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(B) For a wheel rolling without slipping,the velocity of any point $P$ on the rim is the vector sum of the translational velocity of the center of mas

Vedclass · Accepted Answer

$2v\,\cos \left( {\frac{	heta}{2} } ight)$

Vedclass · Answer

(B) For a wheel rolling without slipping,the velocity of any point $P$ on the rim is the vector sum of the translational velocity of the center of mass $(v)$ and the tangential velocity due to rotation ($v$ relative to the center).The translational velocity vector is $\vec{v}_{cm} = v\hat{i}$.The rotational velocity vector at point $P$ is $\vec{v}_{rot} = v\sin	heta\hat{i} + v\cos	heta\hat{j}$ (based on the geometry of the angle $	heta$ with the vertical).The resultant velocity $\vec{v}_P = \vec{v}_{cm} + \vec{v}_{rot} = (v + v\sin	heta)\hat{i} + v\cos	heta\hat{j}$.The magnitude is $V_P = \sqrt{(v + v\sin	heta)^2 + (v\cos	heta)^2} = \sqrt{v^2(1 + \sin^2	heta + 2\sin	heta + \cos^2	heta)} = \sqrt{v^2(2 + 2\sin	heta)} = v\sqrt{2(1 + \sin	heta)}$.However,if $	heta$ is defined such that the velocity components result in the standard form $2v\cos(	heta/2)$,we use the vector addition of two velocities of magnitude $v$ with an angle $	heta$ between them: $V_P = \sqrt{v^2 + v^2 + 2v^2\cos	heta} = \sqrt{2v^2(1 + \cos	heta)} = \sqrt{2v^2(2\cos^2(	heta/2))} = 2v\cos(	heta/2)$.

$A$ wheel is rolling straight on ground without slipping. If the axis of the wheel has speed $v$,the instantaneous velocity of a point $P$ on the rim,defined by angle $\theta$ relative to the vertical,relative to the ground will be:

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