A.4 Rigid Body Mechanics
Angular quantities, 𝛳, 𝜔, ⍺
Angular displacement is a measure of the angle a line from the center to a point on the outside edge sweeps through as the object rotates - measured in radians (rad)
Angular velocity is a measure of the rate of change of the angular position (rad s-1)
Angular acceleration is the rate of change of angular velocity
θ
s
r
r
Rotational Motion Relationships
Moments
You can find the moment (turning effect of a force) of a force acting on a body
The turning motion caused by a force is dependant on:
The magnitude of the force
The distance the force is from the pivot point
(For example, the further you push a door from the hinge, the less effort is required to close it.)
To calculate the total moment about a point:
Moment = Force x Perpendicular distance
5N
3m
C
Moments are measured in Newton-metres
→ You must always include the direction of the moment
(either clockwise or anticlockwise)
→ The distance must always be perpendicular from the pivot to the force itself…
Moments
You can find the moment of a force acting on a body
The turning motion caused by a force is dependant on:
The magnitude of the force
The distance the force is from the pivot point
(For example, the further you push a door from the hinge, the less effort is required to close it.)
To calculate the total moment about a point:
Moment = Force x Perpendicular distance
4N
2m
F
Calculate the moment of the force about point F
Moments
You can find the moment of a force acting on a body
The turning motion caused by a force is dependant on:
The magnitude of the force
The distance the force is from the pivot point
(For example, the further you push a door from the hinge, the less effort is required to close it.)
To calculate the total moment about a point:
Moment = Force x Perpendicular distance
9N
4m
A
Calculate the moment of the force about point A
→ Draw a triangle to find the perpendicular distance!
30°
4Sin30
Moments
You can find the sum of the moment of a set of forces acting on a body
Sometimes you will have a number of moments acting around a single point.
P
5N
3N
4N
2m
1m
1m
Calculate the sum of the moments acting about the point P
→ Start by calculating each moment individually (it might be useful to label them!)
(1)
(3)
(2)
(1)
(2)
(3)
Choosing clockwise as the positive direction…
If we had chosen anticlockwise as the positive direction our answer would have been -8Nm anticlockwise
→ This is just 8Nm clockwise (the same!)
Moments
You can find the sum of the moment of a set of forces acting on a body
Sometimes you will have a number of moments acting around a single point.
P
5N
5N
2m
4m
(1)
(2)
Calculate the sum of the moments acting about the point P
→ Start by calculating each moment individually (it might be useful to label them!)
(1)
(2)
Choosing anticlockwise as the positive direction…
Moments
You can solve problems about bodies resting in equilibrium by equating the clockwise and anticlockwise moments
The diagram to the right shows a uniform rod of length 3m and weight 20N resting horizontally on supports at A and C, where AC = 2m.
Calculate the magnitude of the normal reaction at both of the supports
A
C
B
2m
1m
RA
RC
20N
1.5m
0.5m
As the rod is in equilibrium, the total normal reaction (spread across both supports) is equal to 20N (the total downward force)
Take moments about C (you do not need to include RC as its distance is 0)
(1)
(2)
(1)
(2)
The clockwise and anticlockwise moments must be equal for equilibrium
Divide by 2
Use the original equation to calculate RC
“Uniform rod” = weight is in the centre
This makes sense – as RC is closer to the centre of mass is bearing more of the object’s weight!
Moments
You can solve problems about bodies resting in equilibrium by equating the clockwise and anticlockwise moments
A uniform beam, AB, of mass 40kg and length 5m, rests horizontally on supports at C and D where AC = DB = 1m.
When a man of mass 80kg stands on the beam at E, the magnitude of the reaction at D is double the reaction at C.
By modelling the beam as a rod and the man as a particle, find the distance AE.
A
D
B
1m
1m
RC
RD
1.5m
40g
80g
“Uniform beam” = weight is in the centre
C
E
2RC
The normal reactions must equal the total downward force
Divide by 3
RD is double this
As the reaction at D is bigger, the man must be closer to D than C
40g
80g
1.5m
Moments
You can solve problems about bodies resting in equilibrium by equating the clockwise and anticlockwise moments
A uniform beam, AB, of mass 40kg and length 5m, rests horizontally on supports at C and D where AC = DB = 1m.
When a man of mass 80kg stands on the beam at E, the magnitude of the reaction at D is double the reaction at C.
By modelling the beam as a rod and the man as a particle, find the distance AE.
A
D
B
1m
1m
1.5m
40g
80g
(2)
(3)
C
E
(4)
(1)
40g
80g
Let us call the required distance x (from A to E)
(we could do this around any point, but this will make the algebra easier)
x
(2)
(3)
(4)
(1)
Equilibrium so anticlockwise = clockwise
1.5m
Group terms
Cancel g’s
Calculate
So the man should stand 3.25m from A!
Moments
You can solve problems about non-uniform bodies by finding or using the centre of mass
The mass of a non-uniform body can be modelled as acting at its centre of mass
Sam and Tamsin are sitting on a non-uniform plank AB of mass 25kg and length 4m.
The plank is pivoted at M, the midpoint of AB, and the centre of mass is at C where AC = 1.8m.
Tamsin has mass 25kg and sits at A. Sam has mass 35kg. How far should Sam sit from A to balance the plank?
A
B
25g
25g
35g
C
M
RM
Let Sam sit ‘x’ m from the midpoint
Take moments about M (this way we don’t need to know RM)
1.8m
0.2m
x
(1)
(2)
(3)
(1)
(2)
(3)
The rod is in equilibrium so anticlockwise = clockwise
Group terms
Cancel g’s
Divide by 35
Sam should sit 3.57m from A (or 0.43m from B)
→ Make sure you always read where the distance should be measured from!
Moments
You can solve problems about non-uniform bodies by finding or using the centre of mass
A rod AB is 3m long and has weight 20N. It is in a horizontal position resting on supports at points C and D, where AC = 1m and AD = 2.5m.
The magnitude of the reaction at C is three times the magnitude of the reaction at D.
Find the distance of the centre of mass of the rod from A.
C
D
1m
1.5m
0.5m
RC
RD
A
B
20N
RC = 3RD
Estimate where the centre of mass is on your diagram
3RD
5N
15N
Divide by 4
Moments
You can solve problems about non-uniform bodies by finding or using the centre of mass
A rod AB is 3m long and has weight 20N. It is in a horizontal position resting on supports at points C and D, where AC = 1m and AD = 2.5m.
The magnitude of the reaction at C is three times the magnitude of the reaction at D.
Find the distance of the centre of mass of the rod from A.
C
D
1m
1.5m
0.5m
A
B
20N
5N
15N
Now take moments about A, calling the required distance ‘x’
(You’ll find it is usually easiest to do this from the end of the rod!)
(1)
(3)
(2)
x
(1)
(2)
(3)
Equilibrium so anticlockwise = clockwise
Group terms
Calculate
The centre of mass is 1.38m from A
Moments
Moments
Moments
Figure 1
A uniform beam AB has mass 20 kg and length 6 m. The beam rests in equilibrium in a horizontal position on two smooth supports. One support is at C, where AC = 1 m, and the other is at the end B, as shown in Figure 1. The beam is modelled as a rod.
(a) Find the magnitudes of the reactions on the beam at B and at C.
A boy of mass 30 kg stands on the beam at the point D. The beam remains in equilibrium. The magnitudes of the reactions on the beam at B and at C are now equal. The boy is modelled as a particle.
(b) Find the distance AD.
Moments
Torque
For rotational motion to happen beyond turning a beam or object with a fixed pivot these conditions must be met:
the axis of rotation must be identified (r in the diagram opposite)
the line of action of the force must be identified and the components established. The component perpendicular to the axis of rotation is the component of the force we use (F sin 𝛳) in the diagram opposite).
Torque 𝛕 = F sin 𝛳 r
Note - units are N m as this is a force multiplied by a distance (cf moment and work done)
Torque examples
For rotational motion to happen beyond turning a beam or object with a fixed pivot these conditions must be met:
Couples
Rotation is often caused by pairs of equal and opposite forces. For example, when the steering wheel of a car is turned, the left and the right hand gripping the wheel both apply equal forces in opposite directions. Such pairs of forces are called couples.
A couple is a pair of parallel forces of equal magnitude but with different lines of action and acting in opposite directions.
As the forces in a couple are equal and opposite, there is no resultant translational force acting on the object, and no translational motion. Couples cause rotation only.
What is the total torque?
Equilibrium?
Stating something is in equilibrium is simply saying the forces are balanced - rotational as well.
What is T assuming the system is in equilibrium
Rotational Equilibrium
In rotational equilibrium, the sum of all the torques is equal to zero. In other words, there is no net torque on the object.
There is no angular acceleration, α = 0.
The object is either not rotating or it is rotating at a constant speed, 𝛕net = 0.
Linear Equilibrium
In linear equilibrium, the sum of all the forces is equal to zero. In other words, there is no net force on the object.
There is no linear acceleration, a = 0.
The object is either not moving linearly or it is moving at a constant velocity.
Total Equilibrium
In total equilibrium, both net force and net torque are equal to zero. In other words, there is no net force or net torque on the object.
Fnet = 0 and 𝛕net = 0
a = 0 and α = 0
Worked example
A torque of 10 Nm is applied to an object whose moment of inertia is 5.0kgm2. What is the angular acceleration?
α = 𝛕/I = 10/5.0 = 2.0 rad s−2
How does F=ma look now?
I = moment of inertia, replaces mass
Moment of inertia
Newton's First Law states that an object remains at rest or moving at constant velocity unless acted on by unbalanced forces.
Newton's First Law is most naturally cast in terms of inertia and so is sometimes referred to as the Law of Inertia. Inertia is a property of an object which determines its ability to resist any changes in its velocity, whether it be changes to speed or direction. Objects with large mass have large inertia - objects with small mass have small inertia.
Moment of inertia
The moment of inertia depends on the distribution of mass throughout the object relative to the axis of rotation. The closer the mass is to the axis of rotation, the smaller the moment of inertia. This helps to understand why, in the race, the hollow objects, where the mass is all distributed on the surface, accelerate slower down the slope, as they possess a larger moment of inertia.
A4 guidance: “The equation for the moment of inertia of a specific mass distribution will be provided when necessary” pg 38.
Moment of inertia
A hoop has a diameter of 70cm and a mass of 300g.
What is the moment of inertia of the hoop about the central axis?
From last slide, the equation for the moment of inertia
of a hoop is I = mr2. So
I = 0.3 × 0.352 = 0.04 kg m2 to 2 s. f.
Rotating systems
When two rotating objects share a common axis of rotation, then their individual moments of inertia are added together to give a moment of inertia for the combined system.
For example, a rotating carousel will have a moment of inertia of ½ M R2
If a child jumps on (mass of m, distance from center of r), the moment of inertia of the combination is given by:
This holds for any combination provided they rotate about a common axis.
cf question 17 in your textbook (page 128)
Try this
A park roundabout has been left rotating at 30 rpm. A passer-by slows it down and stops it by applying a tangential braking force of 200N. The moment of inertia of the roundabout is 500 kg m2 and its radius is 2.0 m. Calculate:
a. the initial angular velocity
b. the braking torque being applied
c. the angular deceleration of the roundabout
d. the time taken to stop and how many revolutions this will take.
Angular acceleration
Newton’s Second Law for Translation: For the linear acceleration of the center of mass, we have:
�mgsinθ−μdN=ma
Angular acceleration
Newton’s Second Law for Rotation: The torque due to frictional force is μdN x R, since torque is also equal to I⍺, we get:
(μdN)R = I⍺
Use this to find a relationship between alpha and R.
Rotational kinetic energy
½ mv2
v=𝝎r
so rotational kinetic energy?
Worked example
A disc is spun up initially to 100 rpm and then later to 300 rpm. It has a radius of 50 cm and a mass of 10 kg. Find Ek in each case.
Worked example
A disc is spun up initially to 100 rpm and then later to 300 rpm. It has a radius of 50 cm and a mass of 10 kg. Find Ek in each case.
For the disc, the moment of inertia is ½ mr2, (1.25 kg m2).
@ 100 rpm = 100 × 2π/60 = 10.5rads−1, we have Ek = 68.9 J.
@ 300 rpm = 300 × 2π/60 = 31.4rads−1, we have Ek = 616 J. That is quite a dramatic increase.
Worked example
A disc is spun up initially to 100 rpm and then later to 300 rpm. It has a radius of 50 cm and a mass of 10 kg. Find Ek in each case.
The same mass is now made into a hoop, with everything else being kept the same. Calculate the new value for Ek in each case. Comment on the answers.
For the hoop, the moment of inertia is mr2, so it will be 2.5 kgm2.
Try this
A solid steel cylindrical rotor is being tested. The rotor has a mass of 272 kg and a radius of 38.0 cm. During the test, the rotor reaches an angular speed of 14000 rev min−1 before breaking.
a. Calculate the angular velocity when this happens.
b. How much energy had the rotor stored? What happened to this energy when the rotor broke?
Assuming the Earth to be a sphere with moment of inertia ⅖MR2, where M is the mass of the Earth (6.0 × 1024kg) and R is the radius of the Earth (6.4 × 106m), calculate the moment of inertia of the Earth.
b. What is its rotational kinetic energy?
More problems
Page 128 of your textbook (answers on canvas)
Work and Power in Rotating Systems
A rigid body turns through an angle θ about an axis when a force F is applied. The perpendicular distance from the axis to the line of action of the force is R. As the rigid body rotates about the axis, the force moves along an arc of length s = Rθ.
The work done is the product of the force and the distance moved:
W = FRθ
often written as,
W = 𝛕θ
Work and Power in Rotating Systems
Power is the rate of doing work, so what would this look like?
Example question including energy
Derive an equation for the acceleration of the mass.
Resolving forces for the falling mass;
F = mg-T
Example question including energy
Derive an equation for the acceleration of the mass.
Resolving forces for the falling mass;
F = mg-T
To find tension we can investigate the large cylinder:
Angular momentum
Conservation of angular momentum
Just as linear momentum is the product of mass and linear velocity, angular momentum of a rotating object is the product of its moment of inertia and its angular velocity.
It is defined as follows: L = Iω
The total angular momentum of a system is constant provided that no external torques are
acting on it.
Conservation of angular momentum
A solid metal disc of mass 960 g and radius 8.8 cm is rotating at 4.7 rad s−1.
Calculate the moment of inertia of the disc.
Calculate the new angular velocity after a mass of 500 g is dropped quickly and carefully on to the disc at a distance of 6.0 cm from the centre.
Conservation of angular momentum
A solid metal disc of mass 960 g and radius 8.8 cm is rotating at 4.7 rad s−1.
Calculate the moment of inertia of the disc.
Calculate the new angular velocity after a mass of 500 g is dropped quickly and carefully on to the disc at a distance of 6.0 cm from the centre.
I = 0.5 × 0.96 × (8.8 × 10−2)2 = 3.7 × 10−3 kg m2
Moment of inertia of added mass = mr2 = 0.5 × (6.0 × 10−2)2
= 1.8 × 10−3 kg m2
using, L = Iω = constant
(3.7 × 10−3) × 4.7 = ((3.7 × 10−3) + (1.8 × 10−3)) × ω
ω = 3.2 rad s−1
Homework
watch:
Angular Momentum (try at home if possible)
Rotating Chair and Weights Experiment
Exam question
Exam question
Exam question
Review