Question

The tangential acceleration of a particle in a circular motion of radius $2\text{m}$ is $a_t=\alpha t{\text{m/s}}^2$ (where $\alpha$ is a constant) . Initially, the particle is at rest. Net acceleration vector of the particle makes ${45}^{\circ }$ with the radial acceleration after 2 sec. The value of constant $\alpha$ is:

• A. $\frac{1}{2}{\text{m/s}}^3$
• B. $1{\text{m/s}}^3$
• C. $2{\text{m/s}}^2$
• D. $1{\text{m/s}}^2$

Answer 1

The correct option is B $1{\text{m/s}}^3$

Tangential acceleration is the rate of change of speed along circular path.
$a_t=\frac{dv}{dt}$ ...(i)
Given:-
$a_t=\alpha t{\text{m/s}}^2$
$\Rightarrow \frac{dv}{dt}=\alpha t$
Integrating both sides:
${\int }_{v=0}^{v=v}dv=\alpha {\int }_{t=0}^{t=2}tdt$
$\Rightarrow v=\alpha {\left[\frac{t^2}{2}\right]}_0^2=2\alpha$
$\therefore$ Speed at $t=2\text{s}$ is $v=2\alpha$

At $t=2\text{s}$, the acceleration vectors can be represented as:


From vector triangle at $t=2\text{s}$ :
$\tan {45}^{\circ }=\frac{a_t}{a_c}$
$\therefore a_t=a_c$
$\Rightarrow \left(\alpha t\right)=\frac{v^2}{R}\Rightarrow \left(2\alpha \right)=\frac{(2\alpha {)}^2}{2}$
$\therefore \alpha =1{\text{m/s}}^3$