Question

A particle moves with deceleration along the circle of radius R so that at any moment of time its tangential and normal accelerations are equal in magnitude. At the initial moment t = 0 the speed of the particle equals $v_0$, then :

(i) the speed of the particle as a function of the distance covered s will be

(ii) the total acceleration of the particle as function of velocity.

• A. v = $v_0$ $e^{-s/R}$
• A. a = \sqrt{2}$$\frac{v^2}{R}
• B. v = $v_0$ $e^{s/R}$
• B. a = $\frac{v^2}{R}$
• C. v = $v_0$ $e^{-R/s}$
• C. a = $\frac{2v^2}{R}$
• D. v = $v_0$ $e^{R/s}$
• D. a = $\frac{2\sqrt{2}v^2}{R}$

Answer 1

Answer: (A), (A)

Both (i) and (ii) are (A)

Answer 2

The problem involves analyzing the motion of a particle in non-uniform circular motion. We are given that the particle moves with deceleration, and the magnitudes of its tangential and normal accelerations are equal.

Part (i): Speed of the particle as a function of the distance covered (s)

1. Normal Acceleration ($a_n$): This acceleration is always directed towards the center of the circle and its magnitude is given by:

   $a_n = \frac{v^2}{R}$

   where $v$ is the instantaneous speed and $R$ is the radius of the circle.

2. Tangential Acceleration ($a_t$): This acceleration is responsible for changing the magnitude of the speed. It can be expressed as:

   $a_t = \frac{dv}{dt}$

   Since the particle is decelerating, $a_t$ is negative (opposite to the direction of velocity). Also, using the chain rule, $a_t = \frac{dv}{ds} \frac{ds}{dt} = v \frac{dv}{ds}$, where $s$ is the distance covered.

3. Given Condition: The magnitudes of tangential and normal accelerations are equal: $|a_t| = |a_n|$. Since the particle is decelerating, its speed is decreasing, so $a_t = -a_n$. Therefore, we can write:

   $v \frac{dv}{ds} = -\frac{v^2}{R}$

4. Solve the Differential Equation: Assuming $v \neq 0$ (the particle is moving), we can divide both sides by $v$:

   $\frac{dv}{ds} = -\frac{v}{R}$

   This is a separable differential equation. Separate the variables:

   $\frac{dv}{v} = -\frac{ds}{R}$

5. Integrate: Integrate both sides from the initial conditions ($s=0$, $v=v_0$) to the current conditions ($s$, $v$):

   $\int_{v_0}^{v} \frac{dv'}{v'} = \int_{0}^{s} -\frac{ds'}{R}$

   $[\ln v']_{v_0}^{v} = [-\frac{s'}{R}]_{0}^{s}$

   $\ln v - \ln v_0 = -\frac{s}{R} - 0$

   $\ln\left(\frac{v}{v_0}\right) = -\frac{s}{R}$

6. Exponentiate: To find $v$, take the exponential of both sides:

   $\frac{v}{v_0} = e^{-s/R}$

   $v = v_0 e^{-s/R}$

Part (ii): Total acceleration of the particle as a function of velocity

1. Total Acceleration ($a$): In circular motion, the tangential acceleration ($\vec{a_t}$) and normal (centripetal) acceleration ($\vec{a_n}$) are always perpendicular to each other. The magnitude of the total acceleration is given by the Pythagorean theorem:

   $a = \sqrt{a_t^2 + a_n^2}$

2. Apply Given Condition: We are given that the magnitudes of tangential and normal accelerations are equal, i.e., $|a_t| = |a_n|$. We know $a_n = \frac{v^2}{R}$. Therefore, $|a_t| = \frac{v^2}{R}$.

3. Substitute into Total Acceleration Formula:

   $a = \sqrt{\left(\frac{v^2}{R}\right)^2 + \left(\frac{v^2}{R}\right)^2}$

   $a = \sqrt{2 \left(\frac{v^2}{R}\right)^2}$

   $a = \sqrt{2} \frac{v^2}{R}$

The final answer is $\boxed{\text{Both (i) and (ii) are (A)}}$

Explanation of the solution: (i) Tangential acceleration $a_t = v \frac{dv}{ds}$ and normal acceleration $a_n = \frac{v^2}{R}$. Given $|a_t| = |a_n|$ and deceleration, $v \frac{dv}{ds} = -\frac{v^2}{R}$. Solving the differential equation $\frac{dv}{v} = -\frac{ds}{R}$ by integrating from $(v_0, 0)$ to $(v, s)$ yields $\ln(v/v_0) = -s/R$, which simplifies to $v = v_0 e^{-s/R}$. (ii) Total acceleration $a = \sqrt{a_t^2 + a_n^2}$. Given $|a_t| = |a_n|$ and $a_n = \frac{v^2}{R}$, substitute these into the formula: $a = \sqrt{\left(\frac{v^2}{R}\right)^2 + \left(\frac{v^2}{R}\right)^2} = \sqrt{2 \left(\frac{v^2}{R}\right)^2} = \sqrt{2} \frac{v^2}{R}$.