Question

A rotating fan completes 1130 revolutions every minute. Consider the tip of a blade, at a radius of 19.0 cm. (a) Through what distance does the tip move in one revolution? What are (b) the tip's speed and (c) the magnitude of its acceleration? (d) What is the period of the motion?

Answer 1

a)1.1932 m

b)22.47 m/s

c)2657.8\ m/s^22657.8 m/s

2

d)0.053 s

Answer 2

a)1.1932 m

b)22.47 m/s

c)
`2657.8\ m/s^2`

d)0.053 s

Step-by-step explanation:

Given:

• frequency of the fan , f=1130 rev/min
• Radius of the tip, R=0.19 m

a) Distance moved by tip in on revolution ,

`d=2\pi R\\\\=2*3.14*0.19\\=1.1932\ m`

b)Angular velocity of the the fan is given by
`\omega\\`

`\omega=2\pi f\\=2\pi * (1130)/(60)\\=118.2\ \rm m/s`

Velocity of the tip is given by,

`v=\omega r\\=118.2*0.19\\=22.47\ rad/s`

c)The acceleration is given by,

`a=(v^2)/(r)\\a=(22.47^2)/(0.19)\\a=2657.8\ \rm m/s^2`

d)Time period of the motion is given T,

`T=(2\pi)/(\omega)\\\\T=(2\pi)/(118.2)\\\\T=0.053\ s`