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

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

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A circus acrobat is performing a somersault in mid-air. With reference to the conservation of angular momentum, explain how the acrobat controls their rotation speed during the performance.

When the acrobat leaves the springboard, no external torque acts about their CM (gravity produces no torque about the CM); hence the angular momentum L = Iω is conserved throughout the flight.
By tucking the body into a compact position, the acrobat reduces their moment of inertia I significantly; since Iω = constant, the angular velocity ω increases proportionally, allowing rapid spins.
Before landing, the acrobat extends arms and legs to increase I, which decreases ω and slows the rotation, enabling a controlled, balanced landing.
This is a direct application of conservation of angular momentum: I1ω1 = I2ω2, where the subscripts denote the tucked and extended positions.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A see-saw on a children's playground is 3 m long and is balanced at its centre. A child of mass 30 kg sits 1.2 m from the fulcrum on one side. Where must a child of mass 40 kg sit on the other side to balance it? Use the principle of moments.

For rotational equilibrium, the net torque about the fulcrum must be zero: clockwise torque = anticlockwise torque.
Torque by 30 kg child: τ1 = 30g × 1.2 = 36g N m (anticlockwise from the heavier child's perspective).
For balance: 40g × d = 36g, where d is the distance of the 40 kg child from the fulcrum; solving gives d = 36/40 = 0.9 m.
The 40 kg child must sit 0.9 m from the fulcrum on the opposite side; this distance is less than 1.2 m, consistent with the heavier child needing to sit closer to the pivot.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 3

Applied Concept

A flywheel of mass 20 kg and radius 0.2 m is mounted on a frictionless axle. A cord wrapped around its rim is pulled with a steady force of 25 N. Compute the angular acceleration and find the kinetic energy when 2 m of cord has been unwound (start from rest).

Torque: τ = FR = 25 × 0.2 = 5 N m. Moment of inertia of flywheel (solid disc): I = MR²/2 = 20 × 0.04/2 = 0.4 kg m².
Angular acceleration: α = τ/I = 5.0/0.4 = 12.5 rad s−2.
Angular displacement when 2 m cord unwinds: θ = 2/0.2 = 10 rad. Using ω² = 2αθ = 2 × 12.5 × 10 = 250 rad² s−2.
Kinetic energy = (1/2)Iω² = 0.5 × 0.4 × 250 = 50 J. This equals the work done by the pull (25 N × 2 m = 50 J), confirming the work-energy theorem for rotation.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

With reference to a potter's wheel, explain the physical significance of a large moment of inertia and how the potter maintains steady rotation with minimal continuous effort.

A potter's wheel is a heavy stone disc with a large moment of inertia I about its vertical axis; once set spinning, it strongly resists changes in angular velocity because τ = Iα implies that a large I requires a large torque to produce any given angular acceleration.
The potter kicks the wheel periodically to counteract friction, but between kicks the large I keeps ω nearly constant, providing a steady platform for shaping clay.
From the rotational work-energy theorem, the large I stores significant kinetic energy K = (1/2)Iω², which is gradually released to overcome frictional losses between successive kicks.
This is exactly the principle of a flywheel: large I smooths out variations in speed, ensuring uniform rotation essential for the delicate work of pottery.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A child sits still at one end of a long trolley moving uniformly at speed V on a frictionless floor. The child then runs about on the trolley. What happens to the speed of the centre of mass of the system? Justify your answer.

The child running on the trolley constitutes internal motion — the forces between child and trolley floor are internal forces of the system.
The floor is frictionless, so there is no external horizontal force on the system (trolley + child). By Newton's second law for the CM: MA = Fext = 0.
Since the external force is zero, the acceleration of the CM is zero, meaning the velocity of the CM remains constant at V regardless of any internal motion.
The child and trolley may exchange momenta internally, but the total momentum MV is unchanged — the CM continues at speed V throughout.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

An engine needs to maintain a rotor at a uniform angular speed of 200 rad s−1 and transmits a torque of 180 N m. Calculate the power required. Why is torque needed even for uniform angular speed?

Power in rotational motion: P = τω = 180 × 200 = 36000 W = 36 kW.
Uniform angular velocity implies zero angular acceleration, so by τ = Iα, the net torque on the rotor should be zero for a purely ideal system.
In practice, frictional forces at bearings, air resistance, and load resistance create opposing torques; the engine must supply torque to exactly counter these frictional torques to maintain constant ω.
The applied torque is thus not causing acceleration but overcoming dissipative forces — the power supplied (36 kW) is entirely converted into heat and other loss mechanisms.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

A uniform disk of radius R has a circular hole of radius R/2 cut from it, with the hole's centre at R/2 from the disc's centre. Describe the method to find the centre of gravity of the remaining body.

Treat the original complete disc and the removed disc as separate objects. The original disc has mass M and its CM at centre O; the removed circular disc (radius R/2) has mass M/4 (since area ∝ r²) and its CM at distance R/2 from O.
The remaining body's CM can be found using the formula: M × XCMR = M × 0 − (M/4) × (R/2), where 0 is the CM of the full disc and R/2 is the CM of the removed portion.
Solving: XCMR = −(M/4)(R/2)/M = −R/8; the negative sign indicates the CM shifts away from the side where the hole was made.
The centre of gravity of the remaining flat body is at R/6 from the original centre on the side opposite to where the hole was cut.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

A child stands at the centre of a turntable (I = I0) rotating at 40 rev/min. The child folds arms back reducing moment of inertia to (2/5)I0. Find the new angular speed. Show that kinetic energy increases and explain the source of this extra energy.

By conservation of angular momentum (no external torque on frictionless turntable): I0 ω0 = (2/5 I0) ω. Solving: ω = (5/2)ω0 = (5/2) × 40 = 100 rev/min.
Initial KE: K0 = (1/2)I0ω0². Final KE: K = (1/2)(2I0/5)(5ω0/2)² = (1/2)(2I0/5)(25ω0²/4) = (5/4)(1/2)I0ω0² = (5/4)K0.
Final KE is 5/4 times initial KE — it has increased. The extra energy came from the internal muscular work done by the child in pulling the arms inward against the centrifugal tendency.
The child does positive work against the outward pseudo-force when contracting arms; this metabolic energy is converted into additional rotational kinetic energy of the system.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

A binary star system consists of two stars of equal mass M. Describe the motion of the two stars and their centre of mass when no external force acts on the system.

Since there is no external force on the binary system, by MA = Fext = 0, the CM of the system moves with constant velocity (uniform motion in a straight line).
The gravitational forces between the two stars are internal forces; they cause the stars to orbit around their common CM but do not alter the CM's uniform motion.
In the CM reference frame (where the CM is at rest), the two equal-mass stars orbit diametrically opposite each other in circles around the CM, maintaining equal distances from it at all times.
In the lab frame, each star's trajectory is a combination of the uniform translation of the CM and its circular orbit around the CM — resulting in a helical or wave-like path.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. A force of 30 N is applied to the rope. Find the angular and linear acceleration of the cylinder. (Assume no slipping.)

For a hollow cylinder (thin-walled), I = MR² = 3 × (0.4)² = 0.48 kg m².
Torque: τ = F × R = 30 × 0.4 = 12 N m. Angular acceleration: α = τ/I = 12/0.48 = 25 rad s−2.
Linear acceleration of the rope (tangential): a = αR = 25 × 0.4 = 10 m s−2.
This result shows the direct application of τ = Iα; the hollow cylinder (all mass at radius R) has a larger I than a solid cylinder of the same M and R (I = MR²/2), so it has a smaller angular acceleration for the same torque.

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Chapter 6: Systems of Particles and Rotational Motion | Application Questions

A circus acrobat is performing a somersault in mid-air. With reference to the conservation of angular momentum, explain how the acrobat controls their rotation speed during the performance.

A see-saw on a children's playground is 3 m long and is balanced at its centre. A child of mass 30 kg sits 1.2 m from the fulcrum on one side. Where must a child of mass 40 kg sit on the other side to balance it? Use the principle of moments.

A flywheel of mass 20 kg and radius 0.2 m is mounted on a frictionless axle. A cord wrapped around its rim is pulled with a steady force of 25 N. Compute the angular acceleration and find the kinetic energy when 2 m of cord has been unwound (start from rest).

With reference to a potter's wheel, explain the physical significance of a large moment of inertia and how the potter maintains steady rotation with minimal continuous effort.

A child sits still at one end of a long trolley moving uniformly at speed V on a frictionless floor. The child then runs about on the trolley. What happens to the speed of the centre of mass of the system? Justify your answer.

An engine needs to maintain a rotor at a uniform angular speed of 200 rad s−1 and transmits a torque of 180 N m. Calculate the power required. Why is torque needed even for uniform angular speed?

A uniform disk of radius R has a circular hole of radius R/2 cut from it, with the hole's centre at R/2 from the disc's centre. Describe the method to find the centre of gravity of the remaining body.

A child stands at the centre of a turntable (I = I0) rotating at 40 rev/min. The child folds arms back reducing moment of inertia to (2/5)I0. Find the new angular speed. Show that kinetic energy increases and explain the source of this extra energy.

A binary star system consists of two stars of equal mass M. Describe the motion of the two stars and their centre of mass when no external force acts on the system.

A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. A force of 30 N is applied to the rope. Find the angular and linear acceleration of the cylinder. (Assume no slipping.)

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