Question

A particle P is moving on a circular path of radius r = 1 m. The angular location $\theta$ of the particle is measured as shown in the Figure. The motion of the particle is described by $\theta =2$ sin(t). Find the magnitude of the total acceleration (in m/s${}^2$) of the particle at time $t=\pi /3$ seconds______.

• A. 2

Answer 1

The correct option is A 2
$\theta =2sint$

$\omega =\frac{d\theta }{dt}=2cost$

$\propto =\frac{d^2\theta }{d{t}^2}=-2sint$

$a_t=r\alpha =r\times -2\times sint$

$=1\times -2sin\left(\frac{\pi }{3}\right)=-\sqrt{3}$

$a_a=r{\omega }^2=r\times 4co{s}^{2}t=t\times 4\times {\left(\frac{1}{2}\right)}^2=1$

$a=\sqrt{a_t^2+a_n^2}=\sqrt{(-\sqrt{3}{)}^2+1}=2m/s^2$