Question

When a fan is turned off, its angular speed decreases from 10 rad/s to 6.3 rad/s in 5.0 s. What is the magnitude of the average angular acceleration of the fan

Answer 1

Answer:

$a=-0.74rad/sec^2$

Explanation:

We have final angular speed of the fan $\omega _f=6.3rad/sec$

Initial angular velocity of the fan $\omega _i=10rad/sec$

The angular speed of the is changes from 410 rad/sec to 6.3 rad/sec in 5 sec

So time taken t = 5 sec

From first equation of motion we know that $\omega _f=\omega _i+\alpha t$ putting the values

$6.3=10+\alpha \times 5$

$a=-0.74rad/sec^2$ negative sign of the acceleration shows that speed of the fan is decreasing in nature which is given in the question  

Answer 2

The magnitude of the average angular acceleration will be "-0.74 rad/sec²".

Given:

Final angular speed,

• $\omega_f = 6.3 \ rad/sec$

Initial angular speed,

• $\omega_i = 10 \ rad/sec$

Time,

• t = 5 s

By using the equation of motion, we get

→ $\omega_f = \omega_i +\alpha t$

By substituting the values, we get

→ $6.3=10+\alpha\times 5$

→ $6.3-10 = \alpha\times 5$

→      $-3.7 = \alpha 5$

            $\alpha = - \frac{3.7}{5}$

                $= - 0.74 \ rad/sec^2$

Thus the above response is right.

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