Question 2.SP.35: The 100-mm-diameter pulley on a generator is being turned by......

In Fig. 2-25 point A on the pulley has the same velocity as the belt with which it coincides at the instant. Hence, the angular velocity \omega of the pulley (and also of the fan keyed to the same shaft) is

\omega=\frac{v}{r}=\frac{20}{0.05}=400  \mathrm{rad} / \mathrm{s}

The linear velocity of the fan tip is

v_{B}=(0.075)(400)=30 \mathrm{~m} / \mathrm{s}

The tangential component of the linear acceleration of point A is equal to the acceleration of the belt; that is,

a_{t}=r \alpha \quad \text { or } \quad 6=0.05  \alpha

From this, the angular acceleration \alpha of the system is 120  \mathrm{rad} / \mathrm{s}^{2}. Then the tangential component of the acceleration of point B is

a_{t}=0.075 \times 120=9 \mathrm{~m} / \mathrm{s}^{2}

It has a normal component that equals

a_{n}=0.075 \times 400^{2}=12000 \mathrm{~m} / \mathrm{s}^{2}

Hence, the magnitude of the linear acceleration is

a=\sqrt{(12000)^{2}+(9)^{2}}=12000 \mathrm{~m} / \mathrm{s}^{2}