Chapter 16 - Circular Motion

Question 1

define the radian

Answer 1

“A radian is the angle subtended by a circular arc with length equal to the radius of the circle”

Question 2

what is the equation for the radian

Answer 2

Angle (radians) = arc length/radius

Question 3

define angular velocity

Answer 3

“the angular velocity, (lowercase omega w), of an object moving in a circular path is defined as the rate of change of angle”

Question 4

what are the three equations for angular velocity

Answer 4

w = delta theta/ time

w = 2(Pi)/T
(T is period)

w = 2f(Pi)
(f is frequency)

Question 5

what are the units for angular velocity

Answer 5

radS^-1

usually

Question 6

why is something moving in a circular path accelerating and what does this mean

Answer 6

• when moving on a circular path at a constant speed your velocity is still changing because the direction of travel changes and velocity is a vector
• this means there is an acceleration
• this means there must be a resultant force

Question 7

what is the name of this resultant force acting when an object moves on a circular path

Answer 7

the centripetal force

Question 8

what are the key features of the centripetal force

Answer 8

• it acts perpendicular to velocity
• it acts towards the centre of the circle
• it has no ‘horizontal’ component in the same direction as the object’s tangental speed so the speed of the object remains constant

Question 9

how can we derive the equations for linear velocity and angular velocity

Answer 9

speed = distance/time
distance in this case = 2r(pi)
and time = T (period)
because it’s the circumference of the circle so

v = 2r(pi)/T

as w = 2(pi)/T this gives us
V = wr

Question 10

what is the equation linking linear and angular velocity

Answer 10

v = wr

Question 11

what is the derivation for the equation using acceleration and radius

Answer 11

• imagine a small section of a circle, radius r
• the angle is deltatheta, the arc length is delta(x), the velocity at one end of the arc is Va and Vb at the other

By definition delta(theta) = delta(x)/r

As Va and Vb act in different directions there must have been an acceleration, this is given by
a = delta(v)/delta(t)

as delta(x) is small, we can model it as a straight line so 
delta(x) = v x delta(t)

substituting gives
delta(theta) = vdelta(t)/r
as delta(theta) also = delta(v)/v
this means

vdelta(t)/r = delta(v)/v

so v^2/r = delta(v)/delta(r) = a (from part 1)

thus V^2/r = a

Question 12

what are the two equations for acceleration on a circle

Answer 12

a = V^2/r
and
a = w^2 x r

Question 13

how to derive the two centripetal force equations

Answer 13

• we know F = Ma
• we know a = V^2/r and a = (w^2)r

so therefore
F = mv^2/r
F = m(w^2)r

Question 14

what are the three derivations you need to know

Answer 14

1) derivation of v = wr
2) derivation of a = V^2/r and a = w^2r
3) derivation of F = mv^2/r and F = mw^2r

Question 15

what is an experiment we can do to investigate circular motion

Answer 15

• set up equipment where there is a mass on a string, passing through a metal/glass tube, attached to another mass
• the length of the string above the tube is r
• we know that the faster we swing the mass, the more likely the hanging mass is to rise up
• this is because swinging the mass faster means it requires a greater centripetal force, if the weight of the other mass isn’t sufficient then the weight will rise up due to tension in the string
• if the centripetal force required is lower than the weight of the hanging mass then the hanging mass will drop
• this along with our equations allows us to do a variety of experiments and plot different graphs

Question 16

what is a centrifuge and briefly explain how it works

Answer 16

• a centrifuge is a device which spins samples of liquids in tubes in order to separate the components of the liquid
• the heavier components of the liquid will travel to the outside (bottom of the tube) because they will require a greater centripetal force to keep them in place
• the lighter components of the liquid will remain on top

Question 17

why do banked surfaces effectively increase the centripetal force

Answer 17

• a component of the normal contact force acts horizontally towards the centre

Question 18

how do banked surfaces allow objects/riders to corner faster

Answer 18

• they increase the effective centripetal force acting on the object
• this means they can travel faster before the required centripetal force is too high

Question 19

how can normal contact forces be changed due to centripetal forces

Answer 19

• where there is a centripetal force, the normal contact force acting on an object will be lower/higher depending on if it is acting in the same direction as the weight or the opposite direction to the weight
• e.g. the centripetal force acting on an object at the equator will decrease it’s mass as it acts ‘up’ on the object
• but a centripetal force acting on an object at the bottom of a loop will increase the effective normal contact force

Question 20

what is a conical pendulum

Answer 20

• this is a pendulum which swings around in a circle rather than back and forth
• it used to be used in smooth timing mechanisms

Question 21

derive the suitable equations for the conical pendulum to show velocity and angle are independent of mass

Answer 21

f = ma = mv^2/r
F = Ftsin(theta)
where Ft is the tension in the string and theta is the angle between the string and the centre line

thus Mv^2/r = Ftsin(theta)

we also know that the weight of the pendulum bob is being conteracted

so Ftcos(theta) = mg

divide first by second and you get

tan(theta) = v^2/rg

Question 22

For an object moving on a banked surface at angle theta, what is the equation for the vertical component of it’s normal reaction force and what is this equivalent to

Answer 22

N(vertical) = Ncos(theta) = W = mg

Question 23

what is the easiest way to calculate the centripetal force for the bung experiment

Answer 23

W = mg = Fc

Question 24

what to remember on questions about planes flying at angles

Answer 24

• the centripetal force is the HORIZONTAL component of lift
• the weight of the plane is usually the VERTICAL component of lift
• trig will likely be needed to calculate these

Question 25

when might an object travelling on a vertical circular path experience weightlessness

Answer 25

• at the top of the circle

- IF the centripetal force Fc = weight of the object W