#### Answer

(a) The tangential acceleration is $-2.6~m/s^2$
(b) The crankshaft rotates through 31 revolutions as it stops.

#### Work Step by Step

(a) We can convert 2500 rpm to units of rad/s
$\omega = (2500~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$
$\omega = 261.8~rad/s$
We can find the angular acceleration.
$\alpha = \frac{\omega-\omega_0}{t}$
$\alpha = \frac{0-261.8~rad/s}{1.5~s}$
$\alpha = -174.5~rad/s^2$
We can find the tangential acceleration of a point on the surface.
$a_t = \alpha ~r$
$a_t = (-174.5~rad/s^2)(0.015~m)$
$a_t = -2.6~m/s^2$
The tangential acceleration is $-2.6~m/s^2$
(b) $\theta = \omega_0~t+\frac{1}{2}\alpha~t^2$
$\theta = (261.8~rad/s)(1.5~s)+\frac{1}{2}(-174.5~rad/s^2)(1.5~s)^2$
$\theta = 196.4~rad$
We can convert the angle from radians to revolutions.
$\theta = (196.4~rad)(\frac{1~rev}{2\pi~rad})$
$\theta = 31~rev$
The crankshaft rotates through 31 revolutions as it stops.