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Rotation flash cards

Master Rotation through 89 JEE Advanced-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Rotation, question and answer

15 of this chapter's 89 cards, laid out open so you can read straight through. The remaining 74 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.Define moment of inertia of a particle and of a system about an axis.

    For a particle, I=mr2I = mr^2 where rr is the perpendicular distance from the axis. For a system, I=miri2I = \sum m_i r_i^2, and for a continuous body I=r2dmI = \int r^2\,dm. It is the rotational analogue of mass.

    Hint: Sum of mr2mr^2 about the axis.

  2. 2.State the parallel axis theorem.

    I=Icm+Md2I = I_{cm} + Md^2, where IcmI_{cm} is the moment of inertia about an axis through the centre of mass and dd is the perpendicular distance between the two parallel axes.

    Hint: Shift from CM by Md2Md^2.

  3. 3.State the perpendicular axis theorem and its restriction.

    For a planar (2D) lamina, Iz=Ix+IyI_z = I_x + I_y, where x,yx,y lie in the plane and zz is perpendicular to it. It applies only to flat laminae, not 3D bodies.

    Hint: Lamina only; zz-axis is normal to the plane.

  4. 4.Moment of inertia of a thin uniform rod (length LL, mass MM) about a perpendicular axis through its centre?

    I=112ML2I = \frac{1}{12}ML^2. By parallel axis theorem, about one end it becomes I=13ML2I = \frac{1}{3}ML^2.

    Hint: Centre gives twelfth; end gives third.

  5. 5.Moment of inertia of a uniform ring (mass MM, radius RR) about its central axis and about a diameter?

    About the central (symmetry) axis: I=MR2I = MR^2. About a diameter, by perpendicular axis theorem 2Id=MR22I_d = MR^2, so Id=12MR2I_d = \frac{1}{2}MR^2.

    Hint: All mass at radius RR; diameter is half.

  6. 6.Moment of inertia of a uniform disc (mass MM, radius RR) about its central axis and about a diameter?

    Central axis: I=12MR2I = \frac{1}{2}MR^2. About a diameter: Id=14MR2I_d = \frac{1}{4}MR^2 (from perpendicular axis theorem).

    Hint: Half for axis, quarter for diameter.

  7. 7.Moment of inertia of a solid sphere (mass MM, radius RR) about a diameter, and about a tangent?

    About a diameter: I=25MR2I = \frac{2}{5}MR^2. About a tangent: I=25MR2+MR2=75MR2I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2.

    Hint: Two-fifths; tangent adds MR2MR^2.

  8. 8.Moment of inertia of a thin hollow (spherical shell) sphere about a diameter and about a tangent?

    About a diameter: I=23MR2I = \frac{2}{3}MR^2. About a tangent: I=23MR2+MR2=53MR2I = \frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2.

    Hint: Two-thirds for shell; tangent adds MR2MR^2.

  9. 9.Moment of inertia of a solid cylinder (mass MM, radius RR) about its own (longitudinal) axis?

    I=12MR2I = \frac{1}{2}MR^2, independent of length LL. Same as a disc because it is a stack of discs about the common axis.

    Hint: Length doesn't matter for the long axis.

  10. 10.Moment of inertia of a hollow cylinder / thin cylindrical shell about its longitudinal axis?

    I=MR2I = MR^2 for a thin shell. For a thick hollow cylinder with radii R1,R2R_1, R_2: I=12M(R12+R22)I = \frac{1}{2}M(R_1^2 + R_2^2).

    Hint: Thin shell behaves like a ring.

  11. 11.Define radius of gyration kk.

    kk is defined by I=Mk2I = Mk^2, so k=I/Mk = \sqrt{I/M}. It is the distance from the axis at which the whole mass could be concentrated to give the same moment of inertia.

    Hint: I=Mk2I = Mk^2.

  12. 12.What is the radius of gyration of a solid sphere about its diameter?

    I=25MR2=Mk2k=25R=R0.40.632RI = \frac{2}{5}MR^2 = Mk^2 \Rightarrow k = \sqrt{\frac{2}{5}}\,R = R\sqrt{0.4} \approx 0.632R.

    Hint: Take 2/5\sqrt{2/5} of RR.

  13. 13.Moment of inertia of a uniform rectangular plate (sides a,ba,b, mass MM) about an axis through its centre perpendicular to the plate?

    I=112M(a2+b2)I = \frac{1}{12}M(a^2 + b^2), obtained by adding 112Ma2\frac{1}{12}Ma^2 and 112Mb2\frac{1}{12}Mb^2 via the perpendicular axis theorem.

    Hint: Add the two in-plane contributions.

  14. 14.Define torque as a vector and give its magnitude.

    τ=r×F\vec{\tau} = \vec{r}\times\vec{F}, magnitude τ=rFsinθ=Fd\tau = rF\sin\theta = F\,d, where d=rsinθd = r\sin\theta is the moment arm (perpendicular distance from axis to line of action of the force).

    Hint: Force times perpendicular lever arm.

  15. 15.State the rotational form of Newton's second law about a fixed axis.

    τnet=Iα\tau_{net} = I\alpha, where α\alpha is the angular acceleration. It holds about a fixed axis or about the centre of mass even if the CM accelerates.

    Hint: Rotational F=maF = ma.

Open the interactive deck for the other 74 cards, with self-grading so the ones you keep missing come back.

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