Question

The blades of a fan running at low speed turn at 26.2 rad/s. When the fan is switched to high speed, the rotation rate increases uniformly to 36.5 rad/s in 5.75 seconds. What is the magnitude of the fan's angular acceleration

Answer 1

The angular acceleration is given by 1.8 rad/s^2.

Step-by-step explanation:

initial angular speed, wo = 26.2 rad/s

final angular velocity, w = 36.5 rad/s

time, t = 5.75 seconds

The first equation of motion is

`w = wo + \alpha t\\\\36.5 = 26.2 + 5.75\alpha\\\\\alpha = 1.8 rad/s^2`

Answer 2

Answer:
`1.79\ rad/s^2`

Step-by-step explanation:

Given

Initial angular speed is
`\omega_1=26.2\ rad/s`

Final angular speed is
`\omega_2=36.5\ rad/s`

Time period
`t=5.75\ s`

Magnitude of the fan's acceleration is given by

`\Rightarrow \alpha=(\omega_2-\omega_1)/(t)`

Insert the values

`\Rightarrow \alpha=(36.5-26.2)/(5.75)\\\\\Rightarrow \alpha=(10.3)/(5.75)\\\\\Rightarrow \alpha=1.79\ rad/s^2`

Thus, fan angular acceleration is
`1.79\ rad/s^2`