Question

You spin a globe at $2.5\,rads/sec$ and then give it a push to speed it up to $3\,rads/ sec$. If it takes 0.2 secs to change the speed of the globe, what is the angular acceleration in \[rad{\text{ }}se{c^{ - 1}}\].

Answer 1

Hint:In order to solve this problem we will use the second equation of angular motion of uniform angular acceleration. On substituting the initial and final angular velocity with time we will get the required angular acceleration of the globe.

Formula used:
\[\omega = {\omega _o} + \alpha t\]
Here, \[\omega \]= final angular velocity, \[{\omega _o}\]= initial angular velocity and \[\alpha \]= angular acceleration.

Complete step by step answer:
Angular velocity may be divided into two categories. The pace at which a point object circles around a given origin, or the temporal rate at which its angular location changes relative to the origin, is referred to as orbital angular velocity. In contrast to orbital angular velocity, spin angular velocity relates to how quickly a rigid body rotates with regard to its centre of rotation and is independent of the choice of origin.

In general, angular velocity has an angle per unit time dimension (angle replacing distance from linear velocity with time in common). Because the radian is a dimensionless quantity, the SI unit of angular velocity is radians per second (\[{s^{ - 1}}\]). The symbol omega ($\Omega $, sometimes ) is used to denote angular velocity. Positive angular velocity denotes counter-clockwise rotation, whereas negative denotes clockwise rotation.

Now, \[\omega = {\omega _o} + \alpha t\]
\[\Rightarrow \omega \]= 3
\[\Rightarrow {\omega _o}\]= 2.5
$\Rightarrow t = 0.2$
Now \[\omega = {\omega _o} + \alpha t\]
\[3 = 2.5 + \alpha (0.2)\]
\[\Rightarrow {{\alpha = }}\dfrac{{{\text{3 - 2}}{\text{.5}}{\text{ }}}}{{{\text{0}}{\text{.2}}}}{\text{ = 2}}{\text{.5 rad}}{\text{ }}{{\text{s}}^{{\text{ - 2}}}}\]
\[ \therefore {{\alpha = 2}}{\text{.5 rad}}{\text{ }}{{\text{s}}^{{\text{ - 2}}}}\]

Hence, the angular acceleration is $2.5\,rad/s^2$.

Note:This equation is only valid in an inertial frame. When we move to a non inertial frame then we cannot apply this formula because acceleration will not be constant. It will vary and hence we will get different acceleration values at different points of time. One should also take care of unit conversion while solving these types of problems.