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
[tex]\alpha = \dfrac{d\omega}{dt}[/tex]
[tex]\alpha = \dfrac{d}{dt}( 5 t² + 2)[/tex]
[tex]\alpha =10 t[/tex]
[tex]\alpha =10\times 0.5[/tex]
α = 5 rad/s²
velocity
v = ω r
v = 3.25 x 0.8
v = 2.6 m/s
tangential acceleration
[tex]a_t = \alpha r[/tex]
[tex]a_t =5 \times 0.8[/tex]
[tex]a_t =4\ m/s^2[/tex]
normal acceleration
[tex]a_n = \omega^2\ r[/tex]
[tex]a_n = 3.25^2\times 0.8[/tex]
[tex]a_n =8.45 \m/s^2[/tex]
[tex]a = \sqrt{a_n^2 + a_t^2}[/tex]
magnitude of the acceleration
[tex]a = \sqrt{8.45^2 + 4^2}[/tex]
a = 9.35 m/s²
