A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal…

Question

A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t=0 is 4 m s^(-1), the time taken to complete the first revolution will be 1/(α)1-e^(-2π) s, where α=______.

MARKS Content Team · Accepted Answer

Correct answer: 8. Given: a→_c=a→_t ⇒v^2r=dv/dt ⇒∫4v(dvv^2=∫0tdt)/r ⇒(-1)/v4v=t/r ⇒(-1)/v+1/4=2t ⇒v=4/(1-8t)=ds/dt ⇒4∫0tdt/(1-8t)=∫0sds r=0.5 m, distance covered s=2πr=π ⇒4×([ln(1-8t)]0t)/(-8)=π ⇒ℓn(1-8t)=-2π ⇒1-8t=e^(-2π) ⇒t=1-e^(-2π)1/8 s So, α=8.

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