B.1 – Rigid bodies and rotational dynamics

Description

Nature of science:

Modelling: The use of models has different purposes and has allowed scientists to identify, simplify and analyse a problem within a given context to tackle it successfully. The extension of the point particle model to actually consider the dimensions of an object led to many groundbreaking developments in engineering. (1.2)

Understandings:

Torque
Moment of inertia
Rotational and translational equilibrium
Angular acceleration
Equations of rotational motion for uniform angular acceleration
Newton’s second law applied to angular motion
Conservation of angular momentum

Applications and skills:

Calculating torque for single forces and couples
Solving problems involving moment of inertia, torque and angular acceleration
Solving problems in which objects are in both rotational and translational equilibrium
Solving problems using rotational quantities analogous to linear quantities
Sketching and interpreting graphs of rotational motion
Solving problems involving rolling without slipping

Guidance:

Analysis will be limited to basic geometric shapes
The equation for the moment of inertia of a specific shape will be provided when necessary
Graphs will be limited to angular displacement–time, angular velocity–time and torque–time

Data booklet reference:

Theory of knowledge:

Models are always valid within a context and they are modified, expanded or replaced when that context is altered or considered differently. Are there examples of unchanging models in the natural sciences or in any other areas of knowledge?

Utilization:

Structural design and civil engineering rely on the knowledge of how objects can move in all situations

Aims:

Aim 7: technology has allowed for computer simulations that accurately model the complicated outcomes of actions on bodies

Directly related questions

20N.3.SL.TZ0.7b: Explain the changes to the rotational kinetic energy in the person-turntable system.
20N.3.SL.TZ0.7b: Explain the changes to the rotational kinetic energy in the person-turntable system.
20N.3.SL.TZ0.b: Explain the changes to the rotational kinetic energy in the person-turntable system.
17N.3.SL.TZ0.7b:
Show that the linear acceleration a of the hoop is given by the equation shown.

a = $\frac{{g \times \sin q}}{2}$
17N.3.SL.TZ0.7b:
Show that the linear acceleration a of the hoop is given by the equation shown.

a = $\frac{{g \times \sin q}}{2}$
17N.3.SL.TZ0.b:
Show that the linear acceleration a of the hoop is given by the equation shown.

a = $\frac{{g \times \sin q}}{2}$
17N.3.SL.TZ0.7c: Calculate the acceleration of the hoop when θ = 20°. Assume that the hoop continues to roll...
17N.3.SL.TZ0.7c: Calculate the acceleration of the hoop when θ = 20°. Assume that the hoop continues to roll...
17N.3.SL.TZ0.c: Calculate the acceleration of the hoop when θ = 20°. Assume that the hoop continues to roll...
17N.3.SL.TZ0.7d:
State the relationship between the force of friction and the angle of the incline.
17N.3.SL.TZ0.7d:
State the relationship between the force of friction and the angle of the incline.
17N.3.SL.TZ0.d:
State the relationship between the force of friction and the angle of the incline.
17N.3.SL.TZ0.6a: Explain what is meant by proper length.
17N.3.SL.TZ0.6a: Explain what is meant by proper length.
17N.3.SL.TZ0.a: Explain what is meant by proper length.
18M.3.SL.TZ2.6a.ii:
In moving from point A to point B, the centre of mass of the wheel falls through a vertical distance of 0.36 m. Show that the translational speed of the wheel is about 1 m s^–1 after its displacement.
18M.3.SL.TZ2.6a.ii:
In moving from point A to point B, the centre of mass of the wheel falls through a vertical distance of 0.36 m. Show that the translational speed of the wheel is about 1 m s^–1 after its displacement.
18M.3.SL.TZ2.a.ii:
In moving from point A to point B, the centre of mass of the wheel falls through a vertical distance of 0.36 m. Show that the translational speed of the wheel is about 1 m s^–1 after its displacement.
18M.3.SL.TZ2.6b.ii:
Describe the effect of F on the angular speed of the wheel.
18M.3.SL.TZ2.6b.ii:
Describe the effect of F on the angular speed of the wheel.
18M.3.SL.TZ2.b.ii:
Describe the effect of F on the angular speed of the wheel.
18M.3.SL.TZ2.6a.iii:
Determine the angular velocity of the wheel at B.
18M.3.SL.TZ2.6a.iii:
Determine the angular velocity of the wheel at B.
18M.3.SL.TZ2.a.iii:
Determine the angular velocity of the wheel at B.
18M.3.SL.TZ1.6b.ii:
Calculate, for the merry-go-round after one revolution, the angular momentum.
18M.3.SL.TZ1.6b.ii:
Calculate, for the merry-go-round after one revolution, the angular momentum.
18M.3.SL.TZ1.b.ii:
Calculate, for the merry-go-round after one revolution, the angular momentum.
18M.3.SL.TZ1.6d.i:
Explain why the angular speed will increase.
18M.3.SL.TZ1.6d.i:
Explain why the angular speed will increase.
18M.3.SL.TZ1.d.i:
Explain why the angular speed will increase.
18N.3.SL.TZ0.6c.ii: At the instant the rod becomes vertical calculate the angular momentum of the rod.
18N.3.SL.TZ0.6c.ii: At the instant the rod becomes vertical calculate the angular momentum of the rod.
18N.3.SL.TZ0.c.ii: At the instant the rod becomes vertical calculate the angular momentum of the rod.
18N.3.SL.TZ0.6a: Calculate the force the support exerts on the rod.
18N.3.SL.TZ0.6a: Calculate the force the support exerts on the rod.
18N.3.SL.TZ0.a: Calculate the force the support exerts on the rod.
18N.3.SL.TZ0.6b.ii:
After time t the rod makes an angle θ with the horizontal. Outline why the equation $\theta = \frac{1}{2}\alpha {t^2}$ cannot be used to find the time it takes θ to become $\frac{\pi }{2}$ (that is for the rod to become vertical for the first time).
18N.3.SL.TZ0.6b.ii:
After time t the rod makes an angle θ with the horizontal. Outline why the equation $\theta = \frac{1}{2}\alpha {t^2}$ cannot be used to find the time it takes θ to become $\frac{\pi }{2}$ (that is for the rod to become vertical for the first time).
18N.3.SL.TZ0.b.ii:
After time t the rod makes an angle θ with the horizontal. Outline why the equation $\theta = \frac{1}{2}\alpha {t^2}$ cannot be used to find the time it takes θ to become $\frac{\pi }{2}$ (that is for the rod to become vertical for the first time).
18N.3.SL.TZ0.6b.i:
Calculate, in rad s^–2, the initial angular acceleration $\alpha$ of the rod.
18N.3.SL.TZ0.6b.i:
Calculate, in rad s^–2, the initial angular acceleration $\alpha$ of the rod.
18N.3.SL.TZ0.b.i:
Calculate, in rad s^–2, the initial angular acceleration $\alpha$ of the rod.
19M.3.SL.TZ2.8a: Outline why the normal force acting on the ladder at the point of contact with the wall is equal...
19M.3.SL.TZ2.8a: Outline why the normal force acting on the ladder at the point of contact with the wall is equal...
19M.3.SL.TZ2.a: Outline why the normal force acting on the ladder at the point of contact with the wall is equal...
19M.3.SL.TZ2.9a:
Show that the total kinetic energy E_k of the sphere when it rolls, without slipping, at speed v is ${E_{\text{K}}} = \frac{7}{{10}}m{v^2}$ .
19M.3.SL.TZ2.9a:
Show that the total kinetic energy E_k of the sphere when it rolls, without slipping, at speed v is ${E_{\text{K}}} = \frac{7}{{10}}m{v^2}$ .
19M.3.SL.TZ2.a:
Show that the total kinetic energy E_k of the sphere when it rolls, without slipping, at speed v is ${E_{\text{K}}} = \frac{7}{{10}}m{v^2}$ .
19M.3.SL.TZ2.9b:
A solid sphere of mass 1.5 kg is rolling, without slipping, on a horizontal surface with a speed of 0.50 m s^-1. The sphere then rolls, without slipping, down a ramp to reach a horizontal surface that is 45 cm lower.

Calculate the speed of the sphere at the bottom of the ramp.
19M.3.SL.TZ2.9b:
A solid sphere of mass 1.5 kg is rolling, without slipping, on a horizontal surface with a speed of 0.50 m s^-1. The sphere then rolls, without slipping, down a ramp to reach a horizontal surface that is 45 cm lower.

Calculate the speed of the sphere at the bottom of the ramp.
19M.3.SL.TZ2.b:
A solid sphere of mass 1.5 kg is rolling, without slipping, on a horizontal surface with a speed of 0.50 m s^-1. The sphere then rolls, without slipping, down a ramp to reach a horizontal surface that is 45 cm lower.

Calculate the speed of the sphere at the bottom of the ramp.
19N.3.SL.TZ0.5b(i): Identify the physical quantity represented by the area under the graph.
19N.3.SL.TZ0.5b(i): Identify the physical quantity represented by the area under the graph.
19N.3.SL.TZ0.b(i): Identify the physical quantity represented by the area under the graph.
19N.3.SL.TZ0.5b(ii): Show that the angular velocity of the flywheel at t = 5.00 s is 200 rad s–1.
19N.3.SL.TZ0.5b(ii): Show that the angular velocity of the flywheel at t = 5.00 s is 200 rad s–1.
19N.3.SL.TZ0.b(ii): Show that the angular velocity of the flywheel at t = 5.00 s is 200 rad s–1.
19N.3.SL.TZ0.5c(ii):
At t = 5.00 s the flywheel is spinning with angular velocity 200 rad s^–1. The support bearings exert a constant frictional torque on the axle. The flywheel comes to rest after 8.00 × 10³ revolutions. Calculate the magnitude of the frictional torque exerted on the flywheel.
19N.3.SL.TZ0.5c(ii):
At t = 5.00 s the flywheel is spinning with angular velocity 200 rad s^–1. The support bearings exert a constant frictional torque on the axle. The flywheel comes to rest after 8.00 × 10³ revolutions. Calculate the magnitude of the frictional torque exerted on the flywheel.
19N.3.SL.TZ0.c(ii):
At t = 5.00 s the flywheel is spinning with angular velocity 200 rad s^–1. The support bearings exert a constant frictional torque on the axle. The flywheel comes to rest after 8.00 × 10³ revolutions. Calculate the magnitude of the frictional torque exerted on the flywheel.
19N.3.SL.TZ0.5a: State the torque provided by the force W about the axis of the flywheel.
19N.3.SL.TZ0.5a: State the torque provided by the force W about the axis of the flywheel.
19N.3.SL.TZ0.a: State the torque provided by the force W about the axis of the flywheel.
19N.3.SL.TZ0.5c(i): The flywheel is in translational equilibrium. Distinguish between translational equilibrium and...
19N.3.SL.TZ0.5c(i): The flywheel is in translational equilibrium. Distinguish between translational equilibrium and...
19N.3.SL.TZ0.c(i): The flywheel is in translational equilibrium. Distinguish between translational equilibrium and...
17N.3.SL.TZ0.7a: On the diagram, draw and label the forces acting on the hoop.
17N.3.SL.TZ0.7a: On the diagram, draw and label the forces acting on the hoop.
17N.3.SL.TZ0.a: On the diagram, draw and label the forces acting on the hoop.
17N.3.SL.TZ0.7e:
The angle of the incline is slowly increased from zero. Determine the angle, in terms of the coefficient of friction, at which the hoop will begin to slip.
17N.3.SL.TZ0.7e:
The angle of the incline is slowly increased from zero. Determine the angle, in terms of the coefficient of friction, at which the hoop will begin to slip.
17N.3.SL.TZ0.e:
The angle of the incline is slowly increased from zero. Determine the angle, in terms of the coefficient of friction, at which the hoop will begin to slip.
18M.3.SL.TZ1.6a:
Show that the angular acceleration of the merry-go-round is 0.2 rad s^–2.
18M.3.SL.TZ1.6a:
Show that the angular acceleration of the merry-go-round is 0.2 rad s^–2.
18M.3.SL.TZ1.a:
Show that the angular acceleration of the merry-go-round is 0.2 rad s^–2.
18M.3.SL.TZ1.6b.i:
Calculate, for the merry-go-round after one revolution, the angular speed.
18M.3.SL.TZ1.6b.i:
Calculate, for the merry-go-round after one revolution, the angular speed.
18M.3.SL.TZ1.b.i:
Calculate, for the merry-go-round after one revolution, the angular speed.
18M.3.SL.TZ1.6c:
Calculate the new angular speed of the rotating system.
18M.3.SL.TZ1.6c:
Calculate the new angular speed of the rotating system.
18M.3.SL.TZ1.c:
Calculate the new angular speed of the rotating system.
18M.3.SL.TZ1.6d.ii:
Calculate the work done by the child in moving from the edge to the centre.
18M.3.SL.TZ1.6d.ii:
Calculate the work done by the child in moving from the edge to the centre.
18M.3.SL.TZ1.d.ii:
Calculate the work done by the child in moving from the edge to the centre.
18M.3.SL.TZ2.6a.i:
The moment of inertia of the wheel is 1.3 × 10^–4 kg m². Outline what is meant by the moment of inertia.
18M.3.SL.TZ2.6a.i:
The moment of inertia of the wheel is 1.3 × 10^–4 kg m². Outline what is meant by the moment of inertia.
18M.3.SL.TZ2.a.i:
The moment of inertia of the wheel is 1.3 × 10^–4 kg m². Outline what is meant by the moment of inertia.
18N.3.SL.TZ0.6c.i:
At the instant the rod becomes vertical show that the angular speed is ω = 2.43 rad s^–1.
18N.3.SL.TZ0.6c.i:
At the instant the rod becomes vertical show that the angular speed is ω = 2.43 rad s^–1.
18N.3.SL.TZ0.c.i:
At the instant the rod becomes vertical show that the angular speed is ω = 2.43 rad s^–1.
19M.3.SL.TZ2.8b:
Calculate F.
19M.3.SL.TZ2.8b:
Calculate F.
19M.3.SL.TZ2.b:
Calculate F.
19N.3.SL.TZ0.5b(iii):
Calculate the maximum tension in the string.
19N.3.SL.TZ0.5b(iii):
Calculate the maximum tension in the string.
19N.3.SL.TZ0.b(iii):
Calculate the maximum tension in the string.
20N.3.SL.TZ0.6a:
Show that the final angular velocity of the bar is about $3 rad s^{- 1}$ .
20N.3.SL.TZ0.6a:
Show that the final angular velocity of the bar is about $3 rad s^{- 1}$ .
20N.3.SL.TZ0.a:
Show that the final angular velocity of the bar is about $3 rad s^{- 1}$ .
20N.3.SL.TZ0.6b:
Draw the variation with time $t$ of the angular displacement $θ$ of the bar during the acceleration.
20N.3.SL.TZ0.6b:
Draw the variation with time $t$ of the angular displacement $θ$ of the bar during the acceleration.
20N.3.SL.TZ0.b:
Draw the variation with time $t$ of the angular displacement $θ$ of the bar during the acceleration.
20N.3.SL.TZ0.6c:
Calculate the torque acting on the bar while it is accelerating.
20N.3.SL.TZ0.6c:
Calculate the torque acting on the bar while it is accelerating.
20N.3.SL.TZ0.c:
Calculate the torque acting on the bar while it is accelerating.
20N.3.SL.TZ0.6d:
The torque is removed. The bar comes to rest in $30$ complete rotations with constant angular deceleration. Determine the time taken for the bar to come to rest.
20N.3.SL.TZ0.6d:
The torque is removed. The bar comes to rest in $30$ complete rotations with constant angular deceleration. Determine the time taken for the bar to come to rest.
20N.3.SL.TZ0.d:
The torque is removed. The bar comes to rest in $30$ complete rotations with constant angular deceleration. Determine the time taken for the bar to come to rest.
20N.3.SL.TZ0.7a:
Explain the direction in which the person-turntable system starts to rotate.
20N.3.SL.TZ0.7a:
Explain the direction in which the person-turntable system starts to rotate.
20N.3.SL.TZ0.a:
Explain the direction in which the person-turntable system starts to rotate.
20N.3.SL.TZ0.8:
A solid sphere of radius $r$ and mass $m$ is released from rest and rolls down a slope, without slipping. The vertical height of the slope is $h$ . The moment of inertia $I$ of this sphere about an axis through its centre is $\frac{2}{5} m r^{2}$ .

Show that the linear velocity $v$ of the sphere as it leaves the slope is $\sqrt{\frac{10 g h}{7}}$ .
20N.3.SL.TZ0.8:
A solid sphere of radius $r$ and mass $m$ is released from rest and rolls down a slope, without slipping. The vertical height of the slope is $h$ . The moment of inertia $I$ of this sphere about an axis through its centre is $\frac{2}{5} m r^{2}$ .

Show that the linear velocity $v$ of the sphere as it leaves the slope is $\sqrt{\frac{10 g h}{7}}$ .
23M.3.HL.TZ1.8a: The moment of inertia of the rod about the axis is 0.180 kg m2. Show that the moment of inertia...
23M.3.SL.TZ1.6a: The moment of inertia of the rod about the axis is 0.180 kg m2. Show that the moment of inertia...
23M.3.HL.TZ1.8a: The moment of inertia of the rod about the axis is 0.180 kg m2. Show that the moment of inertia...
23M.3.HL.TZ1.a: The moment of inertia of the rod about the axis is 0.180 kg m2. Show that the moment of inertia...
23M.3.SL.TZ1.6a: The moment of inertia of the rod about the axis is 0.180 kg m2. Show that the moment of inertia...
23M.3.SL.TZ1.a: The moment of inertia of the rod about the axis is 0.180 kg m2. Show that the moment of inertia...
23M.3.HL.TZ2.8a:
Show that the net torque on the system about the central axis is approximately 30 N m.
23M.3.SL.TZ2.6a:
Show that the net torque on the system about the central axis is approximately 30 N m.
23M.3.HL.TZ2.8a:
Show that the net torque on the system about the central axis is approximately 30 N m.
23M.3.HL.TZ2.a:
Show that the net torque on the system about the central axis is approximately 30 N m.
23M.3.SL.TZ2.6a:
Show that the net torque on the system about the central axis is approximately 30 N m.
23M.3.SL.TZ2.a:
Show that the net torque on the system about the central axis is approximately 30 N m.

Sub sections and their related questions

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B.1 – Rigid bodies and rotational dynamics

Description

Directly related questions

Sub sections and their related questions

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