Circular Motion

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Last updated 10:10 PM on 5/12/26
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28 Terms

1
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What is the equation for critical speed for circular motion in a vertical plane?

v = √(gr)

Critical speed = √(acceleration of free fall Ɨ radius)

2
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How do you derive equation for critical speed for circular motion in a vertical plane?

Weight = centripetal force, W = mv2/r

mg = mv2/r

g = v2/r

v2 = gr

v = √(gr)

3
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For circular motion in a vertical plane, what is the centripetal force equal to at the top of the circle when at the critical velocity?

The weight is equal to the centripetal force

4
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Why are some sharp bends in roads banked?

  • When a vehicle hoes round a bend on a flat road, the centripetal force is provided by friction between tyres and road.

  • Cars must slow down to ensure the available friction is enough to provide the required centripetal force

  • Additional centripetal force can be provided by banking the road so that there’s a horizontal component of the normal contact force which acts towards the centre of the bend. This allows cars to travell faster round the bend

5
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Why do aeroplanes bank their wings to turn?

When šœƒ is the angle of banking

  • The horizontal component of the lift šæš‘ š‘–š‘›šœƒ is unbalanced and provides a centripetal force so the plane will follow a horizontal circular path.

  • The vertical component šæš‘š‘œš‘ šœƒ is balanced by the weight.

6
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What is centripetal force?

  • The force required to maintain circular motion

  • It magnitude is constant and it acts perpendicular to velocity and towards the centre of the circle

7
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Definition of Angular velocity

The rate of change of angular displacement of an object moving in a circular path

8
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Derive ω = 2Ļ€/T

Angular velocity is the rate of change of angle, so ω = Īø/t

In a time t equal to one period T, the object will move through an angle Īø equal to 2Ļ€ radians.

Therefore ω = 2Ļ€/T

9
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Derive ω = 2Ļ€f

ω = 2Ļ€/T and f = 1/T, so ω = 2Ļ€f

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Derivation v = rω

  • For a constant speed v = distance travelled / time taken

  • When distance traveled = circumference of circle = 2Ļ€r, t = time period = T

  • v = 2Ļ€r/T = r Ɨ 2Ļ€/T

  • Angular velocity = 2Ļ€/T, so v = rω

11
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How can a body moving at a constant speed be accelerating?

A change in direction of motion changes velocity so causes acceleration (even at constant speed as velocity is a vector)

12
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What is uniform circular motion?

An obejct moving in a circular path that:

  • Has constant speed

  • Only has centripetal acceleration, so no tangential acceleration

13
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A ball is attached to a string and whirled in a vertical circle. Draw a freebody diagram of the ball at the top of this circle and derive the expression for the Tension at this position.

Centripetal force F = mg + T

T = F - mg = mv²/r - mg

<p>Centripetal force F = mg + T</p><p>T = F - mg = mv²/r - mg</p>
14
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A ball is attached to a string and whirled in a vertical circle. Draw a freebody diagram of the ball at the side of this circle and derive the expression for the Tension at this position.

Centripetal force F = T

T = F = mv²/r

<p>Centripetal force F = T</p><p>T = F = mv²/r</p>
15
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A ball is attached to a string and whirled in a vertical circle. Is the ball’s motion uniform circular motion and why?

  • When ball is at side of circle, there is no focre to balance the ball’s weight

  • As such there is a tangential acceleration so the ball’s speed must be changing

  • This means the ball does not have constant speed, so is not moving in uniform circular motion

<ul><li><p>When ball is at side of circle, there is no focre to balance the ball’s weight</p></li><li><p>As such there is a tangential acceleration so the ball’s speed must be changing</p></li><li><p>This means the ball does not have constant speed, so is not moving in uniform circular motion</p></li></ul><p></p>
16
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A ball is attached to a string and whirled in a vertical circle. Draw a freebody diagram of the ball at the bottom of this circle and derive the expression for the Tension at this position.

Centripetal force F = T - mg

T = F + mg = mv²/r + mg

<p>Centripetal force F = T - mg</p><p>T = F + mg = mv²/r + mg</p>
17
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<p><span>A plane is flying in a vertical loop. Draw a freebody diagram of the plane at the top of this loop and derive the expression for the lift force at this position.</span></p>

A plane is flying in a vertical loop. Draw a freebody diagram of the plane at the top of this loop and derive the expression for the lift force at this position.

Centripetal force F = mg + L

L = F - mg = mv²/r - mg

<p>Centripetal force F = mg + L</p><p>L = F - mg = mv²/r - mg</p>
18
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<p><span>A plane is flying in a vertical loop. Draw a freebody diagram of the plane at the side of this loop and derive the expression for the lift force at this position.</span></p>

A plane is flying in a vertical loop. Draw a freebody diagram of the plane at the side of this loop and derive the expression for the lift force at this position.

Centripetal force F = L

L = F = mv²/r

<p>Centripetal force F = L</p><p>L = F = mv²/r</p>
19
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<p>A plane is flying in a vertical loop. Is the plane’s motion <strong>uniform </strong>circular motion and why?</p>

A plane is flying in a vertical loop. Is the plane’s motion uniform circular motion and why?

  • When plane is at the side of the circle, there is no focre to balance the plane’s weight

  • As such there is a tangential acceleration so the plane’s speed must be changing

  • This means the plane does not have constant speed, so is not moving in uniform circular motion

<ul><li><p>When plane is at the side of the circle, there is no focre to balance the plane’s weight</p></li><li><p>As such there is a tangential acceleration so the plane’s speed must be changing</p></li><li><p>This means the plane does not have constant speed, so is not moving in uniform circular motion</p></li></ul><p></p>
20
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<p>A plane is flying in a vertical loop. Draw a free body diagram of the plane at the bottom of this loop and derive the expression for the lift force at this position.</p>

A plane is flying in a vertical loop. Draw a free body diagram of the plane at the bottom of this loop and derive the expression for the lift force at this position.

Centripetal force F = L - mg

L = F + mg = mv²/r + mg

<p>Centripetal force F = L - mg</p><p>L = F + mg = mv²/r + mg</p>
21
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<p><span>A person stands on a Ferris wheel. Draw a freebody diagram of the person at the top of this loop and derive the expression for the normal contact force at this position.</span></p>

A person stands on a Ferris wheel. Draw a freebody diagram of the person at the top of this loop and derive the expression for the normal contact force at this position.

Centripetal force F = mg - N

Normal contact N = mg - F

<p>Centripetal force F = mg - N</p><p>Normal contact N = mg - F</p>
22
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<p>A person stands on a Ferris wheel. Draw a freebody diagram of the person at the left side of this loop and derive the expression for the normal contact force at this position.</p>

A person stands on a Ferris wheel. Draw a freebody diagram of the person at the left side of this loop and derive the expression for the normal contact force at this position.

  • Normal contact = weight = mg

  • Centripetal force provide by friction between person and ferris wheel floor

<ul><li><p>Normal contact = weight = mg</p></li><li><p>Centripetal force provide by friction between person and ferris wheel floor</p></li></ul><p></p>
23
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<p>A person stands on a Ferris wheel which spins. Is the person’s motion <strong>uniform </strong>circular motion and why?</p>

A person stands on a Ferris wheel which spins. Is the person’s motion uniform circular motion and why?

  • At all points on the circle, there is not tangentiall acceleration

  • For example, at the side of the circle the weight is balanced by the normal contact force, so no tangential force or acceleration

  • Therefore speed of person is constant, and they only experience centripetal acceleration, so person’s motion is uniform circular motion

<ul><li><p>At all points on the circle, there is not tangentiall acceleration</p></li><li><p>For example, at the side of the circle the weight is balanced by the normal contact force, so no tangential force or acceleration</p></li><li><p>Therefore speed of person is constant, and they only experience centripetal acceleration, so person’s motion <strong>is </strong>uniform circular motion</p></li></ul><p></p>
24
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<p><span>A person stands on a Ferris wheel. Draw a freebody diagram of the person at the bottom of this loop and derive the expression for the normal contact force at this position.</span></p>

A person stands on a Ferris wheel. Draw a freebody diagram of the person at the bottom of this loop and derive the expression for the normal contact force at this position.

Centripetal force F = N - mg

Normal contact N = F + mg

<p>Centripetal force F = N - mg</p><p>Normal contact N = F + mg</p>
25
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Angular displacement (Īø)

The angle, Īø, swept out in radians.

26
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Radian

The angle subtended by a circular arc that is equal in length to the radius of the circle

27
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Centripetal acceleration
Acceleration towards the centre of the circular path.
28
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Why does the centripetal force not do work on the object?

  • It acts perpedicular to the direction of motion so does not change the object’s speed and therefore energy

  • No energy change so centripetal force doesn’t do work on object