Name: Inertial Frames of Reference
Uploaded: 2019-07-19
Duration: 7 min 10 s
Description: Imagine if you are on a bus or a train, can you say you are in an inertial frame of reference? If yes, how can you predict it? Let us explore more in this vi...

Question

A point moves with deceleration along the circle of radius $R$ so that at any moment of time, its tangential and normal accelerations are equal in moduli. At the initial moment $t = 0$ , the velocity of the point equals $v_{0}$ . Find,
$(a)$ the velocity of the point as a function of time and as a function of the distance covered $s$ .
$(b)$ the total acceleration of the point as a function of velocity and the distance covered.

Accepted Answer

Here, we know that tangential acceleration $w_{t}$ is given by,

$w_{t} = \frac{d v}{d t} = \frac{v d v}{d s}$

And, normal acceleration $w_{n}$ is given by,

$w_{n} = \frac{v^{2}}{R}$

$(a)$ Let the speed of the particle be $v = v$ at time $t = t$ .

Given, at $t = 0$ , $v = v_{0}$ .

As given in the question:

The magnitude of tangential acceleration $=$ The magnitude of radial acceleration

So, we have,

$\Rightarrow |w_{t}| = |w_{n}| . . . (1)$

Let us calculate the velocity $v$ as a function of time $t$ .

Using the equation $(1)$ , we get,

$- \frac{d v}{d t} = \frac{v^{2}}{R}$

(We have used negative sign in tangential acceleration because the point is decelerating)

$t = 0$

Integrating within proper limits, we get,

$\int_{v_{0}}^{v} \frac{d v}{v^{2}} = - \frac{1}{R} \int_{0}^{t} d t$

$\Rightarrow {[\frac{v^{- 2 + 1}}{- 2 + 1}]}_{v_{0}}^{v} = - \frac{t}{R}$

$\Rightarrow {[- \frac{1}{v}]}_{v_{0}}^{v} = - \frac{t}{R}$

$\Rightarrow [\frac{1}{v} - \frac{1}{v_{0}}] = \frac{t}{R}$

$\Rightarrow v = \frac{R v_{0}}{t v_{0} + R} \Rightarrow v = \frac{v_{0}}{\frac{t v_{0} + R}{R}} \Rightarrow v = (\frac{v_{0}}{1 + \frac{v_{0} t}{R}}) . . . (2)$

Now let us calculate the velocity as a function of distance $s$ .

From the equation $(1)$ , we have,

$w_{t} = w_{n}$

$\Rightarrow - v \frac{d v}{d s} = \frac{v^{2}}{R}$

$\Rightarrow - \frac{d v}{v} = \frac{1}{R} d s$

Integrating within proper limits, we get,

$\int_{v_{0}}^{v} \frac{d v}{v} = - \frac{1}{R} \int_{0}^{s} d s$

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

$∴ v = v_{0} e^{- \frac{s}{R}} . . . (3)$

$(b)$ Total acceleration $w$ is given by,

$w = \sqrt{w_{t}^{2} + w_{n}^{2}} \Rightarrow w = \sqrt{2 w_{n}^{2}} [using (1)]$

$\Rightarrow w = \sqrt{2} w_{n} \Rightarrow w = \frac{\sqrt{2} v^{2}}{R} . . . (4)$

Clearly, the equation $(4)$ represents acceleration as a function of velocity.

Now, putting $v$ from the equation $(3)$ into equation $(4)$ , we get acceleration $w$ as a function of distance $s$ covered.

$∴ w = \frac{\sqrt{2}}{R} v_{0}^{2} e^{- \frac{2 s}{R}}$

$\Rightarrow w = \frac{\sqrt{2} v_{0}^{2}}{R e^{\frac{2 s}{R}}}$

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