Question

3. The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if it spins at 1200 rev/min? (b) What is the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac? (c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of g . 14. An ordinary workshop grindstone has a

Answer 1

Answer:

(a) Angular velocity will be 125.6 rad/sec

(b) Linear velocity will be 144.44 m /sec

(c) Centripetal acceleration = 1849.3031 g

Explanation:

We have given diameter d = 2.30 m

So radius r = $\frac{d}{2}=\frac{2.30}{2}=1.15m$

(a) Speed is given as 1200 rev/min

We know that angular velocity is given by $\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 1200}{60}=125.6rad/sec$

(b) Linear speed is given by $v=\omega r=125.6\times 1.15=144.44m/sec$

(c) Centripetal acceleration is given by $a_c=\frac{v^2}{r}=\frac{144.44^2}{1.15}=18141.664m/sec^2$

We know that $g=9.81m/sec^2$

So $18141.66m/sec^2=\frac{18141.664}{9.81}=1849.3031g$