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Byju's Answer

Standard XII

Physics

Pure Rolling

A disc is rol...

Question

A disc is rolling (without slipping) on a frictionless surface. C is its center and Q and P are two points equidistant from C. Let, $V_{P}, V_{Q}$ and $V_{C}$ be the magnitudes of velocity of points P, Q and C respectively, then which of the following is true ?

V_{Q} > V_{C} > V_{P}

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V_{Q} < V_{C} < V_{P}

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V_{Q} = V_{P}, V_{C} = \frac{1}{2}

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V_{Q} < V_{C} > V_{P}

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Solution

The correct option is A $V_{Q} > V_{C} > V_{P}$
${\to V}_{Q} = \to V + ω R$
${\to V}_{C} = \to V$
${\to V}_{P} = ω R - \to V$
$∴ V_{Q} > V_{C} > V_{P} .$
Above velocities are vectorially added so due to reverse sign there will be differnce in magnitudes.

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A disc is rolling (without slipping) on a frictionless surface. C is its center and Q and P are two points equidistant from C. Let, VP,VQ and VC be the magnitudes of velocity of points P, Q and C respectively, then which of the following is true ?

The correct option is A VQ>VC>VP→VQ=→V+ωR→VC=→V→VP=ωR−→V∴VQ>VC>VP.Above velocities are vectorially added so due to reverse sign there will be differnce in magnitudes.

A disc is rolling (without slipping) on a frictionless surface. C is its center and Q and P are two points equidistant from C. Let, $V_{P}, V_{Q}$ and $V_{C}$ be the magnitudes of velocity of points P, Q and C respectively, then which of the following is true ?

The correct option is A $V_{Q} > V_{C} > V_{P}$
${\to V}_{Q} = \to V + ω R$
${\to V}_{C} = \to V$
${\to V}_{P} = ω R - \to V$
$∴ V_{Q} > V_{C} > V_{P} .$
Above velocities are vectorially added so due to reverse sign there will be differnce in magnitudes.