Question

A hoop is placed on the rough surface such that it has an angular velocity $\omega =4$ rad/s and an angular deceleration $\alpha =5rad/s^2$ ,Also, its centre has a velocity of $V_0=5m/s$ and a deceleration $a_0=2m/s^2$. Determine the magnitude of acceleration of point $B$ at this instant.

Answer 1

${\overrightarrow{a}}_B={\overrightarrow{a}}_0+{\overrightarrow{a}}_{B/0}$
Here,${\overrightarrow{a}}_{B/0}$ has two components $a_t$ (tangential acceleration) and $a_n$ (normal acceleration)
$a_t=r\alpha =\left(0.3\right)\left(5\right)=1.5m/s^2$
$a_n=r{\omega }^2=\left(0.3\right)(4{)}^2=4.8m/s^2$
and $a_0=2m/s^2$
$\therefore a_B=\sqrt{(\sum a_x{)}^2+(\sum a_y{)}^2}$
$=\sqrt{(2+4.8\cos {45}^{\circ }-1.5\cos {45}^{\circ }{)}^2+(4.8\sin {45}^{\circ }+1.5\sin {45}^{\circ }{)}^2}$
$=6.21m/s^2$