KVPY SX SB Exam 05-Nov-2017 Question Paper

Question : 91 of 120

Marks: +1, -0

Physics

One end of a rod of length L=1 m is fixed to a point on the circumference of a wheel of radius

R = 1/ √3

m. The other end is sliding freely along a straight channel passing through the center O of the wheel as shown in the figure below. The wheel is rotating with a constant angular velocity ω about O.

The speed of the sliding end P when θ = 60° is

${2ω }/3$
${ ω }/3$
${2ω }/√3$
${ ω }/√3$

Validate

Solution:

\tan 60º = L/R=1/{1\/√3}=√3=\tan θ

∴

x=√{1+1/3}=2/ √3

\cos θ ={R ^2+x^2- L^2 }/{2Rx}

⇒

R^2 + x^2 -L^2 = 2Rx \cos θ

⇒

2x{dx}/{dt}=2R[x(-\sinθ)+\cosθ{dx}/{dt}]

⇒

{dx}/{dt}[x-R\cosθ]=- Rx\sinθ{dθ}/{dt}

∴

-{dx}/{dt}=v \ &\ {dθ}/{dt}=ω

⇒

v= {Rx \sinθω}/{x- R\cosθ}

={1/√3 2/√3.√3/2.ω}/{2/√3-1/√3 . 1/2}={ω/√3}/{1/√3 3/2}={2ω}/3

v= {2ω}/3

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