Question

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 4.40 rev/s; 60.0 revolutions later, its angular speed is 15.0 rev/s. Calculate (a) the angular acceleration (rev/s2), (b) the time required to complete the 60.0 revolutions, (c) the time required to reach the 4.40 rev/s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 4.40 rev/s angular speed.

Answer 1

Answer:

The answer is below

Explanation:

a) Using the formula:

$\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=15\ rev/s,w_o=initial\ velocity=4.4\ rev/s, \\\theta=distance=60\ rev\\\\Substituting:\\\\15^2=4.4^2+2(60)\alpha\\\\2(60)\alpha=15^2-4.4^2\\\\2(60)\alpha=205.64\\\\\alpha=1.71\ rev/s^2$

b) The disk is initially at rest. Using the formula:

$\theta=\omega_ot+\frac{1}{2}\alpha t^2 \\\\but\ \omega_o=0(rest), \theta=60\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\60=0+\frac{1}{2}(1.71) t^2\\\\t=8.4\ s$

c)

$\omega=\omega_o+\alpha t \\\\but\ \omega_o=0(rest), \omega=4.4 \ rev/s\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\4.4=1.71t\\\\t=2.6\ s$

d)

$\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=4.4\ rev/s,w_o=initial\ velocity=0\ rev/s, \\\alpha=1.71\ rev/s^2\\\\Substituting:\\\\4.4^2=0+2(1.71)\theta\\\\19.36=3.42\theta\\\\\theta=5.66\ rev$