Rotational Mechanics and Rigid Bodies (JEE Advanced)

Question 1

A uniform thick spherical shell is made by removing a concentric solid sphere of radius R/2 from a uniform solid sphere of radius R. The mass of the remaining shell is M. Its moment of inertia about a

Accepted Answer

C. Let the density be ρ. For a solid sphere of radius a, mₐ=4/3πρ a³ and Iₐ=2/5mₐ a²∝ a⁵.
Removing the inner sphere: M∝ R³−(R/2)³=7/8R³ and I_(cm)=2/5·4/3πρ (R⁵−R⁵/32).
So I_(cm)=2/5M (R⁵(31/32))/(R³(7/8))=2/5·31/28MR²=31/70MR².
A tangent to the outer surface is parallel to a diameter at distance R, so by the parallel-axis theorem I=31/70MR²+MR²=101/70MR².
Trap: using 2/5(R²+r²)-type averages or subt

Question 2

A thin uniform rectangular plate of mass M has sides a and 2a. Its moment of inertia about the axis along one of its diagonals is ______. Exactly one option is correct.

Accepted Answer

A. Take the centre as origin, x along the side a, y along the side 2a. Then Σ mx²=(Ma²)/12, Σ my²=(M(2a)²)/12=(Ma²)/3 and Σ mxy=0 by symmetry.
For an in-plane axis through the centre along (cosφ,sinφ), the distance of an element from the axis is d=xsinφ−ycosφ, so I=sin²φΣ mx²+cos²φΣ my²=(Ma²)/12sin²φ+(Ma²)/3cos²φ.
For the diagonal, tanφ=2: sin²φ=4/5, cos²φ=1/5, so I=(Ma²)/15+(Ma²)/15=2/15Ma².
(Check

Question 3

A force F⃗=(2î−ĵ+3k̂) N acts at the point with position vector (î+2ĵ−k̂) m. Consider the line through the point P(0,1,1) m parallel to the vector î+ĵ. The magnitude of the torque of F⃗ about thi

Accepted Answer

D. Relative position: r⃗−r⃗_P=(1,1,−2) m.
τ⃗_P=(1,1,−2)×(2,−1,3)=big(1·3−(−2)(−1), (−2)(2)−1·3, 1·(−1)−1·2big)=(1,−7,−3) N m, with magnitude √59 N m.
Unit vector along the line: n̂=(î+ĵ)/√2. Torque about the line: τ⃗_P·n̂=(1−7)/√2=−3√2 N m.
Magnitude =3√2 N m ≈ 4.24 N m (the same for any point P on the line).
Trap: √59 N m is the torque about the point, not about the axis.
Difficulty: Hard

Rotational Mechanics and Rigid Bodies: questions with solutions

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