Question

The distance of a particle moving on a circular path of radius $12\text{m}$ measured from a fixed point on the circle and measured along the circle is given by $S=2t^3$ (in metres). The ratio of its tangential to centripetal acceleration at $t=2\text{s}$ is

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $3:1$

Answer 1

The correct option is B $1:2$

Tangential acceleration is given by,

$a_t=\frac{dv}{dt}=\frac{d}{dt}\left(\frac{ds}{dt}\right)$

$S=2t^3$

$a_t=\frac{d}{dt}\left(6t^2\right)$

or $a_t=12t$ ....(i)

The centripetal acceleration is given by,

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

or $a_c=\frac{{\left(\frac{ds}{dt}\right)}^2}{R}=\frac{(6t^2{)}^2}{R}$

$\Rightarrow a_c=\frac{36t^4}{12}$

or $a_c=3t^4$ ......(ii)

The ratio of $a_t$ to centripetal acceleration $\left(a_c\right)$ will be,

$\frac{a_t}{a_c}=\frac{12t}{3t^4}$

$\Rightarrow \frac{a_t}{a_c}=\frac{4}{t^3}$

Putting $t=2\text{s}$ we get,

$\frac{a_t}{a_c}=\frac{4}{8}$

$\therefore \frac{a_t}{a_c}=\frac{1}{2}$

Why this Question? Caution: Beware!!! In problems involving variable quantities do not take ratio by substituting the values in function. Rather simplify down the ratio to last stage then substitute the value of parameter like time.