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Rotational Inertia

A disc of rad...

Question

A disc of radius $R$ rolls without slipping at speed $v$ along positive x-axis. Velocity of point $P$ at the instant shown in figure is:

{\to V}_{p} = (v + \frac{v r s i n θ}{R})^i + \frac{v r c o s θ}{R}^j

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{\to V}_{p} = (v + \frac{v r s i n θ}{R})^i - \frac{v r c o s θ}{R}^j

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{\to V}_{p} = \frac{v r s i n θ}{R}^i + \frac{v r c o s θ}{R}^j

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{\to V}_{p} = \frac{v r s i n θ}{R}^i - \frac{v r c o s θ}{R}^j

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Solution

The correct option is A ${\to V}_{p} = (v + \frac{v r s i n θ}{R})^i - \frac{v r c o s θ}{R}^j$
Here, $ω = \frac{v}{R}$
$V_{P_{x}} = (v + \frac{v}{R} r s i n θ)^i$
$V_{P_{y}} = - (\frac{v}{R} r c o s θ)^j$
$∴ {\to V}_{P} = (v + \frac{v r s i n θ}{R})^i - \frac{v r c o s θ}{R}^j$

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A disc of radius R rolls without slipping at speed v along positive x-axis. Velocity of point P at the instant shown in figure is:

The correct option is A →Vp=(v+vrsinθR)^i−vrcosθR^jHere, ω=vRVPx=(v+vRrsinθ)^iVPy=−(vRrcosθ)^j∴→VP=(v+vrsinθR)^i−vrcosθR^j

A disc of radius $R$ rolls without slipping at speed $v$ along positive x-axis. Velocity of point $P$ at the instant shown in figure is:

The correct option is A ${\to V}_{p} = (v + \frac{v r s i n θ}{R})^i - \frac{v r c o s θ}{R}^j$
Here, $ω = \frac{v}{R}$
$V_{P_{x}} = (v + \frac{v}{R} r s i n θ)^i$
$V_{P_{y}} = - (\frac{v}{R} r c o s θ)^j$
$∴ {\to V}_{P} = (v + \frac{v r s i n θ}{R})^i - \frac{v r c o s θ}{R}^j$