Rigid Body Mechanics

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शिक्षकों के लिए: Rigid Body Mechanics (Physics, HL) के लिए इस्तेमाल के लिए तैयार लेसन स्लाइड्स, रिवीज़न नोट्स — इन्हें अपने लेसन में इस्तेमाल करें, या टॉपिक को एक इंटरैक्टिव क्लास एक्टिविटी की तरह चलाएं जिसे आपके शिक्षार्थी लाइव गेम की तरह खेलें।

लेसन नोट्स

Torque & Couples

A moment is the turning effect of a force around a pivot: Moment = Force × perpendicular distance from the pivot (N m).
For a non-perpendicular force, resolve the force or use τ = Fr sin θ, where θ is the angle between the force and the line from the pivot.
A couple is a pair of equal and opposite coplanar forces that produce rotation only; the forces are equal in magnitude, opposite in direction, and perpendicular to the distance between them.
The moment of a couple is Force × perpendicular distance between the lines of action; it does not depend on a pivot.
A couple produces zero resultant force (no linear acceleration) but a net torque causing angular acceleration.
For a couple on a wheel of radius r, the net torque is τ = 2Fr sin θ; when θ = 90°, τ = 2Fr.
Torque is maximum when the force is applied perpendicular to the lever arm (θ = 90°); at θ = 0° or 180°, torque is zero.

Rotational Equilibrium

A body is in rotational equilibrium if the resultant torque acting on it is zero; it remains at rest or rotates with constant angular velocity.
The condition for rotational equilibrium is: sum of clockwise torques = sum of anticlockwise torques (principle of moments).
A resultant (unbalanced) torque causes angular acceleration; the direction of angular acceleration is the same as the net torque direction.
When solving equilibrium problems, you can take torques about any point; choose a point that simplifies the calculation (e.g. where unknown forces act).
For a beam balanced on a pivot, the weight acts at the centre of mass; torques are calculated using τ = Fr for perpendicular forces.

Angular Displacement, Velocity & Acceleration

Angular displacement Δθ is the change in angle through which a body rotates, measured in radians.
Linear displacement is related to angular displacement by s = rΔθ.
Angular velocity ω is the rate of change of angular displacement: ω = Δθ / Δt, measured in rad s⁻¹.
Linear speed is related to angular speed by v = rω; also ω = 2πf = 2π/T and v = 2πfr.
Angular acceleration α is the rate of change of angular velocity: α = Δω / Δt, measured in rad s⁻².
Linear acceleration is related to angular acceleration by a = rα.
Graphs: angular displacement is the area under the angular velocity–time graph; angular velocity is the gradient of the angular displacement–time graph and the area under the angular acceleration–time graph; angular acceleration is the gradient of the angular velocity–time graph.

Angular Acceleration Formula (Rotational Kinematics)

The linear kinematic equations can be rewritten for uniform angular acceleration:
ωf = ωi + αt
Δθ = ωi t + ½ αt²
ωf² = ωi² + 2αΔθ
Δθ = (ωi + ωf) t / 2
These equations are used when angular acceleration is constant; identify the known and unknown quantities to select the appropriate equation.
When converting from RPM to rad s⁻¹, use ω = 2π × (RPM / 60).

Moment of Inertia

Moment of inertia I is the resistance of a body to a change in rotational motion; it depends on the mass distribution about the axis of rotation.
Moment of inertia is measured in kg m².
For a point mass, I = mr², where r is the distance from the axis of rotation.
The total moment of inertia of a system is the sum of the moments of inertia of its parts: I_total = Σ mr².
The moment of inertia of a rigid body depends on its shape, density, and orientation relative to the axis of rotation.
Common moments of inertia (given in exams): solid cylinder/disc I = ½MR², solid sphere I = ⅖mr², hollow sphere I = ⅔mr², thin rod about centre I = 1/12 mL².
You are not expected to memorise moments of inertia of different shapes; they will be provided where needed.

Newton’s Second Law for Rotation

Newton’s second law for rotation: τ = Iα, where τ is torque (N m), I is moment of inertia (kg m²), and α is angular acceleration (rad s⁻²).
This is the rotational analogue of F = ma.
Torque is the rotational equivalent of force; moment of inertia is the rotational equivalent of mass.
For a point mass, I = mr² and α = a/r, leading to τ = Iα.
In problems with a hanging mass and pulley, apply F = ma to the linear motion and τ = Iα to the rotation, linking linear and angular acceleration by a = rα.

Angular Momentum

Angular momentum L is the rotational equivalent of linear momentum: L = Iω, measured in kg m² rad s⁻¹ (or kg m² s⁻¹).
For a point mass, L = mvr, where r is the perpendicular distance from the axis of rotation.
Conservation of angular momentum: the total angular momentum of a system remains constant unless acted upon by a net resultant torque.
For a constant total angular momentum: Iiωi = Ifωf.
Examples: an ice skater spinning faster when arms are pulled in (I decreases, ω increases); a diver tucking; a collapsing star increasing its rotation rate.
Objects travelling in straight lines can have angular momentum about a point not on their line of motion.

Conservation of momentum

Angular Impulse

Angular impulse is the change in angular momentum produced by a torque acting over a time interval: ΔL = τΔt.
Angular impulse is measured in kg m² s⁻¹ or N m s.
The equation requires a constant resultant torque; if torque varies, use an average value.
A small torque acting for a long time can produce the same angular impulse as a large torque acting for a short time.
On a torque–time graph, the area under the graph equals the angular impulse or change in angular momentum.
For a changing torque, ΔL = I(ωf − ωi) can be used with the area under the graph.

Rotational Kinetic Energy

A rotating body has rotational kinetic energy: Ek = ½Iω² (or Ek = L² / 2I).
For an object rolling without slipping, total kinetic energy is the sum of translational and rotational kinetic energy: Ek_total = ½mv² + ½Iω².
Rolling without slipping requires friction; the point of contact has zero velocity, the centre of mass has velocity v = ωr, and the top point has velocity 2v.
When an object rolls down a slope, gravitational potential energy is converted into both translational and rotational kinetic energy: mgΔh = ½mv² + ½Iω².
Using v = ωr and the moment of inertia, the total kinetic energy can be expressed in terms of ω or v; for a solid sphere, mgΔh = 7/10 mω²r².
If an object slides (slips) without rolling, there is no rotational kinetic energy and angular velocity is zero.

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साइन अप करें — फ्री लॉग इन

प्रैक्टिस सवाल

फ्री प्रीव्यू — 61 में से 8 सवाल। सभी देखने के लिए साइन अप करें।

1.Which of the following best defines the moment of a force about a pivot?
Easy
- AThe product of the force and the perpendicular distance from the pivot to the line of action of the force
- BThe product of the force and the distance from the pivot to the point of application of the force
- CThe force divided by the perpendicular distance from the pivot
- DThe product of the force and the time for which it acts
2.Which pair of forces acts as a couple?
Easy
- ATwo forces equal in magnitude, opposite in direction, and perpendicular to the distance between them
- BTwo forces equal in magnitude and in the same direction, separated by a distance
- CTwo forces of different magnitudes acting in opposite directions
- DTwo forces equal in magnitude and opposite in direction, but not perpendicular to the distance between them
3.A force of 50 N is applied at an angle of 30° to a spanner of length 0.20 m. What is the torque produced?
Medium
- A5.0 N m
- B8.7 N m
- C10 N m
- D0.50 N m
4.A uniform beam is pivoted at its centre. Two forces act on it: a 10 N force downwards at 0.30 m to the left of the pivot, and a 15 N force downwards at 0.20 m to the right of the pivot. What is the resultant torque on the beam?
Medium
- A0 N m, the beam is in rotational equilibrium
- B1.0 N m clockwise
- C1.0 N m anticlockwise
- D6.0 N m clockwise
5.A solid cylinder of mass 2.0 kg and radius 0.50 m rotates about its central axis. What is its moment of inertia? (I = ½MR²)
Medium
- A0.25 kg m²
- B0.50 kg m²
- C1.0 kg m²
- D0.125 kg m²
6.A star of mass M and radius R rotates with angular velocity ω. It collapses to a sphere of radius R/4 without losing mass. What is its new angular velocity? (Assume it remains a uniform sphere.)
Hard
- A16ω
- B4ω
- Cω/4
- Dω/16
7.Which of the following statements about a couple are correct? (Select all that apply.)
Medium
- AThe resultant force of a couple is zero.
- BA couple produces a net torque.
- CThe moment of a couple depends on the choice of pivot.
- DA couple causes angular acceleration.
- EThe forces in a couple must act along the same line of action.
8.Which of the following are units of angular momentum? (Select all that apply.)
Medium
- Akg m² s⁻¹
- BN m s
- Ckg m s⁻¹
- DJ s
- Ekg m² s⁻²

और सवाल

9.The moment of inertia of a rigid body depends only on its mass, not on the axis of rotation.
Easy
True or false?
10.A body in rotational equilibrium must be at rest.
Easy
True or false?
11.When a diver tucks their arms and legs in during a dive, their moment of inertia decreases and their angular velocity increases.
Medium
True or false?
12.Match each rotational quantity with its defining equation or relationship.
Medium
- Torque
- Angular velocity
- Angular acceleration
- Moment of inertia (point mass)
- τ = Fr sin θ
- ω = Δθ/Δt
- α = Δω/Δt
- I = mr²
13.Place the following steps in the correct order to find the final angular velocity of a rotating object given its initial angular velocity, angular acceleration, and time.
Medium
- Identify the known quantities: initial angular velocity, angular acceleration, and time.
- Choose the appropriate rotational kinematic equation: ωf = ωi + αt.
- Substitute the known values into the equation.
- Calculate the final angular velocity.
14.Which of the following are required for a pair of forces to form a couple? (Select all that apply.)
Medium
- AThey are equal in magnitude
- BThey act in opposite directions
- CThey act along the same line of action
- DThey are perpendicular to the distance between them
- EThey produce a net linear force
15.A rotating turntable decelerates uniformly to rest. Which of the following statements are correct for this motion? (Select all that apply.)
Medium
- AThe angular acceleration is opposite in direction to the angular velocity
- BThe angular displacement is equal to the area under the angular velocity–time graph
- CThe angular acceleration is equal to the gradient of the angular displacement–time graph
- DThe final angular velocity is zero
- EThe angular acceleration is constant and non-zero
16.A torque–time graph shows a constant torque of 4.0 N m acting for 3.0 s on a wheel of moment of inertia 2.0 kg m² that starts from rest. What is the final angular velocity of the wheel?
Medium
- A6.0 rad s⁻¹
- B2.7 rad s⁻¹
- C12 rad s⁻¹
- D0.67 rad s⁻¹
17.A solid sphere rolls without slipping down a slope. Its moment of inertia is I = ⅖mr². Which expression gives its total kinetic energy at the bottom in terms of m, r and its angular velocity ω?
Medium
- A7/10 mr²ω²
- B1/2 mr²ω²
- C2/5 mr²ω²
- D9/10 mr²ω²
18.A flywheel of moment of inertia 0.50 kg m² spins at 8.0 rad s⁻¹. What is its rotational kinetic energy?
Medium
- A16 J
- B4.0 J
- C32 J
- D2.0 J
19.The graph of angular velocity against time for a rotating body is a straight line passing through the origin with a positive gradient. Which quantity does the gradient of this graph represent?
Medium
- AAngular acceleration
- BAngular displacement
- CMoment of inertia
- DAngular momentum
20.A body in rotational equilibrium can be rotating with a constant angular velocity.
Easy
True or false?
21.When a hollow sphere rolls without slipping, the point of the sphere in contact with the ground has a linear velocity of 2v, where v is the velocity of the centre of mass.
Medium
True or false?
22.Match each rotational quantity with its correct SI unit.
Medium
- Torque
- Angular momentum
- Moment of inertia
- Angular impulse
- kg m² s⁻¹
- N m
- N m s
- kg m²
23.Match each linear quantity with its rotational equivalent in Newton's second law and kinematics.
Medium
- Mass m
- Force F
- Linear velocity v
- Linear acceleration a
- Angular acceleration α
- Angular velocity ω
- Moment of inertia I
- Torque τ
24.Place the following steps in the correct order to calculate the maximum height reached by a solid sphere rolling without slipping up a ramp from an initial angular velocity ω.
Medium
- Write the total kinetic energy at the bottom as ½mv² + ½Iω² and substitute v = ωr and I = ⅖mr²
- Set the total kinetic energy equal to the gravitational potential energy mgΔh at maximum height
- Rearrange to make the height Δh the subject and evaluate
- Identify the initial angular velocity ω and radius r of the sphere
25.Place the following steps in the correct order to determine the net torque on a wheel with several forces acting on it.
Medium
- Identify each force and its perpendicular distance from the axis of rotation
- Calculate the torque of each force using τ = Fr sin θ
- Sum the clockwise torques and the anticlockwise torques separately
- Subtract the smaller total from the larger to find the net torque and state its direction
26.A figure skater spins on the spot with arms outstretched, then pulls her arms in. Which of the following statements are correct? (Select all that apply.)
Hard
- AHer angular momentum stays constant
- BHer moment of inertia decreases
- CHer angular velocity increases
- DHer rotational kinetic energy stays constant
- EA net external torque acts on her
27.A bicycle pedal is at the bottom of its rotation. The distance from the pedal to the axis of rotation is 24 cm. A downward force of 90 N is applied to the pedal. What is the magnitude of the torque produced about the axis?
Medium
- A0 N m
- B21.6 N m
- C45.0 N m
- D90.0 N m
28.A force of 50.0 N is applied at an angle of 43° to a spanner of length 25 cm. What is the torque produced?
Medium
- A8.5 N m
- B9.1 N m
- C12.5 N m
- D13.7 N m
29.A couple consists of two forces of magnitude 175 N each, applied at an angle of 75.0° to a bar of length 23.5 cm. What is the torque provided by the couple?
Medium
- A39.7 N m
- B41.1 N m
- C79.4 N m
- D82.2 N m
30.A block of mass 10 g hangs from a string wrapped around a solid cylindrical pulley of mass 5.0 kg and radius 0.30 m. The pulley's moment of inertia is I = ½MR². The block accelerates downwards at 2.4 m s⁻². What is the magnitude of the tension in the string?
Hard
- A0.074 N
- B0.098 N
- C0.147 N
- D0.245 N
31.A hollow sphere of mass 0.80 kg and radius 30 cm rolls without slipping. Its moment of inertia is I = ⅔mr². What is its total kinetic energy when its centre of mass moves at 3.0 m s⁻¹?
Hard
- A6.0 J
- B3.6 J
- C5.4 J
- D7.2 J
32.A solid sphere (I = ⅖mr²) rolls without slipping down a slope. It starts from rest at height h. Which expression gives its linear speed v at the bottom?
Hard
- Av = √(10gh/7)
- Bv = √(2gh)
- Cv = √(5gh/7)
- Dv = √(7gh/10)
33.A torque–time graph for a rotating object shows a triangular positive pulse of peak torque 10 N m lasting 3 s, followed by a constant negative torque of 5 N m lasting 2 s. The object has moment of inertia 5.0 kg m² and initial angular velocity 2.0 rad s⁻¹ clockwise. Taking anticlockwise as positive, what is the angular velocity at t = 5 s?
Medium
- A1.0 rad s⁻¹ anticlockwise
- B1.0 rad s⁻¹ clockwise
- C3.0 rad s⁻¹ clockwise
- D0 rad s⁻¹
34.A flywheel of mass M and radius R rotates at angular velocity ω. Its rotational kinetic energy is EK. A second flywheel of mass ½M and radius ½R is placed on top and they rotate together. The new angular velocity is (2/3)ω. What is the new rotational kinetic energy?
Hard
- AEK/2
- BEK/4
- CEK/8
- DEK/24
35.A solid sphere and a hollow sphere of the same mass and radius are released from rest at the top of the same slope and roll without slipping. Which of the following statements are correct? (Select all that apply.)
Hard
- AThe solid sphere reaches the bottom first.
- BThe hollow sphere reaches the bottom first.
- CBoth spheres have the same total kinetic energy at the bottom.
- DThe hollow sphere has a larger fraction of its total kinetic energy as rotational kinetic energy.
- EThe solid sphere has a larger moment of inertia than the hollow sphere.
36.A wheel is acted on by several forces. Which of the following statements about the resultant torque and the wheel's motion are correct? (Select all that apply.)
Medium
- AIf the sum of clockwise torques equals the sum of anticlockwise torques, the wheel has zero angular acceleration.
- BA net torque always causes the wheel to speed up.
- CA net torque causes an angular acceleration in the direction of the net torque.
- DIf the resultant torque is zero, the wheel must be at rest.
- EThe direction of angular acceleration is opposite to the net torque.
37.For a body rolling without slipping, the point of contact with the ground has zero linear velocity.
Easy
True or false?
38.The moment of inertia of a body depends only on its mass, not on how the mass is distributed relative to the axis of rotation.
Medium
True or false?
39.If the resultant torque on a rotating body is zero, its angular momentum is conserved.
Medium
True or false?
40.Match each rotational quantity with the area or gradient it corresponds to on a graph.
Medium
- Angular displacement
- Angular velocity
- Angular acceleration
- Area under the angular velocity–time graph
- Gradient of the angular displacement–time graph
- Gradient of the angular velocity–time graph
41.What is the SI unit of torque?
Easy
42.What is the name of the quantity defined as the resistance to a change in rotational motion?
Medium
43.A wheel of radius 0.25 m is rotating at 4.0 rad s⁻¹. Calculate the linear speed of a point on its rim.
Medium
44.A disc starts from rest and accelerates uniformly at 2.0 rad s⁻² for 5.0 s. Calculate its angular displacement in radians.
Medium
45.Complete the following statements about rotational motion.
Medium
The rate of change of angular displacement is called ____. The rate of change of angular velocity is called ____. The product of moment of inertia and angular velocity is called ____.
46.Complete the equation for the torque of a couple.
Medium
For a couple consisting of two forces F separated by a perpendicular distance d, the torque is given by τ = ____.
47.A thin rod of mass 0.60 kg and length 0.40 m rotates about its centre. The moment of inertia of a thin rod about its centre is I = (1/12)mL². Calculate the moment of inertia of the rod.
Medium
48.Complete the following equations for rotational motion.
Easy
Torque is given by τ = I____ and angular momentum is given by L = I____.
49.Name the three factors that determine the moment of inertia of a rigid body.
Medium
The moment of inertia of a body depends on its ____, its ____, and its ____ relative to the axis of rotation.
50.What name is given to the rotational equivalent of linear momentum?
Medium
51.What is the name of the quantity defined as the product of a force and the perpendicular distance from a pivot?
Medium
52.A couple is formed by two forces of magnitude 20 N whose lines of action are separated by a perpendicular distance of 0.30 m. Calculate the torque of the couple.
Medium
53.Explain why a couple does not produce a net linear force but does produce a net torque.
Hard
54.Explain why the moment of inertia of a single object can take different values.
Hard
55.State and explain whether the torque produced by a fixed force applied to a door changes when the force is applied at an angle θ rather than perpendicular to the door.
Hard
56.A vase is acted on by two equal and opposite forces of magnitude 5 N applied at its ends, perpendicular to its length of 30 cm. Explain whether these forces produce translation, rotation, or both.
Hard
57.A person applying a force to a spanner changes their grip to apply a larger torque while keeping the magnitude of the force the same. State and explain the angle at which the force should be applied to exert the greatest torque.
Hard
58.A wheel of radius 0.25 m is rotating at 4.0 rad s⁻¹. Calculate the linear speed of a point on its rim.
Medium
- A0 m/s
- B2 m/s
- C1 m/s
- D3 m/s
59.A disc starts from rest and accelerates uniformly at 2.0 rad s⁻² for 5.0 s. Calculate its angular displacement in radians.
Medium
- A25 rad
- B31 rad
- C22 rad
- D28 rad
60.A thin rod of mass 0.60 kg and length 0.40 m rotates about its centre. The moment of inertia of a thin rod about its centre is I = (1/12)mL². Calculate the moment of inertia of the rod.
Medium
- A0.008 kg m²
- B1.008 kg m²
- C-0.992 kg m²
- D2.008 kg m²
61.A couple is formed by two forces of magnitude 20 N whose lines of action are separated by a perpendicular distance of 0.30 m. Calculate the torque of the couple.
Medium
- A6 N m
- B7 N m
- C8 N m
- D5 N m

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