Question

The speed of a particle travelling in a circle of radius $20\text{cm}$ increases uniformly from $6\text{m/s}$ to $8\text{m/s}$ in $4\text{sec}$. The angular acceleration of particle is :

• A. $5{\text{rad/ s}}^2$
• B. $7.5{\text{rad/ s}}^2$
• C. $2.5{\text{rad/ s}}^2$
• D. $2{\text{rad/ s}}^2$

Answer 1

The correct option is C $2.5{\text{rad/ s}}^2$

Since speed increases uniformly, average tangential acceleration will be equal to instantaneous tangential acceleration.

Therefore, instantaneous tangential acceleration is given by,

$a_t=\frac{dv}{dt}=\frac{v_2-v_1}{t_2-t_1}$

$\Rightarrow a_t=\frac{8-6}{4}=0.5{\text{m/s}}^2$

Angular acceleration of the particle will be,

$\alpha =\frac{a_t}{r}$

$\Rightarrow \alpha =\frac{0.5{\text{m/s}}^2}{\left(\frac{20}{100}\text{m}\right)}$

$\therefore \alpha =\frac{5}{2}=2.5{\text{rad/s}}^2$

Alternatve:

$\alpha =\frac{{\omega }_t-{\omega }_o}{t}=\frac{v-u}{rt}$

$\alpha =\frac{8-6}{0.2\times 4}=\frac{2}{0.8}=2.5{\text{rad/s}}^2$

Why this question ? It intends to your understanding of fundamentals of average and instantaneous tangential acceleration in a circular motion. Tip: In case of uniform circular motion the average and instantaneous value of tangential acceleration will be the same.