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Q.Derive the relation between Torque and Angular momentum.

Jammu Kashmir JkboseJammu and Kashmir Board of School Education (Class 11) 2018Subjective· 3mImportance★★★★★
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Differentiating the definition of angular momentum L = r x p with respect to time, and using Newton's second law, directly gives tau = dL/dt.

Definitions: For a particle of position vector r and linear momentum p = m v, its angular momentum about the origin is defined as

L = r x p

and the torque acting on it about the same origin, due to a force F acting on it, is defined as

tau = r x F

Derivation of the relation: Differentiate L = r x p with respect to time, using the product rule for the cross product (which must preserve the order of the two factors):

dL/dt = d(r x p)/dt = (dr/dt) x p + r x (dp/dt)

The first term: dr/dt = v (the velocity), so the first term becomes v x p = v x (m v) = m (v x v). Since the cross product of any vector with itself is zero (v x v = 0), this whole first term vanishes.

The second term: by Newton's second law, dp/dt = F (the net force acting on the particle). So the second term becomes r x F, which is exactly the definition of torque, tau.

Putting the two terms together:

dL/dt = 0 + r x F = tau

So:

tau = dL/dt

…

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