Question 1Multiple choice · 1 markA* challenge
A flywheel of moment of inertia 0.8 kg m2 rotating at 30 rad s−1 is brought uniformly to rest in 10 s.
How many revolutions does it make while it slows down?
- A 12
- B 24
- C 48
- D 150
Show answer
Answer: B
Free A.4 Rigid body mechanics questions for IB Physics (SL & HL), HL only, part of A Space, time & motion. Each question has the answer and the marking points an examiner would look for.
Generate a full A.4 Rigid Body Mechanics worksheet →All A Space, time & motion questionsA flywheel of moment of inertia 0.8 kg m2 rotating at 30 rad s−1 is brought uniformly to rest in 10 s.
How many revolutions does it make while it slows down?
Answer: B
A skater spinning freely on ice pulls in her arms, so that her moment of inertia decreases by a factor of 2. Which statements are correct?
I. Her angular momentum is unchanged.
II. Her angular speed increases by a factor of 2.
III. Her rotational kinetic energy is unchanged.
Answer: A
A uniform beam of length 4.0 m and mass 20 kg is hinged at one end and held horizontal by a vertical string attached to the other end. A 10 kg mass rests on the beam 1.0 m from the hinge.
Calculate the tension in the string.
Calculate the force exerted by the hinge on the beam.
The 10 kg mass is removed and the string is cut. The moment of inertia of the beam about the hinge is ⅓ML2. Calculate the initial angular acceleration of the beam.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 123 N moments about hinge: 4.0T = 196 × 2.0 + 98 × 1.0 ✓ T = 123 N ✓ | 2 |
| (b) | 172 N upwards 294 − 123 ✓ = 172 N upwards ✓ | 2 |
| (c) | 3.7 rad s−2 I = ⅓ × 20 × 4.02 = 107 kg m2 ✓ torque = 196 × 2.0 = 392 N m ✓ α = 3.7 rad s−2 ✓ | 3 |
A flywheel with moment of inertia 0.50 kg m2 is accelerated from rest by a constant torque of 2.0 N m for 6.0 s.
Calculate the angular acceleration.
Calculate the angular velocity after 6.0 s.
Calculate the rotational kinetic energy and the angular momentum after 6.0 s.
A stationary disc of moment of inertia 0.25 kg m2 is dropped onto the flywheel and they rotate together. Determine the final angular velocity and the energy lost.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 4.0 rad s−2 α = 2.0/0.50 = 4.0 rad s−2 ✓ | 1 |
| (b) | 24 rad s−1 ω = 4.0 × 6.0 = 24 rad s−1 ✓ | 1 |
| (c) | 144 J; 12 kg m2 s−1 E = ½ × 0.50 × 242 ✓ = 144 J ✓ L = 0.50 × 24 = 12 kg m2 s−1 ✓ | 3 |
| (d) | 16 rad s−1; 48 J angular momentum conserved: 12 = 0.75ω ✓ ω = 16 rad s−1 ✓ E = ½ × 0.75 × 162 = 96 J so 48 J lost ✓ | 3 |
A solid cylinder of mass 4.0 kg and radius 0.20 m rolls without slipping down a slope from a height of 1.5 m. Moment of inertia I = ½mr2.
Show that the total kinetic energy at the bottom is (3/4)mv2.
Calculate the speed at the bottom.
Explain why a hoop of the same mass and radius reaches the bottom more slowly.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | translational + rotational ½mv2 + ½Iω2 with ω = v/r ✓ ½mv2 + ¼mv2 = ¾mv2 ✓ | 2 |
| (b) | 4.4 m s−1 mgh = ¾mv2 ✓ v = 4.4 m s−1 ✓ | 2 |
| (c) | larger I; more energy rotational hoop has a larger moment of inertia ✓ a greater fraction of energy is rotational, so less translational KE ✓ | 2 |
A flywheel is a uniform disc of mass 20 kg and radius 0.4 m (I = ½Mr2). It starts from rest and a constant torque of 10 N m acts for 6 s.
Calculate the moment of inertia and the angular acceleration.
Calculate the angular speed after this time and the number of revolutions made.
Calculate the rotational kinetic energy and show that it equals the work done by the torque.
A ring of mass 5 kg and radius 0.4 m (I = Mr2) is dropped onto the spinning flywheel and they rotate together. Determine the final angular speed and the energy transferred.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 1.60 kg m2; 6.25 rad s−2 I = 1.60 kg m2 ✓ α = τI = 6.25 rad s−2 ✓ | 2 |
| (b) | 37.5 rad s−1; 17.9 ω = αt = 37.5 rad s−1 ✓ θ = ½αt2 = 113 rad ✓ 17.9 revolutions ✓ | 3 |
| (c) | 1125 J E = ½Iω2 = 1125 J ✓ W = τθ = 1125 J ✓ | 2 |
| (d) | 25.0 rad s−1 angular momentum is conserved ✓ Iring = 0.800 kg m2 ✓ ω = 1.60 × 37.52.40 = 25.0 rad s−1 ✓ energy lost = 375 J ✓ | 4 |
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