Q.
A spot of light S rotates in a horizontal plane with a constant angular velocity of 0.1rad/s. The spot of light P moves along the wall at a distance of 3m from S. The velocity of spot P, where θ=45∘, is
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ManipalManipal 2012Laws of Motion
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Solution:
The situation is shown in figure.
From figure, x=rtanθ ∴ Velocity of P is v=dtdx=rsec2θ(dtdθ)
where, dtdθ= angular velocity of rotation of spot =ω ∴v=ωrsec2θ
At ϕ=45∘, so θ=45∘
Hence, v=0.1×3×sec245∘ =0.1×3×2=0.6m/s