Question

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.250rev/s . The magnitude of the angular acceleration is 0.891rev/s2 . Both the the angular velocity and angular accleration are directed clockwise. The electric ceiling fan blades form a circle of diameter 0.800m .
1. Compute the fan's angular velocity magnitude after time 0.194s has passed. (rev/s)
2. Through how many revolutions has the blade turned in the time interval 0.194s from Part A? (rev)

3. What is the tangential speed vtan(t) of a point on the tip of the blade at time t = 0.194s ? (m/s)

4. What is the magnitude a of the resultant acceleration of a point on the tip of the blade at time t = 0.194s ? (m/s^2)

Answer 1

Answer:

1

The fan's angular velocity $w_f =0.43\ rev /s$

2

The number of turns is $\theta = 0.0652\ rev$

3

  The tangential speed $v_t = 1.06 m/s$

4

The  magnitude a of the resultant acceleration $a_r = 3.6 m/s^2$

Explanation:

From the question we are given that

         The initial angular velocity is $w_{in} = 0.250 \ rev/s$

          The angular acceleration is $\alpha = 0.891 \ rev/s^2$

          The diameter of the ceiling fan blades $d=0.800\ m$

The diagram for this motion is shown on the first uploaded image

The motion involved in this problem statement is rotational motion and this motion can be defined in terms of angular velocity , angular displacement and angular acceleration

           

       Also the angular velocity defines the speed with which a body moves around a fixed axis and this mathematically defined as

             $w =\frac{\Delta \theta}{\Delta t}$

          $w_f = w_{in} + \alpha t$

where  $w_f$ is final angular velocity and

     => $w_f = 0.25 +(0.891)(0.194)$

               $=0.43\ rev /s$

 Now the angular displacement$(\theta)$ is the angle made by a body moving in a circular path and it show us the relationship between the distance moved around the circular path and the distance move at the center of the circular path in terms of angle and this is measured in radians

 The angular displacement can also be expressed as

                 $\theta = w_{in} t + \frac{1}{2} \alpha t^2$

                    $= (0.250)(0.194)+\frac{1}{2}(0.891)(0.194)^2$

                $\theta = 0.0652\ rev$

                       

Angular acceleration is the change of angular velocity with time and this is mathematically defined as

             $\alpha =\frac{\Delta w}{\Delta t}$

The tangential velocity is the velocity that is been measured at an point tangent to the circular path of motion and this mathematically represented as

             $v_t = w_fr = (\frac{d}{2} )w_f$

                            $= \frac{0.800}{2} (0.423)$

                           $= [\frac{0.800}{2} ] (0.423)[\frac{2 \pi \ rad}{1.0 rev} ]$

                          $= 1.06 m/s$

In this motion described in the problem statement we would have two acceleration component which are centripetal acceleration$(a_c)$ and tangential acceleration $(a_t)$

The centripetal acceleration is defined as the rate of change of the tangential velocity and this is mathematically represented as

                 $a_c = \frac{v^2}{r}$

                     $= \frac{(1.063 m/s)^2}{{\frac{0.800m}{2} }}$

                     $=2.83 \ m/s^2$

this is always directed toward the center of the circular path

The tangential acceleration can be defined as the linear acceleration of the body at any point on the circular path of motion and it is mathematically

represented as

                   $a_t = \alpha r$

                       $= [\frac{0.800m}{2} ](0.891 rev /s^2 )$

                       $= [\frac{0.800m}{2} ](0.891 rev/s^2)[\frac{2\pi\ rad}{1.0 rev} ]$

                  $a_t = 2.24 m/s^2$

This directed tangential to the circular path  

Note: As shown in the diagram the two component of the acceleration are always perpendicular to each other.

Between this two component of acceleration is the resultant acceleration and it is mathematically represented as

                $a_r = \sqrt{(a_c^2 + a_t^2)}$

                $a_r = \sqrt{(2.83m/s^2)^2 + (2.24m/s^2)^2 }$

                    $= 3.6 m/s^2$