Question

) The angular acceleration of the disk is defined by ???? = 3t 2 + 12 rad/s, where t is in seconds. If the disk is originally rotating at ????0 = 12 rad/s. a) Determine the angular motion when t = 2 s. b) Determine the magnitude of the velocity of point A on the disk when t = 2 s. c) Determine the normal and tangential components of acceleration of point A on the disk when t = 2 s.

Answer 1

Answer:

a) $\theta = 52\,rad$, b) $v_{A} = 44\cdot r \,\frac{m}{s}$, c) $a_{A,n} = 1936\cdot r\,\frac{m}{s^{2}}$, $a_{A,t} = 24\cdot r\,\frac{m}{s^{2}}$

Explanation:

a) The angular motion is obtained by integrating the angular acceleration function twice:

$\alpha = 3\cdot t^{2} + 12$

$\omega = t^{3} + 12\cdot t + 12$

$\theta = \frac{1}{4}\cdot t^{4} + 6\cdot t^{2} + 12\cdot t$

The angular motion when t = 2 s. is:

$\theta = \frac{1}{4}\cdot (2\,s)^{4} + 6\cdot (2\,s)^{2} + 12\cdot (2\,s)$

$\theta = 52\,rad$

b) Let be $r$ the distance between A and the rotation axis, measured in meters. The magnitude of the angular velocity when t = 2 s. is:

$\omega = (2\,s)^{3} + 12\cdot (2\,s) + 12$

$\omega = 44\,\frac{rad}{s}$

Finally, the magnitude of the velocity is:

$v_{A} = 44\cdot r \,\frac{m}{s}$

c) The angular acceleration of the disk when t = 2 s. is:

$\alpha = 3\cdot (2\,s)^{2} + 12$

$\alpha = 24\,\frac{rad}{s^{2}}$

Lastly, the normal and tangential components at point A are, respectively:

$a_{A,n} = \omega^{2}\cdot r$

$a_{A,n} = \left(44\right)^{2}\cdot r$

$a_{A,n} = 1936\cdot r\,\frac{m}{s^{2}}$

$a_{A,t} = \alpha \cdot r$

$a_{A,t} = 24\cdot r\,\frac{m}{s^{2}}$