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NEET
Exam
NEET
Subject
Physics
Chapter
Rotational Motion

NEET · Physics · Rotational Motion

The torque T on a body about a given point is found to be equal to...

The torque T on a body about a given point is found to be equal to (\mathrm{A} \times \mathrm{L}) where A is constant vector, and LL is the angular momentum of the body about that point. From this it follows that (A) (\frac{\mathrm{dL}}{\mathrm{dt}}) is perpendicular to L at all instants of time (B) the component of L in the direction of A does not change with time. (C) the magnitude of L does not change with time. (D) L does not change with time

Options

  1. AA, B, C
  2. BA, C, D
  3. CB, C, D
  4. DA, B, C, D

Correct answer

A. A, B, C

Step-by-step solution

(Eτ=d L′dt\frac{\mathrm{E}}{\mathrm{\tau}}=\frac{\mathrm{d} \mathrm{~L}^{\prime}}{\mathrm{dt}}) Given that (\hat{t\mathrm{t}}=\ddot{A\mathrm{A}} ×L¨⇒dLdt=A∥×LX\times \ddot{\mathrm{L}} \Rightarrow \frac{\mathrm{dL}}{\mathrm{dt}}=\stackrel{\|}{\mathrm{A}} \times \stackrel{\mathbb{X}}{\mathrm{L}}) (dLdt\frac{\mathrm{dL}}{\mathrm{dt}}) From cross-product rule, (dLdt\frac{d L}{d t}) is always perpendicular to the plane containing AA and L By the dot product definition (\vec{L} . \vec{L}=L²) Differentiating with respect to time (\stackrel{Z\mathrm{Z}}{ L\mathrm{~L}} ⋅dLdt+ LZ⋅dLdt=\cdot \frac{\mathrm{dL}}{\mathrm{dt}}+\stackrel{\mathrm{Z}}{\mathrm{~L}} \cdot \frac{\mathrm{dL}}{\mathrm{dt}}=2  L⋅dLdt\mathrm{~L} \cdot \frac{\mathrm{dL}}{\mathrm{dt}}) (\Rightarrow \quad 2 L ⋅dLdt=\cdot \frac{d L}{d t}=2 L dLdt\frac{d L}{d t}) Since (dLdt\frac{d L}{d t}) is perpendicular to (L \Rightarrow dLdt=\frac{d L}{d t}=0 \Rightarrow L=) constant
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