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← JEE Advanced 2014 · Paper 1
PhysicsInteger+3Question 13 of 60

JEE Advanced 2014 · Paper 1, Physics question 13

A uniform circular disc of mass 1.5 kg and radius 0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude F=0.5F=0.5 N are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in rad s−1\mathrm{rad\,s^{-1}} is

A disc with centre O and an inscribed equilateral triangle XYZ; forces F act along the three sides, all pointing the same way round the triangle (along Z→X at X, X→Y at Y, Y→Z at Z); a radius is drawn from O to Z.
Answer and worked solution

Official answer: 2

Torque. Each side of the inscribed equilateral triangle is at distance r/2r/2 from the centre (the inradius is half the circumradius). All three forces turn the same way:
τ=3Fr2=3×0.5×0.25=0.375 N m.\tau=3F\frac r2=3\times0.5\times0.25=0.375\text{ N m}.
The forces sum to zero, so the centre stays at rest.

Angular acceleration. I=12Mr2=12×1.5×0.25=0.1875I=\tfrac12Mr^2=\tfrac12\times1.5\times0.25=0.1875 kg m², so α=τ/I=2 rad s−2\alpha=\tau/I=2\ \mathrm{rad\,s^{-2}}.

After 1 s. ω=αt=2 rad s−1\omega=\alpha t=2\ \mathrm{rad\,s^{-1}}.

Answer: 2

Topics:Torque and rotational dynamicsMoment of inertia
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