = ? N = 975N W = 650NAM = AM =

AM = 97.5kgSo the scales will register 97.5kg

% Apparent Mass

Question 5(g) A short time later Jim is travelling over a hill on the roller-

coaster. He now feels lighter than usual. Explain why Jim feels lighter than usual. Include a force diagram in your explanation?

Jim’s sensation of weight comes from N and the equation below shows that N will be less than W at the top of the valley so Jim will feel lighter. (note: the calculation takes towards the centre of the circle as the positive direction).

= W – N N = W –

N

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Question 5(h) If the hill had a radius of 12m, what is the maximum speed

that the cart can travel at without leaving the track?

vmax = ? r = 12m

vmax = vmax = vmax = 10.954ms-1

vmax 11ms-1

Question 5(i) How much lighter will Jim feel if the cart is travelling at 3.0

ms-1 remembering that he is 65kg in weight and the curvature of the hill is 12m in radius?

% AW = ? W = 650N N = 601.25 %AW = 100%AW = 100%AW = 92.5%Since the apparent weight is 92.5% Jim will feel 7.5% lighter

N = ? m = 65kg r = 12m v = 3.0ms-1

N = W – N –

N 601.25N

???

Question 6A ride at an amusement park involves passengers ‘looping the loop’, in a car as shown. The car starts from rest at A, which is 22m above C. The car and its passengers have a combined mass of 250kg. The circular part of the tract (with C at the bottom and D at the top) has a radius of 8m.

(a) Calculate the speed of the car at C (Ignore friction and air resistance?

22m

v = ? m = 250kg h = 22m g = 10Nkg-1

Ug Ek

mgh = ½ mv2

gh = ½ v2

10 × 22 = ½ × v2

220 = ½ × v2

440 = v2

20.9762 = v2

v 21 ms-1

8.0m


(b) What will be the normal reaction on the cart at C?

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N = ? m = 250kg r = 8m v = 20.9762 ms-1

N = + W N + 250 10N 16250N

N 1.6 104 N


(c) What is the minimum speed required for the cart to stay in contact with the track at point D if it is not securely attached to the track?

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vmin = ? r = 8m

vmin = vmin = vmin = 8.94427ms-1

vmin 8.9ms-1


(d) What speed will the car be going at D (if the car stays on the track)?

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8.0m

v = ? m = 250kg u = 20.9762 ms-1 h = 16m g = 10Nkg-1

Eki Ek

f+ Ug

½ mu2 = ½ mv2 + mgh

½ u2 = ½ v2 + gh

u2 = v2 + 2gh

20.97622 = v2 + 2 × 10 × 16

440 = v2 + 320

120 = v2

10.954452 = v

v 11 ms-1


(d) What speed will the car be going at D (if the car stays on the track)?

22m8.0m

v = ? m = 250kg h = 6m g = 10Nkg-1

Ug Ek

mgh = ½ mv2

gh = ½ v2

10 × 6 = ½ × v2

60 = ½ × v2

120 = v2

10.954452 = v

v 11 ms-1

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6.0m

ORThis may not be

your first choice to solve this problem but it is the only practical way of

finding the speed after a series of up and downs on a

roller coaster, because you can

ignore the ups and downs and just look at the start and end

of the path.


(e) What will be the normal reaction on the cart at D?

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8.0m

N = ? m = 250kg r = 8m v = 10.95445 ms-1

N = – W N – 250 10N 1249.999N N 1.2 103 N

Question 7A circular humpback bridge is shown below. A sign before the bridge advises: ‘Cars may lose contact with the road surface at the crest if they exceed a speed of 36kmh-1 (=10ms-1).

(a) The bridge radius (at its crest) is 10m. What is the acceleration of the truck if it is travelling at 10ms-1 at the crest.


Question 7A circular humpback bridge is shown below. A sign before the bridge advises: ‘Cars may lose contact with the road surface at the crest if they exceed a speed of 36kmh-1 (=10ms-1).

(b) Explain why the truck is likely to lose contact with the road surface at this speed.


Remainder of The Circular Motion Notes

The solution to the rest of the circular motion notes and examples are still being created.

Circular Motion. The situation below is called uniform circular motion. 20ms -1.

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