The centre of a wheel rolling on a plane surface moves with a speed v0 . A particle on the rim of the wheel at the same level as the centre will be moving at a speed √x v0. Then the value of x is.

Question

The centre of a wheel rolling on a plane surface moves with a speed v0 . A particle on the rim of the wheel at the same level as the centre will be moving at a speed √x v0. Then the value of x is. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/934f6e4a6ae0010f4a903be2b249e54d-.png" /> For no slipping v0=ω R Now vA=vB=√v02+(ω R)2 =√2 v0 ⇒ x=2

Q. The centre of a wheel rolling on a plane surface moves with a speed $v_{0} .$ A particle on the rim of the wheel at the same level as the centre will be moving at a speed $x v_{0}$ . Then the value of $x$ is_______.

For no slipping $v_{0} = ω R$
Now $v_{A} = v_{B} = v_{0}^{2} + (ω R)^{2}$
$= 2 v_{0}$
$\Rightarrow x = 2$

Q. The centre of a wheel rolling on a plane surface moves with a speed v0​. A particle on the rim of the wheel at the same level as the centre will be moving at a speed xv0​​. Then the value of x is_______.

For no slipping v0​=ωR Now vA​=vB​=v02​+(ωR)2​ =2​v0​ ⇒x=2

Q. The centre of a wheel rolling on a plane surface moves with a speed $v_{0} .$ A particle on the rim of the wheel at the same level as the centre will be moving at a speed $x v_{0}$ . Then the value of $x$ is_______.

For no slipping $v_{0} = ω R$
Now $v_{A} = v_{B} = v_{0}^{2} + (ω R)^{2}$
$= 2 v_{0}$
$\Rightarrow x = 2$