NEET-XII-Physics
10: Rotational Mechanics
- #8Let
→Fbe a force acting on a particle having position vector
→r. Let
→Γbe the torque of this force about the origin, then
(a)
r→.Γ→=0 and F→.Γ→=0
(b)
r→.Γ→=0 but F→.Γ→≠0
(c)
r→.Γ→≠0 but F→.Γ→=0
(d)
r→.Γ→≠0 and F→.Γ→≠0digAnsr: aAns : (a) `` \stackrel{\to }{r}.\stackrel{\to }{\,\mathrm{\,\Gamma \,}}=0\,\mathrm{\,and\,}\stackrel{\to }{F}.\stackrel{\to }{\,\mathrm{\,\Gamma \,}}=0``
We have:
`` \stackrel{\to }{\tau }=\stackrel{\to }{r}\times \stackrel{\to }{F}``
Thus, `` \stackrel{\to }{\,\mathrm{\,\tau \,}}`` is perpendicular to `` \stackrel{\to }{r}`` and `` \stackrel{\to }{F}``.
Therefore, we have:
`` \stackrel{\to }{r}.\stackrel{\to }{\,\mathrm{\,\tau \,}}=0\,\mathrm{\,and\,}\stackrel{\to }{F}.\stackrel{\to }{\,\mathrm{\,\tau \,}}=0``
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