6.2 - Circular Motion

Question 1

What units can we measure angles in

Answer 1

Angles can be measured in radians as well as in degrees

Question 2

When may we use radians

Answer 2

When we are measuring rotation, we often use the radian to measure angles

Question 3

How can we calculate the distance/circumference/ arc length an object has travelled on a circular path

Answer 3

The distance the object travels around the circumference (in metres) would be equal to the angle swept out in radians.

angle (in radians ) = arc length/radius of arc

So, theta = s/r

So, for a complete circle, in which the circumference is equal to 2pi x r, the angle swept out will be equal to

= 2 x pi x r/ r = 2 pi radians

Question 4

What is angular displacement

Answer 4

Angular displacement is the vector measurement of the angle through which something has moved. The standard convention is that anti-clockwise rotation is a positive number and clockwise rotation is a negative number.

Question 5

How do we convert from radians to degrees

Answer 5

Radians to degrees = multiply by 180/pi

Question 6

How do we convert from degrees to radians

Answer 6

Multiply by pi/180

Question 7

How does the British army use a system for angle measurements

Answer 7

The British army uses a system for angle measurement in which a complete circle is divided into 6400 ‘mils’. This is an abbreviation for milliradian. The idea is that at a distance/radius of 1km, an angle of 1 mil would represent a distance of 1 metre.

So when aiming artillery fire, a horizontal adjustment of 1 mil in angle of fire should move to a target 1 metre sideways for every kilometre distance away.

Question 8

What is angular velocity

Answer 8

An object moving in a circle sweeps out a certain angle in a certain time, depending upon how fast it is moving. The rate at which angular displacement changes is called the angular velocity, (omega, w)

So w = theta / t

It’s measured in rads^-1

If an object completes a full circle (2pi radians) in a time period, T (it’s capital T now because it’s the time period for a full circle) then the angular velocity can be rearranged and given by

T = 2pi/w

Since in an earlier chapter we found f = 1/T

Therefore, w = 2pi x f
^the frequency of rotation is the reciprocal of the time period.

Question 9

What is the instantaneous velocity

Answer 9

Also known as the actual velocity, we know that v = s/t and since theta = s/r (arc length/radius)

So s = r x theta

Thus, v = r x theta / t

And so, v = r x w

Question 10

Tell me about centripetal acceleration/ velocity

Answer 10

Velocity is a vector, and so it is correctly described by quoting both its magnitude and direction. An acceleration can change either of these, or both. An object moving in a ⭕️ circle may travel at a constant speed (and a constant angular velocity) but the direction it is moving in must constantly change. This means it is constantly accelerating. As this acceleration represents represents the changes in direction around the circle, it is called the centripetal acceleration, a.

Question 11

How can we derive the formula for centripetal acceleration - you need to be able to do this

Answer 11

Check the image on page 24 for it to make sense

In order to determine how to calculate the centripetal acceleration, we must consider how quickly the direction, and therefore, velocity is changing.

At the arbitrary positions of the rotating object, A, and at s time t, later B, we consider the components of the objects velocity, in the x and y directions.

Point C is equal to the midpoint of AB, so as A and B are equal distances above and below point C, the vertical velocity component, v(subscript y), is the same at A and at B v(subscript y) = v x cos theta

So, the vertical acceleration is zero, a(subscript y) = 0

Horizontally, the magnitude of the velocity is equal at both parts but in opposite directions,

At A: v(subscript x) = v x sin theta
At B: v(subscript x) = -v x sin theta

So the acceleration in the horizontal direction, calculated as change in velocity over time

a(subscript x) = 2vsin theta / t

Since v = r theta / t, this can be rearranged to give t = r x theta / v

Here the angle moved in time t, is labelled as 2 theta,

So t = r x 2theta /v

Therefore,

a(subscript x) = v2vsin theta/ r2 theta = v^2 x sin theta / r theta

This must be true for all values of theta, and as we want to find the instantaneous acceleration at any point on the circumference, we must consider the general answer as we reduce theta to zero. In the limit as theta tends to zero,
Sin theta /theta = 1, so sin theta =0

Therefore, we can simplify the equation before to make a = v^2/r

From thus definition v = rw

a is also equal to a = r x w^2

Question 12

Why is the centripetal acceleration always directed towards the centre of the circle

Answer 12

The centripetal acceleration in this case is just the horizontal acceleration, as we considered the object in a position along a horizontal radius. Following a similar derivation at any point around the circle will always have identical components of velocity that are perpendicular to the radius on either side of the point being considered. Thus, the centripetal acceleration is always directed towards the centre of the circle.

Question 13

Define radian

Answer 13

A radian is a unit of angle measurement, equivalent to 57.3 degrees

Question 14

Define angular displacement

Answer 14

Angular displacement is the vector measurement of the angle through which something has moved

Question 15

Define angular velocity

Answer 15

Angular velocity, w is the rate at which the angular displacement changes, unit is radians per second

Question 16

Define centripetal acceleration

Answer 16

Centripetal acceleration, a is the acceleration towards the centre of a circle that corresponds to the changes in direction to maintain an objects motion around that circle

Question 17

How does circular motion work

Answer 17

When a hammer thrower whirls an athletic hammer around in a circle, the hammer has an angular velocity. Yet when the thrower lets the hammer go, it will fly off, following a straight line which is the direction in which it was moving at the instant of release.

The direction is always along the edge of the circle (a tangent) at the point when it was released

Question 18

Tell me about an example with centripetal force - a hammer thrower/ what is centripetal force

Answer 18

As the hammer is whirled at a constant speed, the magnitude of the velocity is always the same. However, the direction of the velocity is constantly working. This means that the vector of velocity is constantly changing, and a change in velocity is an acceleration. Newton’s first law tells us that acceleration will only happen if there is a resultant force. The hammer is constantly being pulled towards the centre of the circle. In this example, the force providing this pull is the tension in the string (or chain).

For any object moving in a circle, there must be a resultant force to cause this acceleration, and it is called centripetal force

Question 19

Are there any special centripetal forces?

Answer 19

No, there is no special centripetal force, any resultant force that makes an object move in a circle is labelled as a centripetal force

Eg tension, weight etc.

Question 20

How can we find the formula for the centripetal force

Answer 20

The formula for the centripetal force on an object moving in a circle can be found from Newton’s second law, and we already derived the formula for centripetal acceleration

F = ma

a = v^2/r

So 
Centripetal force (f) = mv^2/r

Since v = r w

Centripetal force is also equal to

F = mrw^2

Question 21

When will the resultant centripetal force needed be larger

Answer 21

The resultant centripetal force needed will be larger if…

The rotating object has more mass

The object rotated faster

The object is further away from the centre of the circle.

Question 22

Formula for angular displacement

Answer 22

Theta = s/r

Question 23

Formula for angular velocity

Answer 23

w = 2 x pi x f = 2pi/T

Therefore, w = v/r

Question 24

Formula for centripetal acceleration

Answer 24

a = r w^2 = v^2/r

Question 25

What’s the formula for centripetal force

Answer 25

F = mrw^2

Or F = mv^2/r

Question 26

How can we investigate centripetal force practical

Answer 26

This is about verifying the centripetal force equation. Eye protection should be worn during this investigation.

You can investigate the centripetal force equation by spinning a rubber bung on a string around in a circle. The tension In the string, which is the centripetal force, will be provided by the hanging masses at the bottom of the vertical string and thus will be known. Spin the rubber bung around in a circle that keeps a paper clip marker in a constant position near the handle. The paperclip marker allows you to maintain a fixed length of string (radius) which you can measure. You will also need to measure the mass of the rubber bung. Your partner can then time ten revolutions in order to give you the angular velocity. Take angular velocity measurements for different forces (different numbers of hanging masses)

F = mrw^2
Therefore

W^2 = F/mr

A graph of w^2 plotted against F should give a straight line of best fit. The gradient is 1/mr - these are both constants

Question 27

Define centripetal force

Answer 27

Centripetal force is the resultant force towards the centre of the circle to maintain an objects circular motion.