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← JEE Main 2014 · 12 Apr · Shift 1
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JEE Main 2014 · 12 Apr · Shift 1, Physics question 6

A particle is moving in a circular path of radius a, with a constant velocity vv as shown in the figure. The center of circle is marked by 'C'. The angular momentum from the origin O can be written as :

A circle of radius a with centre C on the x-axis; the origin O is on the circle at its leftmost point. A chord from O to the particle's position makes angle θ with the x-axis, and the radius a joins C to that point.
Choose one option
Answer and worked solution

Official answer: A

Take C at (a,0)(a,0). The chord from O at angle θ ends at P=(a(1+cos⁡2θ), asin⁡2θ)P=\left(a(1+\cos2\theta),\,a\sin2\theta\right), and ∠PCx=2θ\angle PCx=2\theta.

The velocity is along the tangent at P, perpendicular to n^=(cos⁡2θ,sin⁡2θ)\hat n=(\cos2\theta,\sin2\theta). The perpendicular distance from O to the line of motion is
OP⃗⋅n^=a(1+cos⁡2θ)cos⁡2θ+asin⁡22θ=a(1+cos⁡2θ).\vec{OP}\cdot\hat n=a(1+\cos2\theta)\cos2\theta+a\sin^22\theta=a(1+\cos2\theta).
So the angular momentum (per unit mass, as the options are written) is va(1+cos⁡2θ)va(1+\cos2\theta).

Topics:Angular momentum and its conservationUniform circular motion
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