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NEET 2003
Exam
NEET
Year
2003
Subject
Physics
Chapter
Motion in Two Dimensions

NEET 2003 · Physics · Motion in Two Dimensions

A particle moves along a circle ((20)/(π)) m with constant...

A particle moves along a circle (20π)m\left(\frac{20}{\pi}\right) \mathrm{m} with constant tangential acceleration. If the velocity of the particle is 80 m/s80 \mathrm{~m} / \mathrm{s} at the end of the second revolution after after motion has begun, the tangential acceleration is:

Options

  1. A40 m/s240 \mathrm{~m} / \mathrm{s}^2
  2. B640πm/s2640 \pi \mathrm{m} / \mathrm{s}^2
  3. C160πm/s2160 \pi \mathrm{m} / \mathrm{s}^2
  4. D40πm/s240 \pi \mathrm{m} / \mathrm{s}^2

Correct answer

A. 40 m/s240 \mathrm{~m} / \mathrm{s}^2

Step-by-step solution

From question r=20πm,v=80 m/s∴θ=2 revolutions =2×2π=4πrad∴ω2=ω02+2αθ∴ω2=2αθ…(i) since ω0=0 Since α=ar and  ω=Vr\begin{aligned}& r=\frac{20}{\pi}m, v=80 \mathrm{~m}/ \mathrm{s}\\ & \therefore \theta=2 \text{ revolutions }=2 \times2 \pi=4 \pi \mathrm{rad}\\ & \therefore \omega^2=\omega_0{ }^2+2 \alpha \theta\\ & \therefore \omega^2=2 \alpha \theta \ldots(\mathrm{i}) \text{ since } \omega_0=0 \\ & \text{ Since } \alpha=\frac{a}{r} \text{\text{ and } } \omega=\frac{\mathrm{V}}{r} \end{aligned}Putting these values in eq. (i)V2r2=2aθr⇒a=V22rθ=80×802×(20π)×4π=40 m/s2\begin{aligned}& \frac{V^2}{r^2}= \frac{2 a \theta}{r}\\ & \Rightarrow a= \frac{V^2}{2 r \theta}=\frac{80 \times 80}{2 \times\left(\frac{20}{\pi}\right) \times 4 \pi}\\ &=40 \mathrm{~m}/ \mathrm{s}^2 \end{aligned}
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