Question:
A particle is moving in a circular path of radius \(a\), with a constant velocity \(v\) as shown in the figure. The centre of circle is marked by '\(C\)'. The angular momentum from the origin \(O\) can be written as: 
A particle is moving in a circular path of radius \(a\), with a constant velocity \(v\) as shown in the figure. The centre of circle is marked by '\(C\)'. The angular momentum from the origin \(O\) can be written as:
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Write the position of the particle from O as OC plus CP, find the angle CP makes with the x-axis in terms of \(\theta\) (exterior angle of the isosceles triangle OCP is \(2\theta\)), and evaluate \(x v_y - y v_x\).
Updated On: Nov 11, 2026
- \(va(1+\cos 2\theta)\)
- \(va(1+\cos\theta)\)
- \(va\cos 2\theta\)
- \(va\)
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The Correct Option is A
Solution and Explanation
Take O as origin, so the centre is C = (a, 0). Let the radius CP make angle \(\phi\) with the +x axis. Position of the particle: \(x = a + a\cos\phi,\ y = a\sin\phi\).
Velocity is tangential (perpendicular to CP): \(v_x = -v\sin\phi,\ v_y = v\cos\phi\).
Angular momentum about O per unit mass (the options contain no mass, so \(m\) is taken as 1): \(L = x v_y - y v_x = (a + a\cos\phi)v\cos\phi + a v\sin^2\phi = av(\cos\phi + \cos^2\phi + \sin^2\phi) = av(1+\cos\phi)\).
In triangle OCP, CO = CP = a, so the base angles at O and P are equal; the angle at O is \(\theta\). The exterior angle at C (between CP and the x-axis) is \(\phi = \theta + \theta = 2\theta\).
Hence \(L = va(1+\cos 2\theta)\).
Answer: (1) \(va(1+\cos 2\theta)\)
Velocity is tangential (perpendicular to CP): \(v_x = -v\sin\phi,\ v_y = v\cos\phi\).
Angular momentum about O per unit mass (the options contain no mass, so \(m\) is taken as 1): \(L = x v_y - y v_x = (a + a\cos\phi)v\cos\phi + a v\sin^2\phi = av(\cos\phi + \cos^2\phi + \sin^2\phi) = av(1+\cos\phi)\).
In triangle OCP, CO = CP = a, so the base angles at O and P are equal; the angle at O is \(\theta\). The exterior angle at C (between CP and the x-axis) is \(\phi = \theta + \theta = 2\theta\).
Hence \(L = va(1+\cos 2\theta)\).
Answer: (1) \(va(1+\cos 2\theta)\)
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