Question

The angular velocity of the disk is defined by ω = ( 5 t 2 + 2 ) r a d / s , where t is in seconds. Determine the magnitudes of the velocity and acceleration of point A on the disk when t = 0.5 s .

Answer 1

Answer

given,

Assume radius of the disk be = 0.8 m

angular velocity of disk

ω = ( 5 t² + 2 ) r a d / s

magnitude of velocity and acceleration = ?

At the instant of time, t = 0.5 s

ω = ( 5 x (0.5)² + 2 ) r a d / s

ω = 3.25 r a d / s

$\alpha = \dfrac{d\omega}{dt}$

$\alpha = \dfrac{d}{dt}( 5 t² + 2)$

$\alpha =10 t$

$\alpha =10\times 0.5$

α = 5 rad/s²

velocity

v = ω r

v = 3.25 x 0.8

v = 2.6 m/s

tangential acceleration

$a_t = \alpha r$

$a_t =5 \times 0.8$

$a_t =4\ m/s^2$

normal acceleration

$a_n = \omega^2\ r$

$a_n = 3.25^2\times 0.8$

$a_n =8.45 \m/s^2$

$a = \sqrt{a_n^2 + a_t^2}$

magnitude of the acceleration

$a = \sqrt{8.45^2 + 4^2}$

a = 9.35 m/s²