Question

A particle begins to move with a tangential acceleration of constant magnitude $0.6\dfrac{m}{{{s^2}}}$ in a circular path. If it slips when its total acceleration becomes $1{\text{ m/}}{{\text{s}}^2}$. Find the angle through which it would have turned before it starts to slip.
A. $\dfrac{2}{3}radians$
B. $\dfrac{2}{3}{\text{degree}}$
C. $\dfrac{4}{3}radian$
D. $\dfrac{4}{3}{\text{degree}}$

Answer 1

In this question, we need to determine the angle through which it would have turned before it starts to slip. For this, we will use the relation between the tangential acceleration and the centripetal acceleration in a circular motion.

Complete step by step answer:
Tangential acceleration of particle ${a_T} = 0.6\dfrac{m}{{{s^2}}}$
Total net acceleration when particle slips ${a_{net}} = 1\dfrac{m}{{{s^2}}}$
We know the net acceleration of a particle in a circular path is given as
$\left| {{a_{net}}} \right| = \sqrt {a_T^2 + a_C^2} - - (i)$
Now substitute the value of tangential acceleration and the net acceleration in the equation (i), we get

$$1 = \sqrt {{{\left( {0.6} \right)}^2} + a_C^2} \\\ \Rightarrow {\left( {0.6} \right)^2} + a_C^2 = 1 \\\ \Rightarrow 0.36 + a_C^2 = 1 \\\ \Rightarrow a_C^2 = 0.64 \\\ \Rightarrow a_C = 0.8 \\\$$

So the centripetal acceleration of the particle ${a_C} = 0.8$
Now by newton’s law of motion we know that ${v^2} = {u^2} + 2aS$, v is the final velocity, u is the initial velocity, a is the acceleration and S is the distance travelled, hence in the case of circular motion we can write
${\omega ^2} = \omega _0^2 + 2{\alpha _T}\theta - - (ii)$
Now since the particle was initially at rest, we can write${\omega _0} = 0$
Hence we can write equation (ii) as
${\omega ^2} = 2{\alpha _T}\theta - - (iii)$
Now we multiply both side of the equation (iii) with R, we get
${\omega ^2}R = 2{\alpha _T}R\theta - - (iv)$
Now we know centripetal acceleration ${a_C}$ is given by the formula ${a_C} = {\omega ^2}R$and tangential acceleration ${a_T}$is given as${a_T} = \alpha R$, hence we can write equation (iv) as

$${\omega ^2}R = 2{\alpha _T}R\theta \\\ \Rightarrow {a_C} = 2{a_T}\theta \\\$$

This can be further written as
$\theta = \dfrac{{{a_C}}}{{2{a_T}}}$
Now substitute the values of tangential acceleration and centripetal acceleration, we get

$$\theta = \dfrac{{{a_C}}}{{2{a_T}}} \\\ \Rightarrow \theta = \dfrac{{0.8}}{{2 \times 0.6}} \\\ \Rightarrow \theta = \dfrac{2}{3}radian \\\$$

Hence the angle through particle would have turned $= \dfrac{2}{3}radian$
Option A is correct.

Note: Centripetal acceleration is directed in inward direction of the circular path and tangential acceleration is in the direction tangent to the circular path and they both are perpendicular to each other.
In a circular motion of the particle net acceleration is given by the formula
$\left| {{a_{net}}} \right| = \sqrt {a_T^2 + a_C^2}$, where ${a_T}$ is the tangential acceleration of the particle in the circular path and ${a_C}$ is the centripetal acceleration of the particle