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Sidana [21]
3 years ago
6

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.
Physics
1 answer:
Salsk061 [2.6K]3 years ago
7 0

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

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