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A.4 Rigid body mechanics

Practice exam-style IB Physics questions for Rigid body mechanics, aligned with the syllabus and grouped by topic.

Verified by Kun
Verified by Kun
Paper
Difficulty
Status
Level
Question 1
HL • Paper 1A
Easy
Calculator Permitted

A force of 40 N40\ \text{N} acts at a point 0.30 m0.30\ \text{m} from a fixed axis. The angle between the force and the radius from the axis is 30∘30^\circ.

What is the magnitude of the torque about the axis?

A.

6.0 N m6.0\ \text{N m}

B.

10 N m10\ \text{N m}

C.

12 N m12\ \text{N m}

D.

3.0 N m3.0\ \text{N m}

Question 2
HL • Paper 1A
Easy
Calculator Permitted

Two equal and opposite forces of magnitude 25 N25\ \text{N} act on a rigid body. Their parallel lines of action are separated by a perpendicular distance of 0.18 m0.18\ \text{m}.

The resultant force and the magnitude of the torque of the couple are

A.

50 N50\ \text{N} and 4.5 N m4.5\ \text{N m}

B.

0 N0\ \text{N} and 4.5 N m4.5\ \text{N m}

C.

0 N0\ \text{N} and 9.0 N m9.0\ \text{N m}

D.

50 N50\ \text{N} and 9.0 N m9.0\ \text{N m}

Question 3
HL • Paper 1A
Easy
Calculator Permitted

A uniform horizontal beam of length 4.0 m4.0\ \text{m} and weight 80 N80\ \text{N} is pivoted at its left-hand end. A downward force of 40 N40\ \text{N} acts at the right-hand end. A vertical upward force FF acts at a point 3.0 m3.0\ \text{m} from the pivot.

The magnitude of FF required for rotational equilibrium is

A simple side-view diagram of a horizontal uniform beam pivoted at its left end. The beam has a downward weight arrow at its centre, a downward applied-force arrow at the right end, and an upward force arrow at a point between the centre and right end. The pivot is clearly marked, and distances along the beam are indicated symbolically without giving the answer.
A.

80 N80\ \text{N}

B.

53 N53\ \text{N}

C.

1.6×102 N1.6\times10^2\ \text{N}

D.

1.1×102 N1.1\times10^2\ \text{N}

Question 4
HL • Paper 1A
Easy
Calculator Permitted

A wheel has an initial angular speed of 3.0 rad s−13.0\ \text{rad s}^{-1} and a uniform angular acceleration of 2.0 rad s−22.0\ \text{rad s}^{-2}.

The angular displacement after 4.0 s4.0\ \text{s} is

A.

44 rad44\ \text{rad}

B.

8.0 rad8.0\ \text{rad}

C.

16 rad16\ \text{rad}

D.

28 rad28\ \text{rad}

Question 5
HL • Paper 1A
Easy
Calculator Permitted

A rigid disc rotates with constant angular speed 12 rad s−112\ \text{rad s}^{-1}. Point P is 0.20 m0.20\ \text{m} from the axis and point Q is 0.40 m0.40\ \text{m} from the axis.

The correct comparison of the centripetal accelerations of P and Q is

A.

aQ=12aPa_Q=\dfrac{1}{2}a_P

B.

aQ=4aPa_Q=4a_P

C.

aQ=2aPa_Q=2a_P

D.

aQ=aPa_Q=a_P

Question 6
HL • Paper 1A
Easy
Calculator Permitted

A flywheel has moment of inertia 0.80 kg m20.80\ \text{kg m}^2. A constant resultant torque of 4.0 N m4.0\ \text{N m} acts on it from rest for 3.0 s3.0\ \text{s}.

The angular speed after 3.0 s3.0\ \text{s} is

A.

1.7 rad s−11.7\ \text{rad s}^{-1}

B.

15 rad s−115\ \text{rad s}^{-1}

C.

5.0 rad s−15.0\ \text{rad s}^{-1}

D.

12 rad s−112\ \text{rad s}^{-1}

Question 7
HL • Paper 1A
Easy
Calculator Permitted

A rotating skater has moment of inertia 6.0 kg m26.0\ \text{kg m}^2 and angular speed 4.0 rad s−14.0\ \text{rad s}^{-1}. The skater pulls in their arms so that the moment of inertia becomes 2.0 kg m22.0\ \text{kg m}^2. No external torque acts.

The final angular speed is

A.

8.0 rad s−18.0\ \text{rad s}^{-1}

B.

1.3 rad s−11.3\ \text{rad s}^{-1}

C.

4.0 rad s−14.0\ \text{rad s}^{-1}

D.

12 rad s−112\ \text{rad s}^{-1}

Question 8
HL • Paper 2
Easy
Calculator Permitted

A student pushes on a door at a point 0.82 m0.82\ \text{m} from the hinge. The force has magnitude 35 N35\ \text{N} and acts at 60∘60^\circ to the line from the hinge to the point of application.

A simple top-view diagram of a hinged door. The hinge is shown at one end and labelled. A point on the door away from the hinge is marked where the force is applied. The distance from hinge to the point is labelled $0.82\ \text{m}$. An arrow labelled $35\ \text{N}$ acts at an angle to the line joining the hinge and point of application, with the angle labelled $60^\circ$. The direction of possible rotation is not shown.
A

Calculate the magnitude of the torque about the hinge.

[2]
Write your answer here...
B

State the condition on the angle for this force to produce maximum torque about the hinge.

[1]
Write your answer here...

0

Question 9
HL • Paper 2
Easy
Calculator Permitted

Two parallel forces of magnitude 18 N18\ \text{N} act on a rigid plate. The forces are opposite in direction and their lines of action are separated by 0.34 m0.34\ \text{m}.

A rectangular rigid plate shown from above. Two equal-length parallel arrows act on different parts of the plate in opposite directions. The perpendicular separation between the two parallel lines of action is labelled $0.34\ \text{m}$. Each arrow is labelled $18\ \text{N}$. No rotation direction is indicated.
A

Explain why this pair of forces can cause rotation without causing translational acceleration of the centre of mass.

[2]
Write your answer here...
B

Calculate the magnitude of the torque of the couple.

[1]
Write your answer here...

0

Question 10
HL • Paper 1A
Medium
Calculator Permitted

Three small masses are fixed to a light rod that rotates about an axis perpendicular to the rod through its centre. The masses are arranged along the rod as follows.

  • 0.20 kg0.20\ \text{kg} at 0.30 m0.30\ \text{m} to the left of the axis
  • 0.40 kg0.40\ \text{kg} at 0.10 m0.10\ \text{m} to the right of the axis
  • 0.10 kg0.10\ \text{kg} at 0.50 m0.50\ \text{m} to the right of the axis

The moment of inertia of the system about the axis is

A.

0.12 kg m20.12\ \text{kg m}^2

B.

0.18 kg m20.18\ \text{kg m}^2

C.

0.047 kg m20.047\ \text{kg m}^2

D.

0.70 kg m20.70\ \text{kg m}^2

Question 11
HL • Paper 1A
Medium
Calculator Permitted

A rigid body of moment of inertia 0.40 kg m20.40\ \text{kg m}^2 is initially at rest. A torque of 0.50 N m0.50\ \text{N m} acts for 6.0 s6.0\ \text{s} in the positive sense. A torque of 0.20 N m0.20\ \text{N m} then acts for 5.0 s5.0\ \text{s} in the opposite sense.

The final angular speed is

A.

10 rad s−110\ \text{rad s}^{-1}

B.

2.5 rad s−12.5\ \text{rad s}^{-1}

C.

7.5 rad s−17.5\ \text{rad s}^{-1}

D.

5.0 rad s−15.0\ \text{rad s}^{-1}

Question 12
HL • Paper 1A
Medium
Calculator Permitted

A body rotates with angular momentum LL. Its moment of inertia decreases from II to I2\dfrac{I}{2} while no external torque acts.

The rotational kinetic energy changes by a factor of

A.

12\dfrac{1}{2}

B.

14\dfrac{1}{4}

C.

22

D.

44

Question 13
HL • Paper 2
Medium
Calculator Permitted

A uniform horizontal beam of weight 120 N120\ \text{N} is supported by a pivot at its left end and a vertical cable at its right end. The beam is 2.4 m2.4\ \text{m} long. A load of weight 80 N80\ \text{N} is placed 0.60 m0.60\ \text{m} from the pivot.

A side-view line diagram of a horizontal beam. The left end rests on a triangular pivot labelled pivot. The right end has a vertical upward cable force labelled $T$. The beam length is labelled $2.4\ \text{m}$. The beam weight $120\ \text{N}$ acts downward at the centre of the beam. A load force $80\ \text{N}$ acts downward closer to the pivot, with distance from the pivot labelled $0.60\ \text{m}$.
A

State the condition for the beam to be in rotational equilibrium.

[1]
Write your answer here...
B

Determine the tension in the cable.

[3]
Write your answer here...

0

Question 14
HL • Paper 2
Medium
Calculator Permitted

A point on the rim of a rotating disc is 0.18 m0.18\ \text{m} from the axis. The disc rotates at 420 rev min−1420\ \text{rev min}^{-1}.

A

Calculate the angular speed of the disc in rad s−1\text{rad s}^{-1}.

[2]
Write your answer here...
B

Calculate the linear speed of the point on the rim.

[1]
Write your answer here...

0

Question 15
HL • Paper 2
Medium
Calculator Permitted

The graph shows the variation of angular velocity ω\omega with time tt for a motor shaft. The angular acceleration is uniform over the interval shown.

Straight-line angular velocity of a motor shaft versus time.
A

State the physical quantity represented by the gradient of the graph.

[1]
Write your answer here...
B

The angular velocity increases from 8.0 rad s−18.0\ \text{rad s}^{-1} to 20 rad s−120\ \text{rad s}^{-1} in 6.0 s6.0\ \text{s}. Determine the angular displacement during this interval.

[3]
Write your answer here...

0

Question 16
HL • Paper 2
Medium
Calculator Permitted

Three small masses are fixed to a light rigid frame that rotates about an axis perpendicular to the frame. Their distances from the axis are shown.

A simple top-view diagram of a light rigid frame with a central rotation axis marked by a cross. Three small point masses are shown at different radial distances from the axis. The masses are labelled $0.20\ \text{kg}$, $0.30\ \text{kg}$ and $0.50\ \text{kg}$, with their respective distances from the axis labelled $0.40\ \text{m}$, $0.25\ \text{m}$ and $0.10\ \text{m}$. The frame itself is labelled light.
A

Calculate the moment of inertia of the system about the axis.

[3]
Write your answer here...

0

Question 17
HL • Paper 2
Medium
Calculator Permitted

A flywheel of moment of inertia 0.72 kg m20.72\ \text{kg m}^2 is initially at rest. A constant resultant torque of 4.5 N m4.5\ \text{N m} acts on it for 8.0 s8.0\ \text{s}.

A

Determine the angular acceleration of the flywheel.

[2]
Write your answer here...
B

Determine the angular speed after 8.0 s8.0\ \text{s}.

[2]
Write your answer here...

0

Question 18
HL • Paper 1B
Medium
Calculator Permitted

A student investigates the torque produced on a light gate by applying the same force at the handle at different angles to the radius line from the hinge.

Annotated top-view diagram of a hinged gate showing the hinge axis, radius line to the handle, applied force, angle between force and radius, and a small table or graph summarizing how the measured torque varies with angle for a fixed force and handle distance.
A

Use the data to determine the torque when the angle between the force and the radius line is 60∘60^\circ.

[1]
Write your answer here...
B

State the angle at which the torque is greatest.

[1]
Write your answer here...
C

Explain why the torque becomes zero when the force is directed through the hinge.

[2]
Write your answer here...

0

Question 19
HL • Paper 1B
Medium
Calculator Permitted

Two equal and opposite tangential forces are applied to a steering wheel to turn it while its centre remains fixed.

Annotated diagram of a circular steering wheel showing two equal and opposite tangential forces applied at different points, the perpendicular separation between their lines of action, and the centre of the wheel.
A

State the resultant force on the steering wheel due to the two applied forces.

[1]
Write your answer here...
B

Calculate the torque of the couple.

[1]
Write your answer here...
C

Explain why the wheel can have angular acceleration even though the resultant force is zero.

[1]
Write your answer here...

0

Question 20
HL • Paper 1A
Medium
Calculator Permitted

Two coaxial discs are free to rotate without external torque. Disc 1 has moment of inertia 0.20 kg m20.20\ \text{kg m}^2 and rotates clockwise at 30 rad s−130\ \text{rad s}^{-1}. Disc 2 has moment of inertia 0.10 kg m20.10\ \text{kg m}^2 and rotates counter-clockwise at 10 rad s−110\ \text{rad s}^{-1}. The discs are then clamped together.

Taking clockwise as positive, the common final angular velocity is

A.

23 rad s−123\ \text{rad s}^{-1} clockwise

B.

17 rad s−117\ \text{rad s}^{-1} counter-clockwise

C.

17 rad s−117\ \text{rad s}^{-1} clockwise

D.

20 rad s−120\ \text{rad s}^{-1} clockwise

Question 21
HL • Paper 1A
Medium
Calculator Permitted

A solid cylinder of mass MM and radius RR rolls without slipping from rest down a slope. The vertical drop of its centre of mass is hh. The moment of inertia of the cylinder about its central axis is 12MR2\dfrac{1}{2}MR^2.

The speed of the centre of mass at the bottom of the slope is

A side-view diagram of a solid cylinder rolling down an inclined plane. The cylinder is shown on the slope, with its centre of mass indicated and the vertical drop of the centre from the starting position to the bottom indicated symbolically as $h$. No numerical values or answer expressions are shown.
A.

2gh\sqrt{2gh}

B.

gh\sqrt{gh}

C.

4gh3\sqrt{\dfrac{4gh}{3}}

D.

2gh3\sqrt{\dfrac{2gh}{3}}

Question 22
HL • Paper 2
Medium
Calculator Permitted

Two coaxial turntables are initially separate. Turntable A has moment of inertia 0.30 kg m20.30\ \text{kg m}^2 and angular velocity +12 rad s−1+12\ \text{rad s}^{-1}. Turntable B has moment of inertia 0.20 kg m20.20\ \text{kg m}^2 and angular velocity −6.0 rad s−1-6.0\ \text{rad s}^{-1}. The turntables are brought into contact and rotate together with common angular velocity Ω\Omega. External torque is negligible.

A pair of coaxial circular turntables shown before contact and after contact. Before contact, turntable A and turntable B are drawn on the same vertical axis but separated, with arrows showing opposite senses of rotation and labels $I_A = 0.30\ \text{kg m}^{2}$, $\omega_A = +12\ \text{rad s}^{-1}$, $I_B = 0.20\ \text{kg m}^{2}$, and $\omega_B = -6.0\ \text{rad s}^{-1}$. After contact, the two are shown clamped together with a single angular velocity label $\Omega$. No numerical calculation is shown.
A

State why angular momentum is conserved during the contact process.

[1]
Write your answer here...
B

Calculate the common angular velocity Ί\Omega after contact.

[3]
Write your answer here...

0

Question 23
HL • Paper 2
Medium
Calculator Permitted

A variable resultant torque acts on a rotor. The graph shows torque τ\tau against time tt. Counter-clockwise torque is taken as positive.

Torque against time showing one positive and one negative triangular pulse.
A

State the quantity represented by the signed area under the torque-time graph.

[1]
Write your answer here...
B

The positive triangular area is 0.36 N m s0.36\ \text{N m s} and the negative triangular area is 0.12 N m s0.12\ \text{N m s}. Determine the change in angular momentum of the rotor.

[2]
Write your answer here...
C

The rotor has moment of inertia 0.080 kg m20.080\ \text{kg m}^2 and is initially at rest. Calculate its final angular velocity.

[1]
Write your answer here...

0

Question 24
HL • Paper 2
Medium
Calculator Permitted

A skater rotates with angular speed 3.0 rad s−13.0\ \text{rad s}^{-1} and moment of inertia 4.8 kg m24.8\ \text{kg m}^2. The skater pulls in their arms so that the moment of inertia becomes 3.2 kg m23.2\ \text{kg m}^2. External torque is negligible.

A

Calculate the final angular speed of the skater.

[2]
Write your answer here...
B

Explain why the rotational kinetic energy changes even though angular momentum is conserved.

[2]
Write your answer here...

0

Question 25
HL • Paper 2
Medium
Calculator Permitted

A rigid block floats on a nearly frictionless air table. Two different horizontal forces of the same magnitude may be applied to the block. Force A acts through the centre of mass. Force B acts along a line that does not pass through the centre of mass.

A top-view diagram of a rectangular block on an air table. The centre of mass is marked by a small dot labelled centre of mass. Force A is drawn as an arrow whose line of action passes through the centre of mass. Force B is drawn as a separate arrow of the same length on the same block, but its line of action misses the centre of mass. The offset distance from the centre of mass to the line of action of B is indicated without giving a numerical value.
A

Compare the translational acceleration of the centre of mass produced by the two forces.

[1]
Write your answer here...
B

Suggest why force B, but not force A, also produces angular acceleration about the centre of mass.

[2]
Write your answer here...

0

Question 26
HL • Paper 1B
Medium
Calculator Permitted

A uniform horizontal beam is hinged to a wall and supported by a cable. A lamp is attached to the beam. The beam is stationary.

Free-body diagram of a horizontal hinged beam with hinge at the wall, cable attached to the outer end at an angle above the beam, weight of the beam acting at its centre, lamp weight acting further from the hinge, and all relevant distances and force labels shown.
A

State the condition for rotational equilibrium of the beam.

[1]
Write your answer here...
B

Calculate the tension in the cable.

[2]
Write your answer here...
C

Explain why the force exerted by the hinge does not appear in the torque equation used in part (b).

[1]
Write your answer here...

0

Question 27
HL • Paper 1B
Medium
Calculator Permitted

The angular speed of a cooling fan is recorded as it speeds up uniformly and then rotates at constant angular speed.

Angular speed of a cooling fan as it speeds up, then stays constant.
A

Determine the angular acceleration during the speeding-up interval.

[1]
Write your answer here...
B

Calculate the angular displacement during the first 8.0 s8.0\ \text{s}.

[2]
Write your answer here...
C

Calculate the linear speed of the indicated point on the blade at the end of the speeding-up interval.

[1]
Write your answer here...
D

Explain why the indicated point still has acceleration when the angular speed is constant.

[1]
Write your answer here...

0

Question 28
HL • Paper 1B
Medium
Calculator Permitted

A skater rotates with negligible external torque while pulling their arms closer to the rotation axis. At t=0t=0, the skater's angular speed is 1.8 rad s−11.8\ \text{rad s}^{-1}. The graph shows how the skater's moment of inertia changes with time.

Moment of inertia of a skater over time.
A

Calculate the final angular speed of the skater.

[2]
Write your answer here...
B

Determine the increase in rotational kinetic energy.

[2]
Write your answer here...
C

Explain why the rotational kinetic energy can increase even though angular momentum is conserved.

[1]
Write your answer here...

0

Question 29
HL • Paper 1B
Medium
Calculator Permitted

A torsion wire exerts a restoring torque on a rigid disc when the disc is rotated through a small angular displacement. The graph shows torque as a function of angular displacement. The moment of inertia of the disc is 0.050 kg m20.050\ \text{kg m}^2.

Torque against angular displacement for a torsion wire.
A

Determine the torsion constant of the wire.

[2]
Write your answer here...
B

Calculate the angular acceleration when the angular displacement is +0.40 rad+0.40\ \text{rad}.

[1]
Write your answer here...
C

Explain why this torque can produce angular simple harmonic motion.

[2]
Write your answer here...

0

Question 30
HL • Paper 2
Medium
Calculator Permitted

A uniform solid cylinder rolls without slipping from rest down a slope. The vertical drop of its centre of mass is 0.75 m0.75\ \text{m}. For a solid cylinder about its central axis, I=12MR2I=\dfrac{1}{2}MR^2. Air resistance is negligible.

A side-view diagram of a solid cylinder rolling down an inclined plane. The cylinder is shown at an upper position and a lower position. The vertical drop of the centre of mass is labelled $0.75\ \text{m}$. The cylinder radius is indicated by a short line from centre to rim labelled $R$. The condition rolling without slipping is stated on the diagram. No final speed is shown.
A

Show that the final speed vv of the centre of mass satisfies gh=34v2gh=\dfrac{3}{4}v^2.

[3]
Write your answer here...
B

Calculate the final speed of the centre of mass.

[1]
Write your answer here...

0

Question 31
HL • Paper 1B
Hard
Calculator Permitted

A variable resultant torque acts on a turntable. The graph shows the torque as a function of time. The turntable has moment of inertia 0.42 kg m20.42\ \text{kg m}^2 and initial angular speed 3.0 rad s−13.0\ \text{rad s}^{-1} in the positive sense.

Torque as a function of time, showing one positive pulse and one negative region.
A

Determine the angular impulse delivered to the turntable over the time interval shown.

[2]
Write your answer here...
B

Calculate the final angular speed of the turntable.

[2]
Write your answer here...
C

Explain the significance of a negative region on the graph.

[1]
Write your answer here...

0

Question 32
HL • Paper 1B
Hard
Calculator Permitted

Four small masses are fixed to a light rotating frame. The frame itself has negligible moment of inertia. A constant resultant torque is then applied about the central axis.

QuantityMass / kgRadius from axis / mResultant torque / N m
Mass 10.200.25—
Mass 20.200.25—
Mass 30.300.10—
Mass 40.100.40—
Applied torque——0.132
A

Calculate the moment of inertia of the system about the central axis.

[2]
Write your answer here...
B

Calculate the angular acceleration produced by the applied torque.

[1]
Write your answer here...
C

The system starts from rest. Calculate its angular speed after 4.0 s4.0\ \text{s}.

[1]
Write your answer here...
D

One mass is moved closer to the axis while the same torque is applied. Suggest the effect on the angular acceleration.

[1]
Write your answer here...

0

Question 33
HL • Paper 1B
Hard
Calculator Permitted

Two coaxial discs rotate freely with opposite senses and are then clamped together. Friction in the clamp is internal to the two-disc system.

DiscMoment of inertia / kg m^2Initial angular velocity / rad s^-1
Disc 10.06024
Disc 20.040-12
A

State which conservation law should be applied during the clamping process.

[1]
Write your answer here...
B

Calculate the common final angular speed, including its rotational sense.

[2]
Write your answer here...
C

Calculate the change in rotational kinetic energy and explain why it occurs.

[2]
Write your answer here...

0

Question 34
HL • Paper 1B
Hard
Calculator Permitted

A motor exerts a net driving torque on a flywheel as it turns through an angular displacement. The flywheel has moment of inertia 0.25 kg m20.25\ \text{kg m}^2 and initial angular speed 16 rad s−116\ \text{rad s}^{-1}.

Net driving torque decreases linearly with angular displacement.
A

Determine the work done on the flywheel over the interval shown.

[2]
Write your answer here...
B

Calculate the final angular speed of the flywheel.

[2]
Write your answer here...
C

Explain why a positive area under this graph corresponds to an increase in rotational kinetic energy.

[1]
Write your answer here...

0

Question 35
HL • Paper 2
Hard
Calculator Permitted

A uniform horizontal beam is hinged to a vertical wall at one end. A cable attached to the other end of the beam makes an angle of 35∘35^\circ above the beam. A box is suspended from the beam 2.2 m2.2\ \text{m} from the hinge.

Side-view line diagram of a horizontal uniform beam hinged to a wall at the left. A cable runs from the right end of the beam up to the wall, making an angle above the beam. A box hangs vertically from a point between the hinge and the right end. Labels show the hinge, beam length, cable angle, beam weight acting at its centre, box weight acting at the suspension point, and cable tension.
A

The beam has length 3.0 m3.0\ \text{m} and mass 18 kg18\ \text{kg}. The box has mass 25 kg25\ \text{kg} and is suspended 2.2 m2.2\ \text{m} from the hinge.

I.

State the expression for the torque of a force about an axis.

[1]
Write your answer here...
II.

Calculate the tension in the cable when the beam is in rotational equilibrium.

[3]
Write your answer here...
B

The cable is reattached so that it makes a smaller angle with the horizontal beam, while the box remains at the same position. Explain the effect on the force exerted by the hinge on the beam.

[3]
Write your answer here...

0

Question 36
HL • Paper 2
Hard
Calculator Permitted

A centrifuge rotor starts from rest. Its angular velocity is increased uniformly, then kept constant, and finally reduced uniformly to rest.

Angular velocity of the centrifuge rotor versus time.
A

During the first 8.0 s8.0\ \text{s} the angular velocity increases uniformly from 00 to 120 rad s−1120\ \text{rad s}^{-1}.

I.

Determine the angular acceleration during this interval.

[2]
Write your answer here...
II.

Determine the angular displacement of the rotor during this interval.

[2]
Write your answer here...
B

The rotor then spins at constant angular velocity for 4.0 s4.0\ \text{s} before being brought uniformly to rest in a further 6.0 s6.0\ \text{s}. Sketch the variation of angular acceleration with time for the complete motion. Numerical values should be shown on the angular acceleration axis.

[2]
Write your answer here...

0

Question 37
HL • Paper 2
Hard
Calculator Permitted

Four identical small masses, each of mass 0.50 kg0.50\ \text{kg}, are attached to a light rigid rectangular frame. The frame rotates about an axis through its centre and perpendicular to the plane of the frame.

Top-view diagram of a light rectangular frame with four equal small masses at the corners. The central rotation axis is marked through the centre of the rectangle, perpendicular to the page. The side lengths of the rectangle are labelled, and arrows show the sense of rotation. A motor torque and a resisting friction torque are indicated as opposing rotational senses.
A

Each mass is 0.50 kg0.50\ \text{kg}. The rectangle has sides 0.80 m0.80\ \text{m} and 0.60 m0.60\ \text{m}. A motor provides a torque of 2.1 N m2.1\ \text{N m} and friction provides an opposing torque of 0.45 N m0.45\ \text{N m}.

I.

Calculate the moment of inertia of the four masses about the axis.

[2]
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II.

Calculate the angular acceleration of the frame.

[2]
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B

The four masses are moved closer to the axis while the motor torque and friction torque remain unchanged. Discuss the effect on the angular acceleration and on the rotational kinetic energy at a given angular speed.

[3]
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0

Question 38
HL • Paper 2
Hard
Calculator Permitted

A small satellite in deep space is initially at rest. Two thrusters fire simultaneously, producing a couple about the centre of mass of the satellite. The moment of inertia of the satellite about this axis is 52 kg m252\ \text{kg m}^2.

Top-view diagram of a rectangular satellite. Two equal and opposite thruster forces act along parallel but separated lines of action on opposite sides of the centre of mass, forming a couple. The perpendicular separation between the force lines is labelled. The centre of mass and rotation axis are marked.
A

Each thruster produces a force of 12 N12\ \text{N}. The perpendicular separation of the two lines of action is 1.8 m1.8\ \text{m}. The thrusters fire for 5.0 s5.0\ \text{s}.

I.

Calculate the torque of the couple.

[2]
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II.

Calculate the angular acceleration of the satellite.

[2]
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III.

Calculate the angular displacement of the satellite while the thrusters fire.

[2]
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B

Explain why the centre of mass of the satellite does not accelerate even though the satellite rotates.

[2]
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0

Question 39
HL • Paper 2
Hard
Calculator Permitted

A disc is suspended from a thin torsion wire. When the disc is twisted and released, the wire exerts a restoring torque on the disc given by τ=−κθ\tau=-\kappa\theta, where θ\theta is the angular displacement from equilibrium. For this question, I=0.084 kg m2I = 0.084\ \text{kg m}^2, κ=0.056 N m rad−1\kappa = 0.056\ \text{N m rad}^{-1}, and the disc is released from rest at θ0=0.40 rad\theta_0 = 0.40\ \text{rad}.

Diagram of a horizontal disc suspended by a vertical torsion wire. The equilibrium position and a twisted position are indicated with an angular displacement label. A restoring torque arrow points back toward equilibrium.
A

The disc has moment of inertia 0.084 kg m20.084\ \text{kg m}^2. The torsion constant is 0.056 N m rad−10.056\ \text{N m rad}^{-1}. The disc is released from rest at angular displacement 0.40 rad0.40\ \text{rad}.

I.

Calculate the initial angular acceleration of the disc, including its direction.

[2]
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II.

Explain why the motion can be described as angular simple harmonic motion.

[2]
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B

Calculate the maximum angular velocity of the disc, assuming no energy is dissipated.

[2]
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0

Question 40
HL • Paper 2
Hard
Calculator Permitted

A small satellite moves in an elliptical orbit around a planet. The planet is much more massive than the satellite. Air resistance is negligible.

Diagram of an elliptical orbit with the planet at one focus. The satellite is shown at perigee and apogee. Radius vectors from the planet to the satellite at both positions are labelled. Tangential velocity arrows at perigee and apogee are shown, with the perigee arrow longer than the apogee arrow.
A

At perigee the orbital radius is 7.0×106 m7.0\times10^6\ \text{m} and the speed is 7.8×103 m s−17.8\times10^3\ \text{m s}^{-1}. At apogee the orbital radius is 1.4×107 m1.4\times10^7\ \text{m}. At both positions the velocity is perpendicular to the radius line.

I.

Explain why the angular momentum of the satellite about the planet is conserved.

[2]
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II.

Calculate the speed of the satellite at apogee.

[2]
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B

Discuss how conservation of angular momentum accounts for the satellite sweeping out equal areas in equal times.

[2]
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0

Question 41
HL • Paper 1B
Hard
Calculator Permitted

Three objects of the same mass and radius roll without slipping down the same ramp. The expression I=kMR2I=kMR^2 may be used, where k=1.0k=1.0 for a hoop, k=0.50k=0.50 for a solid cylinder and k=0.40k=0.40 for a solid sphere. The vertical drop of the centre of mass is 0.35 m0.35\ \text{m}.

RunHoop / m s^-1Solid cylinder / m s^-1Solid sphere / m s^-1
11.842.102.18
21.852.142.21
31.872.182.24
41.892.202.28
A

Calculate the predicted final speed for the solid cylinder.

[2]
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B

State which object is predicted to have the greatest final speed.

[1]
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C

Evaluate whether the data distinguish between the solid cylinder and the solid sphere.

[2]
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0

Question 42
HL • Paper 1B
Hard
Calculator Permitted

A satellite moves in an elliptical orbit around a planet. The gravitational force acts along the line joining the satellite and the centre of the planet. The table gives data for the satellite at the closest and furthest points of the orbit.

PointDistance from planet centre / mSpeed / m s^-1
Closest point7.0 × 10^69.2 × 10^3
Furthest point1.4 × 10^7—
A

Use conservation of angular momentum about the planet centre to calculate the speed at the furthest point.

[2]
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B

Compare the angular speeds at the closest and furthest points.

[2]
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C

Explain why the angular momentum of the satellite about the planet centre is conserved.

[1]
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0

Question 43
HL • Paper 2
Hard
Calculator Permitted

Two coaxial flywheels can rotate freely about the same fixed axle. They are initially separate and are then clamped together. Friction in the clamp acts only between the two flywheels during the joining process.

Diagram of two coaxial flywheels on the same axle. Flywheel A and flywheel B are shown side by side, with arrows indicating opposite initial senses of rotation. A clamp is shown between them. Labels identify the moments of inertia and initial angular velocities of each flywheel.
A

Flywheel A has moment of inertia 0.32 kg m20.32\ \text{kg m}^2 and angular velocity +18 rad s−1+18\ \text{rad s}^{-1}. Flywheel B has moment of inertia 0.18 kg m20.18\ \text{kg m}^2 and angular velocity −10 rad s−1-10\ \text{rad s}^{-1}.

I.

Calculate the common angular velocity immediately after the flywheels are clamped together.

[3]
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II.

Determine the loss of rotational kinetic energy during the clamping process.

[3]
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B

Evaluate the statement: “Because angular momentum is conserved in this process, rotational kinetic energy must also be conserved.”

[2]
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0

Question 44
HL • Paper 2
Hard
Calculator Permitted

A turntable of moment of inertia 0.72 kg m20.72\ \text{kg m}^2 rotates initially with angular velocity 4.0 rad s−14.0\ \text{rad s}^{-1} in the positive sense. A variable external torque is applied.

Torque-time graph with a triangular pulse followed by a constant negative torque.
A

From t=0t=0 to t=4.0 st=4.0\ \text{s}, the torque forms a triangular pulse with maximum value 0.80 N m0.80\ \text{N m}.

I.

Determine the angular impulse delivered during the first interval, 0≤t≤4.0 s0 \le t \le 4.0\ \text{s}.

[2]
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II.

Calculate the angular velocity of the turntable at t=4.0 st=4.0\ \text{s}.

[2]
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B

From t=4.0 st=4.0\ \text{s} to t=8.0 st=8.0\ \text{s}, a constant torque of −0.30 N m-0.30\ \text{N m} acts on the turntable.

I.

Determine the angular velocity at t=8.0 st=8.0\ \text{s}.

[2]
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II.

Explain why the final angular velocity is not equal to the initial angular velocity.

[1]
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0

Question 45
HL • Paper 2
Hard
Calculator Permitted

A solid cylinder rolls without slipping from rest down a straight ramp. The moment of inertia of a solid cylinder about its central axis is I=12MR2I=\frac{1}{2}MR^2.

Side-view diagram of a cylinder rolling down an inclined plane. The vertical height drop of the centre of mass and the distance along the ramp are labelled. An arrow indicates rolling without slipping. A second small inset shows a hoop of the same mass and radius for comparison.
A

The cylinder has mass 1.2 kg1.2\ \text{kg} and radius 0.060 m0.060\ \text{m}. The centre of mass descends through a vertical height of 0.75 m0.75\ \text{m} along a ramp of length 3.0 m3.0\ \text{m}.

I.

Show that the speed of the centre of mass at the bottom of the ramp is about 3.1 m s−13.1\ \text{m s}^{-1}.

[3]
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II.

Calculate the average acceleration of the centre of mass along the ramp, assuming it is uniform.

[2]
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B

hoop of the same mass and radius is released from the same height. For a hoop, I=MR2I=MR^2. Compare the motion of the hoop with that of the cylinder.

[3]
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0

Question 46
HL • Paper 2
Hard
Calculator Permitted

A student stands on a rotating platform and holds two identical dumbbells. The platform rotates freely except for a small constant frictional torque. Treat the student and platform, excluding the dumbbells, as having constant moment of inertia.

Top-view diagram of a student on a circular rotating platform holding two dumbbells symmetrically. Two positions of the dumbbells are indicated: initially farther from the axis and finally closer to the axis. The axis of rotation is marked at the centre, with an arrow showing the initial rotational sense and a frictional torque arrow in the opposite sense.
A

The platform and student have moment of inertia 24 kg m224\ \text{kg m}^2. Each dumbbell consists of two 3.0 kg3.0\ \text{kg} masses. Initially the dumbbells are 1.2 m1.2\ \text{m} from the axis and the angular velocity is 1.6 rad s−11.6\ \text{rad s}^{-1}. The dumbbells are pulled in to 0.40 m0.40\ \text{m} from the axis in 2.0 s2.0\ \text{s}. During this time the frictional torque is 1.8 N m1.8\ \text{N m} opposing the rotation.

I.

Calculate the initial moment of inertia of the system.

[2]
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II.

Determine the final angular velocity of the system.

[3]
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B

Evaluate the claim that “pulling the dumbbells in always conserves angular momentum and therefore must increase the angular speed”.

[3]
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0

Question 47
HL • Paper 2
Hard
Calculator Permitted

A light rigid rod is pivoted at its centre and can rotate freely in a horizontal plane. Two identical small masses are fixed to the ends of the rod. A small piece of clay moves horizontally and sticks to one end of the rod in an inelastic collision.

Top-view diagram of a light rod pivoted at its centre with equal masses at both ends. A small clay mass approaches one end tangentially to the circular path. The distance from the pivot to the impact point is labelled. The pivot is shown at the centre, and the direction of the incoming clay velocity is perpendicular to the rod at the impact point.
A

The rod has length 1.2 m1.2\ \text{m}. Each fixed mass is 0.40 kg0.40\ \text{kg}. The clay has mass 0.20 kg0.20\ \text{kg} and speed 6.0 m s−16.0\ \text{m s}^{-1} immediately before impact. The rod is initially at rest.

I.

Calculate the angular momentum of the clay about the pivot immediately before impact.

[2]
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II.

Calculate the angular velocity immediately after the clay sticks to the rod.

[2]
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B

The collision lasts for a very short time.

I.

Determine the loss of kinetic energy in the collision.

[2]
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II.

Discuss why angular momentum can be conserved during the collision while kinetic energy is not conserved.

[2]
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0

Question 48
HL • Paper 2
Hard
Calculator Permitted

Two masses are connected by a light inextensible string that passes over a pulley. The string does not slip on the pulley. The pulley rotates about a fixed axle with negligible axle friction.

Atwood machine diagram with two hanging masses connected by a string over a pulley. The pulley radius and moment of inertia are labelled. The heavier mass is on one side. Tensions on the two sides of the string are labelled separately, and an arrow shows the angular acceleration of the pulley.
A

The masses are 1.5 kg1.5\ \text{kg} and 2.1 kg2.1\ \text{kg}. The pulley has radius 0.12 m0.12\ \text{m} and moment of inertia 0.018 kg m20.018\ \text{kg m}^2.

I.

Show that the acceleration of the masses is about 1.2 m s−21.2\ \text{m s}^{-2}.

[4]
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II.

Calculate the two tensions in the string.

[2]
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B

Discuss why the two tensions are not equal, even though the string is light.

[2]
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0


A.3 Work, energy and power

A.5 Galilean and special relativity