A disc is rolling without slipping with angular velocity $ω$ . P and Q are two points equidistant from the centre C. The order of magnitude of velocity is [2004]

Question

Smart Achievers · Accepted Answer

(1)

In pure rolling, the point of contact is the instantaneous centre of rotation of all the particles of the disc.

On applying $v = r ω$

We find $ω$ is same for all the particles, then $v \propto r .$

$r_{Q} > r_{C} > r_{P} \Rightarrow v_{Q} > v_{C} > v_{P}$

Solution

Q.
A disc is rolling without slipping with angular velocity $ω$ . P and Q are two points equidistant from the centre C. The order of magnitude of velocity is [2004]

1 $v_{Q} > v_{C} > v_{P}$

2 $v_{P} > v_{C} > v_{Q}$

3 $v_{P} = v_{C}, v_{Q} = \frac{v_{C}}{2}$

4 $v_{P} < v_{C} > v_{Q}$

Ans.
(1)

In pure rolling, the point of contact is the instantaneous centre of rotation of all the particles of the disc.

On applying $v = r ω$

We find $ω$ is same for all the particles, then $v \propto r .$

$r_{Q} > r_{C} > r_{P} \Rightarrow v_{Q} > v_{C} > v_{P}$

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Solution

Q. A disc is rolling without slipping with angular velocity ω. P and Q are two points equidistant from the centre C. The order of magnitude of velocity is [2004]

1 vQ>vC>vP

2 vP>vC>vQ

3 vP=vC, vQ=vC2

4 vP<vC>vQ

Ans. (1) In pure rolling, the point of contact is the instantaneous centre of rotation of all the particles of the disc. On applying v=rω We find ω is same for all the particles, then v∝r. rQ>rC>rP ⇒ vQ>vC>vP