Question

A particle moves on a circle of radius $r$ with centripetal acceleration as a function of time as $a_c=k^{2}r{t}^2$ where $k$ is a positive constant. Find the following quantities as a 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.

Answer 1

Step 1. Given Data,

$a_c=k^{2}r{t}^2$

Step 2. Formula Used,

Centripetal acceleration, $a_c={\displaystyle \frac{v^2}{r}}$

$r$ is the radius

$v$ is the velocity of the particle.

Tangential acceleration, $a_T=\frac{dv}{dt}$

Step. 3 Calculating the speed of the particle,

$a_c=k^{2}r{t}^2$ , $k$ is a constant.

Centripetal acceleration which is $a_c={\displaystyle \frac{v^2}{r}}$.

Equating these two expressions,
${\displaystyle \frac{v^2}{r}}=k^{2}r{t}^2$
$\Rightarrow v^2=k^2{r}^2{t}^2$
$\Rightarrow v=krt$

Thus, the speed of the particle is $krt$.
Step. 4 Calculating the tangential acceleration,
$a_T={\displaystyle \frac{dv}{dt}}=kr{\displaystyle \frac{dt}{dt}}=kr$

Thus, the tangential acceleration is $kr$.
Step. 5 Calculating the resultant acceleration at any instant
$a_R=\sqrt{a_c^2+a_T^2+2a_c{a}_T\cos \theta }=\sqrt{a_c^2+a_T^2}$

The angle between both the acceleration is 90 degrees,

Thus the magnitude is
$a_R=\sqrt{{\left(k^{2}r{t}^2\right)}^2+{\left(kr\right)}^2}=kr\sqrt{k^2{t}^4+1}$

Step. 6 Calculating the angle made by the resultant with tangential direction,
Let's Consider the angle as $\varphi$,
$\tan \varphi ={\displaystyle \frac{a_c}{a_T+a_c\cos \theta }}={\displaystyle \frac{k^{2}r{t}^2}{kr+k^{2}r{t}^2\cos (90)}}={\displaystyle \frac{k^{2}r{t}^2}{kr}}=k{t}^2$.

Therefore, $\varphi ={\tan }^{-1}(k{t}^2)$
Since the individual vectors are perpendicular to each other, so $\varphi ={\tan }^{-1}(k{t}^2)$

Hence, the speed of the particle is $krt$.

Hence the tangential acceleration is $kr$.

Hence the resultant acceleration is $a_R=kr\sqrt{k^2{t}^4+1}$.

Hence the angle made by the resultant with tangential direction is $\varphi ={\tan }^{-1}(k{t}^2)$