Question

A point move in the plane so that its tangential acceleration${\omega _\tau } = a$, and its normal acceleration${\omega _n} = b{t^4}$, where a and b are positive constants, and t is time. At the moment $t = 0$ the point was at rest. Find how the curvature radius R of the point’s trajectory and total acceleration $\omega$ depend on the distance covered s.
A. $R = \dfrac{{{a^3}}}{{2bs}}$
B. $\omega = a\sqrt {1 + {{\left( {\dfrac{{4b{s^2}}}{{{a^3}}}} \right)}^2}}$
C. $R = \dfrac{{{a^3}}}{{4bs}}$
D. $\omega = a\sqrt {1 + {{\left( {\dfrac{{2b{s^2}}}{{{a^3}}}} \right)}^2}}$

Answer 1

Use kinematic equations to determine the final velocity and distance covered by the point. Use the formula for radial acceleration and substitute the obtained expressions of velocity and distance. Equate the value of normal acceleration with the given value of normal acceleration then you will get the curvature radius. Use the formula for total acceleration in the circular motion to determine the total acceleration.

Formula used:
$\omega = \sqrt {\omega _n^2 + \omega _\tau ^2}$
Here, ${\omega _n}$ is the normal acceleration or radial acceleration and ${\omega _\tau }$ is the tangential acceleration.

Complete step by step answer:
We have given, tangential acceleration is a and radial acceleration is $b{t^4}$.
We can calculate the point’s initial linear velocity using kinematic equation as follows,
$v = u + at$
Here, u is the initial velocity, a is the acceleration and t is the time.
We can calculate the distance covered by the point in time t as,
$s = ut + \dfrac{1}{2}a{t^2}$
We have given that the point was at rest at $t = 0$. Therefore, the initial velocity of the point is zero. Therefore, we can write the above equations as follows,
$v = at$ …… (1)
$s = \dfrac{1}{2}a{t^2}$ …… (2)
We have the normal angular acceleration of the point is,
${\omega _n} = \dfrac{{{v^2}}}{R}$
Here, R is the curvature radius of the circular motion.
Using equation (1 and (2)), we can write the above equation as,
${\omega _n} = \dfrac{{{a^2}{t^2}}}{R} = \dfrac{{2as}}{R}$ …… (3)
But we have given that the normal angular acceleration of the point’s trajectory is ${\omega _n} = b{t^4}$.
Therefore, we can write the above equation as,
$\dfrac{{{a^2}{t^2}}}{R} = b{t^4}$
$\Rightarrow R = \dfrac{{{a^2}}}{{b{t^2}}}$
Using equation (1) and (2), we can write the above equation as,
$R = \dfrac{{{a^3}}}{{2bs}}$ …… (4)
We know that the total acceleration for the system is written as,
$\omega = \sqrt {\omega _n^2 + \omega _\tau ^2}$
Using equation (3), we can write the above equation as,
$\omega = \sqrt {{{\left( {\dfrac{{2as}}{R}} \right)}^2} + {a^2}}$
We can substitute equation (4) in the above equation for R.
$\omega = \sqrt {{{\left( {\dfrac{{2as}}{{\dfrac{{{a^3}}}{{2bs}}}}} \right)}^2} + {a^2}}$
$\Rightarrow \omega = \sqrt {{{\left( {\dfrac{{4b{s^2}}}{{{a^2}}}} \right)}^2} + {a^2}}$
$\therefore \omega = a\sqrt {1 + {{\left( {\dfrac{{4b{s^2}}}{{{a^3}}}} \right)}^2}}$

So, the correct answer is “Option A and B”.

Note:
To solve such questions, the key is to remember all kinematic relations. Note that in linear motion, the particle has only linear acceleration but in circular motion, the particle has both radial acceleration acting towards the radius and tangential acceleration.