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Byju's Answer

Standard XIII

Physics

Pure Rolling

A disc is rol...

Question

A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let $v_{P}, v_{Q} a n d v_{C}$ be the magnitude of velocities of points P,Q and C respectively, then

$v_{Q} > v_{C} > v_{P}$

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$v_{Q} < v_{C} < v_{P}$

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$v_{Q} = v_{P}, v_{C} = \frac{1}{2} v_{P}$

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$v_{Q} < v_{C} > v_{P}$

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Solution

The correct option is A
$v_{Q} > v_{C} > v_{P}$

About point O, motion is pure rotational. Since OQ > OC > OP, $v_{Q} > v_{C} > v_{P}$ .

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Similar questions

A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let $v_{P}, v_{Q} a n d v_{C}$ be the magnitude of velocities of points P,Q and C respectively, then

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Q. A disc is rolling without slipping on a horizontal surface with C, as its centre, and Q and P the two points equidistant from
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A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let vP, vQ and vC be the magnitude of velocities of points P,Q and C respectively, then

The correct option is A vQ>vC>vP About point O, motion is pure rotational. Since OQ > OC > OP, vQ>vC>vP.

A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let $v_{P}, v_{Q} a n d v_{C}$ be the magnitude of velocities of points P,Q and C respectively, then

The correct option is A
$v_{Q} > v_{C} > v_{P}$

About point O, motion is pure rotational. Since OQ > OC > OP, $v_{Q} > v_{C} > v_{P}$ .