Question

The propeller of a World War II fighter plane is 2.30 m in diameter. What is its angular velocity in radians per second if it spins at 1200 rev/min? What is the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac? 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.

Answer 1

Answer:

- the angular velocity in radians per second is 125.7 rad/s

- the centripetal acceleration of the propeller tip under these conditions is 144.5 m/s

- the centripetal acceleration of the propeller tip under these conditions in multiplied of g is 1853 g

Explanation:

given information:

diameter. d = 2.30 --> r = d/2 = 2.3/2 = 1.15 m

angular velocity, ω = 1200 rev/min

- angular velocity in radians per second

ω = 1200 rev/min

   = 1200 $\frac{2\pi }{60}$

   = 125.7 rad/s

-  the linear speed of its tip at this angular velocity

v = ω r

  = 125.7 rad/s x 1.15 m

  = 144.5 m/s

- the centripetal acceleration

a = ω^2 r, ω = $\frac{v}{r}$

  = $\frac{v^{2} }{r^{2} } r$

  = $\frac{v^{2} }{r}$

  = $\frac{144.5^{2} }{1.15}$

  = 18160 $m/s^{2}$

covert to multiplei of g, g= 9.8 $m/s^{2}$

a = 18160 /9.8 = 1853 g

Answer 2

Answer:

a) 125.68rad/s b) 144.53m/s c) 18164.78m/s^2 d) 1851.66g

Explanation:

Angular velocity in radian = 1200rev/s* (2pi rad/1rev)* (1/60s) since 1 minute = 60 seconds

Angular velocity in radian = 1200*2*3.142/60 = 125.68 radian/seconds

b) linear velocity = r (radius) * angular velocity = (2.30/2)* 125.68 = 144.53m/s

Centripetal acceleration (acceleration towards the center of a circulatory body) = V^2/r = 144.53*144.53/1.15 = 18164.78m/s^s

In multiples of g (acceleration due to gravity) = 18164.78/ 9.81 = 1,851.66g