Circular Motion Flashcards

(31 cards)

1
Q

Define time period.

A

The time elapsed during one full oscillation.

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2
Q

Define frequency.

A

The number of rotations per unit time.

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3
Q

Define velocity.

A

The rate of change of displacement.

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4
Q

Define angular velocity.

A

The rate of change of angle.

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5
Q

Give an equation for angular velocity based upon time period and theta.

A

omega = theta/T

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6
Q

How do linear and angular velocity vary with radius?

A

Linear velocity decreases as a greater distance (circumference) is covered per unit time.
Angular velocity remains constant.

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7
Q

Give equations relation angular velocity and frequency/time period.

A

omega = 2pif
omega = 2*pi/T

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8
Q

Give an equation relation linear and angular velocity.

A

v=omega*r

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9
Q

How can we make an object describe a circular path of motion?

A

Apply a force directly perpendicular to the direction of it’s velocity.

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10
Q

What is a centripetal force?

A

A force acting towards the centre of a particle’s circular path, perpendicular to the direction of it’s motion.

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11
Q

Why is no work done when an object moves in a circular path?

A

The force applied is perpendicular to the direction of motion - no distance traveled in direction of force

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12
Q

Why is velocity constant in circular motion?

A

No work is done thus Ek is constant.

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13
Q

Why do objects in circular motion accelerate, despite their velocity not changing?

A

The object’s direction is constantly changing. Velocity is a vector quantity thus it is changing, so the object accelerates.

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14
Q

What happens when the force providing the centripetal acceleration of an object is less than the centripetal force required to describe a circular path?

A

The object does not move in a. circular path; it begins to spiral outward.

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15
Q

Give the equation for centripetal acceleration.

A

a=omega^2*r

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16
Q

Give an equation for centripetal force.

A

F=mv^2/r = momega2*r

17
Q

When a car orbits a roundabout, what provides the centripetal force required to do so?

A

Friction due to contact between the tyres and the road.

18
Q

How can we use centripetal forces to explain understeer in a car?

A
  • As speed increases, centripetal force required to maintain circular path rises
  • Friction has a maximum value given by mu*R
  • This will eventually become less than reqd centripetal force
  • The car’s path spirals and you understeer (very dangerous situation to be in!!!)
19
Q

Why is a greater centripetal force generated on a banked road?

A

The horizontal component of friction is added to by the horizontal component of the normal reaction force

20
Q

For a conical pendulum setup, what provides the centripetal force?

A

Horizontal component of tension = Tsin theta

21
Q

For a conical pendulum to remain level, what criteria must be satisfied?

A

Tcos theta = mg so resultant vertical component of force is equal to zero

22
Q

What happens when we increase the speed of a conical pendulum?

A
  • More centripetal force required
  • Tsin theta increases
  • Thus theta increases so cos theta decreases
  • T must increase to maintain Tcos theta = mg for decreasing cos theta
23
Q

Why is a conical pendulum never completely horizontal?

A

If theta = 90 then T cos theta = 0, which can never equal mg given a pendulum bob must have mass!!

24
Q

Describe the resultant force at the top of a Ferris wheel in vertical circular motion.

A

mv^2/r = mg-R

25
Describe the resultant force halfway around a Ferris wheel in vertical circular motion.
mv^2/r, R=mg
26
Describe the resultant force at the bottom of a Ferris wheel in vertical circular motion.
mv^2/r = R-mg
27
How is it possible to become weightless on a ferris wheel?
Where R = 0, we lose contact with the ground, thus you are acted upon by weight = mg completely unopposed.
28
How can we derive the speed at which weightlessness in vertical circular motion occurs?
mv^2/r = mg-R v = sqrt(rg)
29
Describe the resultant force upon a ball on a string at the top of its motion.
mv^2/r = T+mg
30
Describe the resultant force upon a ball on a string at the bottom of its motion.
mv^2/r = T-mg
31
Why can the water in a spinning bucket be made to remain it the bucket as it is whirled vertically?
- The acceleration of the bucket must be equal to or greater than that of the water (=g) - This can be equated to v^2/r to determine the speed at which the bucket must be whirled to achieve this