A particle moves on a circle of radius $r$ with centripetal acceleration as function of time as $a_{c} = k^{2} r t^{2}$ where $k$ is a positive constant. Find the following quantities as function of time at an instant.
a. The speed.of the particle
b. The tangential acceleration of the particle
c. The resultant acceleration, and
d. Angle made by the resultant with tangential direction.

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Solution

a. In the problem, it is given that $a_{c} = k^{2} r t^{2}$
Since $a_{c} = \frac{v^{2}}{r} \Rightarrow \frac{v^{2}}{r} = k^{2} r t^{2}$
$\Rightarrow v^{2} = k^{2} r^{2} t^{2}$
Taking square root on both sides, we get $v = k r t$
b. Tangential acceleration,
$a_{t} = \frac{d v}{d t} = \frac{d}{d t} (k r t) = k r$
c. Resultant acceleration,
$a_{r e s} = \sqrt{a_{c}^{2} + a_{t}^{2}}$
$= \sqrt{(k^{2} r t^{2})^{2} + (k r)^{2}}$
$= k r \sqrt{k^{2} t^{4} + 1}$
d. From Fig., $t a n α = \frac{a_{c}}{a_{1}} = \frac{k^{2} r t^{2}}{k r} = k t^{2}$
$\Rightarrow α = t a n^{- 1} (k t^{2})$

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a. In the problem, it is given that ac=k2rt2Since ac=v2r⇒v2r=k2rt2⇒v2=k2r2t2Taking square root on both sides, we get v=krtb. Tangential acceleration,at=dvdt=ddt(krt)=krc. Resultant acceleration,ares=√a2c+a2t=√(k2rt2)2+(kr)2=kr√k2t4+1d. From Fig., tanα=aca1=k2rt2kr=kt2⇒α=tan−1(kt2)