Question

How long does it take for a rotating object to speed up from 15.0 sto 33.3 rad/s if it has an angular acceleration of 3.45rad/s2?

a) 9.57 s
b) 5.30 s
c) 63.1 s
d) 4.35 s

Answer 1

Answer:

The time taken by the rotating object to speed up from 15.0 s to 33.3 rad/s is 5.30 seconds.        

Explanation:

It is given that,

Initial angular speed of the object, $\omega_o=15\ rad/s$

Final angular speed of the object, $\omega_f=33.3\ rad/s$

Angular acceleration of the object, $\alpha =3.45\ rad/s^2$

Angular acceleration of an object is object is defined as the change in angular velocity per unit time. It is given by :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$t =\dfrac{\omega_f-\omega_i}{\alpha}$

$t =\dfrac{33.3-15}{3.45}$

t = 5.30 seconds

So, the time taken by the rotating object to speed up from 15.0 s to 33.3 rad/s is 5.30 seconds. Hence, this is the required solution.

Answer 2

The time taken will be "5.30 seconds".

According to the question,

Initial angular speed,

• $\omega_i = 15 \ rad/s$

Final angular speed,

• $\omega_f = 33.3 \ rad/s$

Angular acceleration,

• $\alpha = 3.45 \ rad/s^2$

As we know,

→ $\alpha = \frac{\omega_f - \omega_i}{t}$

or,

→ $t = \frac{\omega_f -\omega_i}{\alpha}$

By substituting the values, we get

     $= \frac{33.3-15}{3.45}$

     $= \frac{18.3}{3.45}$

     $= 5.30 \ seconds$

Thus the above answer i.e., "option b" is right.

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