Question

The blades of a ceiling fan have a radius of 0.300 m and are rotating about a fixed axis with an angular velocity of +1.78 rad/s. When the switch on the fan is turned to a higher speed, the blades acquire an angular acceleration of +2.12 rad/s2. After 0.655 s have elapsed since the switch was reset, what is (a) the total acceleration (in m/s2) of a point on the tip of a blade and (b) the angle between the total acceleration and the centripetal acceleration (See Figure b)?

Answer 1

(a)
`a=3.23 m/s^(2)`

(b)
`11.92^(\circ)`

Step-by-step explanation:

Tangential acceleration,
`a_T` is given by

`a_T=r\alpha` where r is fan radius and
`\apha` is angular acceleration

`a_T=0.3*2.12=0.636 m/s^(2)`

Radial acceleration,
`a_R` is given by

`a_R=r\omega_f^(2)=r(\omega_i +\alpha t)^(2)` and substituting the values in the question

`a_R=0.3*(1.78+(2.12*0.655))^(2)=0.3(1.78+1.3886)^(2)=0.3*3.1686^(2)=3.012008`

Total acceleration,
`a=\sqrt {a_R^(2)+a_T^(2)}=\sqrt {3.012008^(2)+0.636^(2)}= 3.231799`

`a=3.23 m/s^(2)`

Angle is given by

`\theta=tan^(-1)(\frac {a_T}{a_r})=tan^(-1)(\frac {0.636}{3.012008})=11.92^(o)`