Step 1 : Understanding the Question:
This problem asks us to determine the net or total linear acceleration of a particle transitioning along a circular trajectory of a fixed radius under a constant angular acceleration. The particle starts from rest, which means its initial angular velocity is zero. As it moves, its speed changes (giving rise to tangential acceleration) and its direction of motion constantly changes (giving rise to radial or centripetal acceleration). We need to find the vector sum of these two components at any given time $t$.
Step 2 : Key Formulas and Approach:
The motion of the particle involves both rotational and translational components, which are linked by the radius of the circle.
The angular velocity after time $t$ is given by the rotational kinematic equation:
\[ \omega = \omega_0 + \alpha t \]
The tangential acceleration ($a_t$) is responsible for changing the speed of the particle and is given by:
\[ a_t = R \alpha \]
The centripetal acceleration ($a_c$) is responsible for keeping the particle in circular motion and is given by:
\[ a_c = \omega^2 R \]
Because these two components are perpendicular to each other, the magnitude of the total linear acceleration is calculated using the Pythagorean theorem:
\[ a_{\text{total}} = \sqrt{a_t^2 + a_c^2} \]
Step 3 : Detailed Explanation:
We start by determining the angular velocity $\omega$ at time $t$ using the initial condition. Since the particle starts from rest, the initial angular velocity $\omega_0$ is equal to zero. Substituting this into our kinematic formula gives $\omega = \alpha t$.
Next, we express the tangential acceleration of the particle. The tangential acceleration depends only on the constant angular acceleration and the radius of the circular path. This gives the constant value $a_t = R \alpha$.
Then, we calculate the centripetal acceleration at time $t$. We substitute the expression for $\omega$ into the centripetal acceleration formula to get $a_c = (\alpha t)^2 R = R \alpha^2 t^2$.
We now combine these two orthogonal acceleration vectors to find the magnitude of the total acceleration. Substituting the individual expressions into the magnitude formula gives $a_{\text{total}} = \sqrt{(R \alpha)^2 + (R \alpha^2 t^2)^2}$.
Expanding the terms inside the square root yields $a_{\text{total}} = \sqrt{R^2 \alpha^2 + R^2 \alpha^4 t^4}$.
To simplify the expression, we factor out the common term $R^2 \alpha^2$ from both terms under the radical, giving $a_{\text{total}} = \sqrt{R^2 \alpha^2 (1 + \alpha^2 t^4)}$.
Extracting the square root of $R^2 \alpha^2$ outside the radical leads to the final simplified expression: $a_{\text{total}} = R \alpha \sqrt{1 + \alpha^2 t^4}$.
Step 4 : Final Answer:
The total acceleration of the particle at time $t$ is given by the expression $R\alpha\sqrt{1 + \alpha^2 t^4}$. This corresponds to Option (C).