Question

A particle moves in a circle of radius $R$ such that its linear speed varies with time $t$ as $v = kt$, where $k$ is a positive constant. The angle $\theta$ between the net acceleration vector and the velocity vector at time $t$ is given by:

• A. \( \tan^{-1}\left(\frac{k^2 t^2}{R}\right) \)
• B. \( \tan^{-1}\left(\frac{kt^2}{R}\right) \)
• C. \( \tan^{-1}\left(\frac{k t}{R}\right) \)
• D. \( \tan^{-1}\left(\frac{R}{k^2 t^2}\right) \)

Hint:

Test the extreme temporal boundary states to clear doubts quickly! At the absolute start (\(t = 0\)), the speed is zero, meaning there is zero centripetal turning force and the particle purely accelerates forward tangentially. Thus, the angle \(\theta\) must be \(0^\circ\). Substituting \(t = 0\) into our options shows that \(\tan^{-1}(0) = 0^\circ\) for Option B, verifying the physical validity of our formula.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the angle \(\theta\) between the net acceleration vector and the velocity vector for a particle in circular motion with speed \(v = kt\).
The velocity vector \(\vec{v}\) of a particle in circular motion is always directed tangentially along the path.
Thus, the angle \(\theta\) is the angle between the net acceleration vector and the tangential direction.
Step 2: Key Formula or Approach:
The total acceleration vector \(\vec{a}\) in circular motion is the vector sum of two perpendicular components:
1. Tangential acceleration (\(a_t\)), which changes the speed:
\[ a_t = \frac{dv}{dt} \]
2. Centripetal acceleration (\(a_c\)), which changes the direction:
\[ a_c = \frac{v^2}{R} \]
The tangent of the angle \(\theta\) between the net acceleration and the tangential acceleration (velocity direction) is:
\[ \tan\theta = \frac{a_c}{a_t} \implies \theta = \tan^{-1}\left(\frac{a_c}{a_t}\right) \]
Step 3: Detailed Explanation:
1. Compute the tangential acceleration component \(a_t\) by differentiating the speed with respect to time:
\[ a_t = \frac{d}{dt}(kt) = k \]
2. Compute the centripetal acceleration component \(a_c\) using the given speed equation:
\[ a_c = \frac{v^2}{R} = \frac{(kt)^2}{R} = \frac{k^2 t^2}{R} \]
3. Substitute both acceleration components into the trigonometric ratio:
\[ \tan\theta = \frac{\left(\frac{k^2 t^2}{R}\right)}{k} = \frac{kt^2}{R} \]
4. Solve for the angle \(\theta\):
\[ \theta = \tan^{-1}\left(\frac{kt^2}{R}\right) \]
This matches Option (B).
Step 4: Final Answer:
The angle \(\theta\) between the net acceleration vector and the velocity vector is \( \tan^{-1}\left(\frac{kt^2}{R}\right) \), which corresponds to Option (B).