Question

A particle moves along a circle of radius $R$ with a constant angular acceleration $\alpha$. If the initial angular velocity is zero, the total acceleration of the particle at time $t$ is:

• A. \( R\alpha \)
• B. \( R\alpha^2 t^2 \)
• C. \( R\alpha\sqrt{1 + \alpha^2 t^4} \)
• D. \( R\alpha t \)

Hint:

To verify your expression instantly during an exam, test the initial boundary condition at \(t = 0\). At the absolute start of the motion, the particle isn't spinning yet (\(\omega = 0\)), so its centripetal component must be zero, leaving only the pure tangential kick-start acceleration (\(a = R\alpha\)). Substituting \(t = 0\) into Option C gives \(R\alpha\sqrt{1+0} = R\alpha\), matching the physical behavior perfectly!

Answer 1

The Correct Option is C

Concept: When a particle moves along a circular path under non-uniform rotational motion (i.e., with an angular acceleration), it experiences two distinct, mutually perpendicular components of acceleration:
• Tangential Acceleration (\(a_t\)): Responsible for changing the magnitude of the linear velocity. It is given by: \[ a_t = R\alpha \]
• Centripetal (or Radial) Acceleration (\(a_c\)): Responsible for changing the direction of the velocity vector, pointing toward the center of the circular path. It is given by: \[ a_c = \omega^2 R \] Since these two components are orthogonal (\(a_t \perp a_c\)), the magnitude of the net total acceleration vector (\(a_{\text{total}}\)) is found using the Pythagorean theorem: \[ a_{\text{total}} = \sqrt{a_t^2 + a_c^2} \]

Step 1: Determining the angular velocity ($\omega$) at time $t$.
We can apply the rotational kinematic equation for constant angular acceleration, analogous to the linear equation \(v = u + at\): \[ \omega = \omega_0 + \alpha t \] Given that the initial angular velocity is zero (\(\omega_0 = 0\)): \[ \omega = 0 + \alpha t = \alpha t \]

Step 2: Calculating the individual acceleration components.

• The tangential acceleration component remains constant over time: \[ a_t = R\alpha \]
• The centripetal acceleration component at time \(t\) is evaluated by substituting our derived \(\omega\): \[ a_c = \omega^2 R = (\alpha t)^2 R = \alpha^2 t^2 R \]

Step 3: Evaluating the net total acceleration magnitude.
Substitute both component expressions into our vector addition layout: \[ a_{\text{total}} = \sqrt{(R\alpha)^2 + (\alpha^2 t^2 R)^2} \] \[ a_{\text{total}} = \sqrt{R^2\alpha^2 + \alpha^4 t^4 R^2} \] Factoring out the shared common terms \(R^2\alpha^2\) inside the square root radical: \[ a_{\text{total}} = \sqrt{R^2\alpha^2 \cdot (1 + \alpha^2 t^4)} \] Taking \(R^2\alpha^2\) cleanly out of the radical yields the final expression: \[ a_{\text{total}} = R\alpha\sqrt{1 + \alpha^2 t^4} \]