Animated Solution for Physics - Kinematics: A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $P$ is such that it sweeps out a length $s=t^3+5$, where $s$ is in metre and $t$ is in second. The radius of the path is $20\text{ m}$. The acceleration of $P$ when $t=2\text{ s}$ is nearly

A Journey on an Arc

Imagine a particle $P$ moving along a circular path of radius $R=20\text{ m}$. It starts from point $A$ and moves counter-clockwise towards point $B$. The distance it travels along the arc is given by the equation $s=t^3+5$.

Because the distance $s$ is a cubic function of time, the speed of the particle is not constant. It is changing with time. This tells us immediately that we are dealing with non-uniform circular motion. In such a motion, the particle will experience two distinct components of acceleration: a tangential component that changes its speed, and a centripetal component that changes its direction.

To understand the motion fully, we first need to find the speed of the particle. Speed is simply the rate of change of distance with respect to time. So, we differentiate the distance equation $s$ with respect to $t$:

This equation gives us the instantaneous speed of the particle at any given time $t$.

Now, let's find the two components of acceleration. The tangential acceleration ($a_t$) is responsible for the change in the magnitude of velocity (speed). We find it by differentiating the speed $v$ with respect to time:

We are asked to find the acceleration at $t=2\text{ s}$. Let's substitute this value into our expressions for speed and tangential acceleration:

At $t=2\text{ s}$:

Next, we calculate the centripetal acceleration ($a_c$). This component is responsible for continuously changing the direction of the particle to keep it on the circular path. It is given by the formula $v^2/R$:

Finally, the net acceleration $\vec{a}$ is the vector sum of the tangential and centripetal accelerations. Since the tangential acceleration is along the tangent and the centripetal acceleration is directed towards the center, they are always perpendicular to each other. Therefore, we can use the Pythagorean theorem to find the magnitude of the net acceleration:

Calculating the square root, we get $a\approx 13.99{\text{ ms}}^{-2}$, which is approximately $14{\text{ ms}}^{-2}$.

Always remember, in non-uniform circular motion, the net acceleration vector is never directed exactly towards the center; it always leans forward or backward depending on whether the particle is speeding up or slowing down!