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Byju's Answer

Standard XII

Mathematics

Equation of Circle with Origin as Center

A disc is rol...

Question

A disc is rolling (without slipping)on a horizontal 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 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} = 1 / 2 V_{P}

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

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Solution

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

$O P < O C < O Q \dots (i)$

Also $V = r ω \dots (i i)$

It follows from (i) and (ii) that

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

or $V_{Q} > V_{C} > V_{P}$

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A disc is rolling (without slipping)on a horizontal surface. C is its center and Q and P are two points equidistant from C. Let VP,VQ and Vc be the magnitudes of velocities of points P, Q and C respectively. Then

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

$O P < O C < O Q \dots (i)$

Also $V = r ω \dots (i i)$

It follows from (i) and (ii) that

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

or $V_{Q} > V_{C} > V_{P}$