Question

ne angular velocity of the disk is w= (5t+ 2) rad/s where t is in seconds. termine the magnitude of the normal component of acceleration at point A (in ) men t = 0.5 seconds. A 0.8 m

Answer 1

The angular velocity at the given time is;

$\omega = 5t^2 + 2$

$\Rightarrow \omega = 5(0.5)^2 + 2$

$\Rightarrow \omega =3.25\;\;rad/s$

NORMAL COMPONENT of acceleration is given as;

$(a_A)_n = \omega^2 r_A$

$\Rightarrow (a_A)_n = (3.25)^2 (0.8)$

$\Rightarrow (a_A)_n =8.45\;\;m/s^2$

$\Rightarrow (a_A)_n =\mathbf{8.45}$

...(Answer)