Chapter 6: Systems of Particles and Rotational Motion Case Study Questions | physics Class 11 CBSE

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Chapter 6: Systems of Particles and Rotational Motion Case Study Questions | physics Class 11 CBSE | ByteNotes.in

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A potter uses a heavy stone wheel of mass 40 kg and radius 0.5 m, pivoted on a frictionless vertical axle. The potter kicks the wheel to set it rotating and then shapes clay on it. Because the wheel has a large moment of inertia, it continues to spin for a long time with minimal effort from the potter. While the wheel is spinning at 120 rpm, the potter presses a lump of clay of mass 2 kg at the rim, momentarily increasing the effective rotating mass. The potter then gradually withdraws the clay toward the centre. Engineers and designers of rotating machinery use exactly this principle when incorporating flywheels into engines to ensure smooth, steady rotation and to prevent jerky motion during operation.

Question 1: What is the moment of inertia of the potter's wheel (treat as a solid disc) about its axle?

Using the formula for a solid disc: I = MR²/2, where M = 40 kg and R = 0.5 m.
I = 40 × (0.5)² / 2 = 40 × 0.25 / 2 = 5 kg m².
This large moment of inertia is what allows the wheel to keep spinning steadily after the initial kick, resisting changes in rotational speed.

Question 2: Why does the wheel slow down slightly when the potter presses the clay lump of mass 2 kg at the rim?

When the clay is pressed at the rim, the total moment of inertia increases: Inew = Iwheel + mR² = 5 + 2×(0.5)² = 5 + 0.5 = 5.5 kg m².
Since no external torque acts on the system (frictionless axle), angular momentum Iω is conserved: I₁ω₁ = I₂ω₂.
With larger I₂, the angular velocity ω₂ must decrease — the wheel slows down slightly when the clay is added at the rim.

Question 3: After adding the clay at the rim, the potter slides the clay inward to 0.1 m from the centre. Calculate the new angular velocity if the wheel was spinning at 120 rpm just before the clay was moved inward. Explain the energy change.

Angular velocity before moving clay inward: ω₁ = 2π×120/60 = 4π rad/s. At rim: I₁ = 5.5 kg m² (as computed above).
After moving clay to r = 0.1 m: I₂ = 5 + 2×(0.1)² = 5 + 0.02 = 5.02 kg m². By conservation: ω₂ = I₁ω₁/I₂ = 5.5×4π/5.02 ≈ 13.77 rad/s ≈ 131.5 rpm.
The wheel speeds up because I decreases; kinetic energy increases (K₂ > K₁) — the extra energy comes from the internal muscular work done by the potter in pulling the clay inward against centrifugal tendency.

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Case Set 2

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During a school science fair, students set up a simple lever experiment using a uniform wooden plank of length 2 m and mass 3 kg, balanced on a fulcrum placed at its centre. A student of mass 30 kg sits at one end. The teacher asks: where should a second student of mass 20 kg sit on the other side so that the plank remains balanced? The students also note that if the second student moves farther from the fulcrum, the plank tips. This experiment illustrates the principle of moments used in designing weighing balances, cranes, and loading ramps. The teacher explains that the mechanical advantage of a lever can be exploited to lift heavy loads with a small effort, and that the position of the centre of gravity of the plank itself plays a role in the calculation.

Question 1: State the principle of moments as applicable to this lever experiment.

The principle of moments states that for a body in rotational equilibrium, the sum of clockwise moments about the fulcrum equals the sum of anticlockwise moments.
Mathematically: load arm × load = effort arm × effort, or d₁F₁ = d₂F₂, where d₁ and d₂ are distances from the fulcrum.
Since the plank is uniform and balanced at its centre, the weight of the plank (3 kg × g) acts at the centre (fulcrum) and contributes zero net torque about the fulcrum.

Question 2: Where must the 20 kg student sit on the opposite side to balance the 30 kg student sitting at the end (1 m from fulcrum)?

For rotational equilibrium about the fulcrum, clockwise moment = anticlockwise moment (plank weight acts at fulcrum and has zero moment).
Moment due to 30 kg student: 30g × 1 m = 30g N m. For 20 kg student at distance d: 20g × d = 30g.
Solving: d = 30/20 = 1.5 m. The 20 kg student must sit 1.5 m from the fulcrum — but the plank is only 1 m long on each side, so the heavier student cannot balance the lighter one within the plank length; the system cannot be balanced in this configuration.

Question 3: The fulcrum is now shifted 0.3 m toward the 30 kg student's end. Find the new positions of both students (each sitting at their respective ends) for the plank to be balanced, accounting for the plank's own weight.

With fulcrum at 0.3 m from centre, the 30 kg student (left end) is 0.7 m from the fulcrum and the 20 kg student (right end) is 1.3 m from the fulcrum.
The plank's CM (at 1 m from each end) is now 0.3 m to the right of the fulcrum, contributing a clockwise moment: 3g × 0.3 = 0.9g N m.
Anticlockwise: 30g × 0.7 = 21g. Clockwise: 20g × 1.3 + 0.9g = 26g + 0.9g = 26.9g. Net clockwise = 26.9g − 21g = 5.9g ≠ 0; the plank does not balance in this arrangement — a quantitative demonstration that fulcrum position and plank weight both matter.

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A gymnast performs a spinning routine on the floor. She begins the spin with her arms and one leg extended outward, rotating at 2 rev/s. Her coach instructs her to bring all limbs tightly against her body midway through the spin to increase her rotation rate, and then extend them again before stopping. The gymnast's moment of inertia with arms and leg extended is estimated at 4 kg m², and with limbs tucked it reduces to 1 kg m². Her performance is a spectacular demonstration of the conservation of angular momentum. Coaches of figure skaters, divers, and ballet dancers use this same principle to choreograph spins and somersaults that require precise control of angular speed without any external assistance.

Question 1: State the physical principle that explains why the gymnast spins faster when she tucks her limbs in.

The principle is conservation of angular momentum: when no external torque acts on a system, its total angular momentum L = Iω remains constant.
When the gymnast tucks her limbs, her moment of inertia I decreases; since Iω = constant, her angular velocity ω must increase proportionally.
The gymnast herself does no external work on the system's angular momentum — the change in speed is purely a consequence of the redistribution of her own mass.

Question 2: Calculate the gymnast's angular velocity when she tucks her limbs in (I reduces from 4 kg m² to 1 kg m²), starting from 2 rev/s.

Using conservation of angular momentum: I₁ω₁ = I₂ω₂, where I₁ = 4 kg m², ω₁ = 2 rev/s, I₂ = 1 kg m².
Solving: ω₂ = I₁ω₁/I₂ = 4×2/1 = 8 rev/s.
The gymnast's angular speed quadruples when she reduces her moment of inertia to one-quarter of its original value — a fourfold increase in spin rate.

Question 3: Compare the gymnast's rotational kinetic energy before and after tucking. Account for the difference and state what happens to kinetic energy when she extends her limbs again.

Before tucking: K₁ = (1/2)I₁ω₁² = (1/2)×4×(2×2π)² = 2×(4π²×4) = 2×157.9 ≈ 315.8 J (in SI); or simply K₁ = (1/2)×4×(4π²×4).
After tucking: K₂ = (1/2)I₂ω₂² = (1/2)×1×(8×2π)² = 0.5×(64×4π²) = 4K₁ — kinetic energy quadruples.
The extra energy comes from the internal muscular work done by the gymnast pulling limbs inward. When she extends again, she does negative work, returning muscles' stored energy; if no friction, K returns to K₁ — in practice, some energy is lost to dissipation.

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In a physics laboratory, students investigate rotational motion by winding a cord around a flywheel (solid disc, mass 20 kg, radius 20 cm) mounted on a frictionless horizontal axle. They attach different masses to the free end of the cord and measure the angular acceleration. The flywheel starts from rest each time. In one trial, a steady pull of 25 N is applied to the cord. Students plot torque versus angular acceleration and find a straight line passing through the origin, confirming the rotational analogue of Newton's second law. They also verify that the work done by the pulling force equals the kinetic energy gained by the flywheel when there is no friction, illustrating the work-energy theorem for rotational motion.

Question 1: Write the rotational analogue of Newton's second law used in this experiment and identify the physical quantities involved.

The rotational analogue of Newton's second law is τ = Iα, where τ is the net external torque, I is the moment of inertia, and α is the angular acceleration.
In this experiment, torque τ = F×R (force × radius of flywheel) and I = MR²/2 for a solid disc; the slope of the torque vs. α graph gives I directly.
This mirrors F = ma exactly: just as a larger mass requires more force for the same acceleration, a larger I requires more torque for the same angular acceleration.

Question 2: Calculate the angular acceleration of the flywheel when a steady pull of 25 N is applied.

Moment of inertia of solid disc: I = MR²/2 = 20×(0.20)²/2 = 20×0.04/2 = 0.4 kg m².
Torque: τ = F×R = 25×0.20 = 5.0 N m.
Angular acceleration: α = τ/I = 5.0/0.4 = 12.5 rad s⁻².

Question 3: When 2 m of cord is unwound from the flywheel (starting from rest, α = 12.5 rad s⁻²), find the final angular velocity and kinetic energy. Verify the work-energy theorem.

Angular displacement: θ = length/radius = 2/0.20 = 10 rad. Using ω² = 2αθ = 2×12.5×10 = 250 rad² s⁻², so ω = √250 ≈ 15.81 rad/s.
Kinetic energy gained: K = (1/2)Iω² = (1/2)×0.4×250 = 50 J.
Work done by pull: W = F×distance = 25×2 = 50 J. Since W = K = 50 J, the work-energy theorem is verified — all work done by the external force is converted into rotational kinetic energy with no frictional loss.

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A construction site uses a crane to lift heavy loads. The crane arm is a uniform steel beam of mass 500 kg and length 10 m, pivoted at one end (the base). A load of 2000 kg hangs from the free end. The crane operates by applying a cable force perpendicular to the beam at 8 m from the pivot. A safety engineer must verify that the crane arm is in rotational equilibrium before authorising lifting operations. She applies the conditions for mechanical equilibrium and calculates the required cable tension. Understanding these principles is critical to ensuring that the net torque about the pivot is zero and that the structure does not rotate under the weight of the load and the beam itself.

Question 1: State the condition for rotational equilibrium of the crane arm and identify the torques acting on it.

Rotational equilibrium requires that the sum of all torques about any chosen axis (here the pivot) is zero: Στ = 0.
Torques acting: (1) weight of beam (500g N) acting downward at its midpoint (5 m from pivot) — clockwise; (2) weight of load (2000g N) acting downward at 10 m from pivot — clockwise; (3) cable tension T acting upward at 8 m from pivot — anticlockwise.
Setting anticlockwise = clockwise: T×8 = 500g×5 + 2000g×10.

Question 2: Calculate the cable tension T required to keep the crane arm in rotational equilibrium.

Taking moments about the pivot: T×8 = (500×9.8×5) + (2000×9.8×10).
T×8 = 24500 + 196000 = 220500 N m.
T = 220500/8 = 27562.5 N ≈ 27.6 kN.

Question 3: If the cable attachment point is moved from 8 m to 5 m from the pivot, find the new required tension. Comment on how the mechanical advantage changes and what this implies for the cable strength requirement.

New moment equation: T′×5 = 500g×5 + 2000g×10 = 220500 N m (same clockwise torques).
T′ = 220500/5 = 44100 N = 44.1 kN — significantly higher than 27.6 kN.
The mechanical advantage decreases when the cable is moved closer to the pivot (shorter effort arm); the cable must be much stronger to provide the same holding effect, illustrating that effort arm length directly determines the mechanical advantage and design requirements of the structure.

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Chapter 6: Systems of Particles and Rotational Motion | Case Study Questions

Passage

Question 1: What is the moment of inertia of the potter's wheel (treat as a solid disc) about its axle?

Question 2: Why does the wheel slow down slightly when the potter presses the clay lump of mass 2 kg at the rim?

Question 3: After adding the clay at the rim, the potter slides the clay inward to 0.1 m from the centre. Calculate the new angular velocity if the wheel was spinning at 120 rpm just before the clay was moved inward. Explain the energy change.

Passage

Question 1: State the principle of moments as applicable to this lever experiment.

Question 2: Where must the 20 kg student sit on the opposite side to balance the 30 kg student sitting at the end (1 m from fulcrum)?

Question 3: The fulcrum is now shifted 0.3 m toward the 30 kg student's end. Find the new positions of both students (each sitting at their respective ends) for the plank to be balanced, accounting for the plank's own weight.

Passage

Question 1: State the physical principle that explains why the gymnast spins faster when she tucks her limbs in.

Question 2: Calculate the gymnast's angular velocity when she tucks her limbs in (I reduces from 4 kg m² to 1 kg m²), starting from 2 rev/s.

Question 3: Compare the gymnast's rotational kinetic energy before and after tucking. Account for the difference and state what happens to kinetic energy when she extends her limbs again.

Passage

Question 1: Write the rotational analogue of Newton's second law used in this experiment and identify the physical quantities involved.

Question 2: Calculate the angular acceleration of the flywheel when a steady pull of 25 N is applied.

Question 3: When 2 m of cord is unwound from the flywheel (starting from rest, α = 12.5 rad s⁻²), find the final angular velocity and kinetic energy. Verify the work-energy theorem.

Passage

Question 1: State the condition for rotational equilibrium of the crane arm and identify the torques acting on it.

Question 2: Calculate the cable tension T required to keep the crane arm in rotational equilibrium.

Question 3: If the cable attachment point is moved from 8 m to 5 m from the pivot, find the new required tension. Comment on how the mechanical advantage changes and what this implies for the cable strength requirement.

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