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The Vault
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The Vault
CBSE Class 11 Physics · 10 questions · 24 marks
Real bodies are not points, so a spinning wheel or a toppling ladder needs a language beyond F = ma. This chapter introduces the centre of mass, which lets an extended body be treated as a single particle for translation, and then builds the rotational parallel of every quantity you already know: torque for force, moment of inertia for mass, angular momentum for momentum. Once the analogy clicks, most rotation problems reduce to equations you have already solved in a linear form.
A solid disc and a thin ring have the same mass and the same radius. Which has the greater moment of inertia about the central axis perpendicular to its plane?
Answer
The ring — for the ring I = MR² while for the disc I = ½MR². All the ring's mass sits at the maximum distance R from the axis, whereas the disc has mass spread over every radius from 0 to R, and moment of inertia weights mass by the square of its distance. Moment of inertia depends only on the mass distribution and the axis, never on the angular speed, so the last option is wrong on principle.
A flywheel of moment of inertia 4 kg·m² is acted on by a constant torque of 12 N·m. Its angular acceleration is:
Answer
3 rad/s² — using the rotational form of Newton's second law, τ = Iα gives α = τ/I = 12/4 = 3 rad/s². Multiplying instead of dividing gives the wrong 48 rad/s², a common slip.
A dancer spinning with her arms outstretched pulls them close to her body. Ignoring friction at the pivot, which quantity remains unchanged?
Answer
Angular momentum — no external torque acts about the vertical axis, so L = Iω is conserved. Pulling the arms in reduces I, so ω must rise. The rotational kinetic energy K = L²/2I actually increases, and that extra energy comes from the muscular work the dancer does in pulling her arms inward against the outward tendency of the rotating mass.
Two particles of masses 2 kg and 6 kg are placed at x = 0 m and x = 8 m on a straight line. The centre of mass lies at:
Answer
x = 6 m — using R = (m₁x₁ + m₂x₂)/(m₁ + m₂) = (2 × 0 + 6 × 8)/(2 + 6) = 48/8 = 6 m. The centre of mass always lies closer to the heavier particle, which is a quick sanity check on the arithmetic.
Assertion (A): A hollow cylinder and a solid cylinder of the same mass and radius, released together from rest at the top of the same incline, reach the bottom at different times. Reason (R): For a body rolling without slipping down an incline, the acceleration depends on the ratio I/MR² and not on the mass or the radius separately.
Answer
Both A and R are true and R is the correct explanation of A — for rolling without slipping the acceleration is a = g sin θ/(1 + I/MR²). For the solid cylinder I/MR² = ½ so a = (2/3)g sin θ, while for the hollow cylinder I/MR² = 1 so a = ½g sin θ. The solid cylinder therefore accelerates faster and arrives first, and the reason given is precisely the relation that produces the difference. Notice that mass and radius cancel out entirely, so a heavy hollow cylinder still loses to a light solid one.
A uniform rod of mass 3 kg and length 2 m rotates about an axis through one end, perpendicular to its length. Find its moment of inertia.
Answer
About the centre, I_cm = ML²/12 = (3)(2)²/12 = 12/12 = 1 kg·m². The end is a distance d = L/2 = 1 m from the centre, so by the parallel axis theorem: I = I_cm + Md² = 1 + (3)(1)² = 4 kg·m². This matches the standard result ML²/3 = (3)(4)/3 = 4 kg·m².
A wheel of moment of inertia 0.5 kg·m² is spinning at 20 rad/s. A constant retarding torque brings it to rest in 8 s. Find the retarding torque and the number of revolutions made before stopping.
Answer
Angular retardation: ω = ω₀ + αt gives 0 = 20 + α(8), so α = −2.5 rad/s². Torque τ = Iα = 0.5 × (−2.5) = −1.25 N·m, i.e. a retarding torque of magnitude 1.25 N·m. Angular displacement: ω² = ω₀² + 2αθ gives 0 = 400 + 2(−2.5)θ, so θ = 400/5 = 80 rad. Number of revolutions = θ/2π = 80/(2π) ≈ 12.7 revolutions.
State the theorem of parallel axes and derive it. Then use it to find the moment of inertia of a solid sphere of mass 5 kg and radius 0.2 m about a tangent to its surface.
Answer
Statement: the moment of inertia of a body about any axis equals its moment of inertia about a parallel axis through the centre of mass plus the product of its mass and the square of the perpendicular distance between the two axes: I = I_cm + Md². Derivation: take the centre of mass at the origin O and let the new axis be parallel to the first, passing through a point at perpendicular distance d. Consider a mass element m_i whose perpendicular distance from the centre-of-mass axis is r_i, with components (x_i, y_i) measured in the plane perpendicular to the axes, so r_i² = x_i² + y_i². Let the shift be along x, so the element's distance from the new axis satisfies r_i′² = (x_i − d)² + y_i² = x_i² + y_i² − 2dx_i + d² = r_i² − 2dx_i + d². Hence I = Σm_i r_i′² = Σm_i r_i² − 2dΣm_i x_i + d²Σm_i. The first term is I_cm and the last is Md². The middle term contains Σm_i x_i, which is M times the x-coordinate of the centre of mass measured from the centre of mass itself, and that is zero by definition. Therefore I = I_cm + Md². Application: for a solid sphere about a diameter, I_cm = (2/5)MR² = (2/5)(5)(0.2)² = 2 × 0.04 = 0.08 kg·m². A tangent is parallel to a diameter at distance d = R = 0.2 m, so I = 0.08 + (5)(0.2)² = 0.08 + 0.20 = 0.28 kg·m². This also equals the standard result (7/5)MR² = 1.4 × 5 × 0.04 = 0.28 kg·m².
A solid sphere of mass 2 kg and radius 0.1 m rolls without slipping down an incline of height 1.4 m, starting from rest (g = 10 m/s²). (a) Find its speed at the bottom. (b) Find the fraction of the total kinetic energy that is rotational. (c) Explain why the answer to (a) is independent of both the mass and the radius of the sphere.
Answer
(a) For rolling without slipping the contact point is instantaneously at rest, so friction does no work and mechanical energy is conserved: mgh = ½mv² + ½Iω², with I = (2/5)mr² and ω = v/r. ½Iω² = ½ × (2/5)mr² × v²/r² = (1/5)mv². So mgh = ½mv² + (1/5)mv² = (7/10)mv². v = √(10gh/7) = √(10 × 10 × 1.4/7) = √(140/7) = √20 ≈ 4.47 m/s. (b) Total kinetic energy = (7/10)mv²; rotational part = (1/5)mv² = (2/10)mv². Fraction = (2/10)/(7/10) = 2/7 ≈ 0.286, that is about 28.6% of the kinetic energy is rotational and 71.4% translational. (c) In the energy equation, every term carries a factor of m, so mass cancels when solving for v. The radius cancels too because the rotational term ½Iω² = ½ × k mr² × (v/r)² = ½ k mv² — the r² in the moment of inertia is exactly undone by the 1/r² from ω = v/r. What survives is only the shape factor k = I/mr², which is 2/5 for every solid sphere regardless of size. That is why a marble and a cannonball of the same shape reach the bottom of the same slope with the same speed.
Read the following and answer the questions that follow: A science club builds a merry-go-round: a uniform horizontal disc of mass 120 kg and radius 2 m, free to rotate about a frictionless vertical axle. It is set spinning at 3 rad/s. A student of mass 40 kg then steps carefully onto the rim and holds on, and may be treated as a point mass. (a) Calculate the moment of inertia of the empty disc. (b) Find the new angular speed after the student is on the rim. (c) Compare the rotational kinetic energy before and after, and account for any difference. (d) The student now walks slowly to the centre. State qualitatively what happens to the angular speed and why.
Answer
(a) For a uniform disc about its central axis, I_disc = ½MR² = ½ × 120 × (2)² = 240 kg·m². (b) The student stepping on exerts no external torque about the axle, so angular momentum is conserved. Treating the student as a point mass on the rim, I_student = mR² = 40 × 4 = 160 kg·m². I₁ω₁ = I₂ω₂ 240 × 3 = (240 + 160)ω₂ 720 = 400ω₂, so ω₂ = 1.8 rad/s. (c) K₁ = ½I₁ω₁² = ½ × 240 × 9 = 1080 J. K₂ = ½I₂ω₂² = ½ × 400 × (1.8)² = ½ × 400 × 3.24 = 648 J. The kinetic energy has fallen by 432 J. This is not a violation of anything: bringing the student up to the rim speed involves rubbing and slipping between shoes and platform, and that friction converts the missing mechanical energy into heat and sound. Angular momentum is still exactly conserved because friction here is an internal force pair producing equal and opposite torques. (d) As the student walks inward, her distance r from the axis decreases, so her contribution mr² to the total moment of inertia falls while angular momentum stays constant. Since L = Iω is fixed, a smaller I forces a larger ω, and the merry-go-round speeds up. The energy for this increase comes from the muscular work the student does walking inward against the outward push she feels in the rotating frame.
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