Question

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude of the angular acceleration is 0.895 rev/s^2 . Both the the angular velocity and angular accleration are directed counterclockwise. The electric ceiling fan blades form a circle of diameter 0.800 m . question1- What is the tangential speed vt of a point on the tip of the blade at time t = 0.200 s ?
question 2- Calculate the magnitude at of the tangential acceleration of a point on the tip of the blade at time t = 0.200 s .
question 3- Calculate the magnitude ar of the radial (or centripetal) acceleration of the point at the end of the fan blade.

Answer 1

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

$\omega_f = \omega_i + \alpha t$

where

$\omega_i = 0.300 rev/s$ is the initial angular velocity

$\alpha = 0.895 rev/s^2$ is the angular acceleration

Substituting t = 0.200 s, we find

$\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s$

Let's now convert it into rad/s:

$\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s$

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

$r=\frac{0.800}{2}=0.400 m$

And so now we can find the tangential speed at t = 0.200 s:

$v=\omega_f r =(3.01)(0.400)=1.2 m/s$

2) $2.25 m/s^2$

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

$a_t = \alpha r$

where $\alpha$ is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

$\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2$

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

$a_t = (5.62)(0.400)=2.25 m/s^2$

3) $3.6 m/s^2$

The centripetal acceleration is given by

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

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

$a=\frac{1.2^2}{0.400}=3.6 m/s^2$

Answer 2

1) The tangential speed of a point on the tip of the blade  = 1.2 m/s

2) The magnitude of the tangential acceleration of a point on the tip of the blade = 2.25 m/s²

3) The magnitude of the radial acceleration of the point at the end of the fan blade = 3.6 m/s²

A) Determine the Tangential speed at a point on the tip of the blade

first step : determine the angular velocity

wf = wi + at  --- ( 1 )

where : wi = 0.300 rev/sec,  a = 0.895 rev/s², t = 0.200 sec

back to equation ( 1 )

wf  = 0.479 rev/s  = 3.01 rad/sec

next step : determine the tangential speed

V = wf * r  --- ( 2)

where : r = 0.800 / 2 = 0.4 m , wf = 3.01 rad/sec

insert values into equation ( 2 )

V = 3.01 * 0.4 = 1.2 m/s

B)  Determine the magnitude of the tangential acceleration of a point on the tip of the blade

Applying the equation below

Tangential acceleration ( at ) = ∝r  ---- ( 3 )

where : ∝ = angular acceleration = 0.895 rev/s² = 5.62 rad/s² , r = 0.4 m

back to equation ( 3 )

Tangential acceleration = 5.62 * 0.4 = 2.25 m/s²

C) Determine the magnitude of the radial acceleration of the point at the end of the fan blade

Given that centripetal acceleration is given as

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

where : v ( tangential speed ) = 1.2 m/s , r = 0.4 m

therefore

a( radial acceleration ) = 1.2² / 0.4

                                     = 3.6 m/s²

Hence we can conclude that the answers to your questions are as listed above

Learn more about Tangential acceleration : https://brainly.com/question/14368661