- #1
Faiq
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Homework Statement
2 A particle P of mass mkg moves on an arc of a circle with centre O and radius a metres. At time t = 0
the particle is at the point A. At time t seconds, angle POA = sin^2 2t.
Find
(i) the value of t when the transverse component of the acceleration of P is first equal to zero
The answer is [tex] \frac{d^2\theta}{dt^2} = 0 [/tex]
Isn't d^2theta/dt^2 equal to radial acceleration. Since angular velocity is rate of change of theta, thus rate of change of change of theta should be angular acceleration aka radial acceleration?
Then why are the equating it as transverse component?