Question

The angular acceleration of the disk is defined by a = 3t 2 + 12 rad>s, where t is in seconds. If the disk is originally rotating at v0 = 12 rad>s, determine the magnitude of the velocity and the n and t components of acceleration of point A on the disk when t = 2 s.

Answer 1

Answer with Explanation:

We are given that

$\alpha=3t^2+12 rad/s^2$

$\omega_0=12 rad/s$

$\omega=\int \alpha dt$

$\omega=\int(3t^2+12)dt=t^3+12t+C$

When t=0 and $\omega_0=12 rad/s$

$12=C$

$\omega=t^3+12t+12$

Substitute t=2

$\omega=(2)^3+12(2)+12=8+24+12=44 rad/s$

$r_A=0.5 m$

$v=\omega r_A=44\times 0.5=22 m/s$

$a_N=\omega^2r_A=(44)^2(0.5)=968m/s^2$

$\alpha=3(2)^2+12=12+12=24 rad/s^2$

$a_t=\alpha r_A=24\times 0.5=12m/s^2$