Question

If motor revolving at $1200\;{\text{rpm}}$ slows down uniformly $900\;{\text{rpm}}$ in $2\;\sec $ calculate the angle axis rotation of the motor and the number of revolution it makes during this time.

Answer 1

Hint: In Kinematics, we studied motion along a straight line and introduced such concepts as displacement, velocity, and acceleration. Two-Dimensional Kinematics dealt with motion in two dimensions. Projectile motion is a special case of two-dimensional kinematics in which the object is projected into the air, while being subject to the gravitational force, and lands a distance away. In this chapter, we consider situations where the object does not land but moves in a curve. We begin the study of uniform circular motion by defining two angular quantities needed to describe rotational motion.

Complete Step by Step Solution:
In this question, if motor revolving at$1200\;{\text{rpm}}$slows down uniformly $900\;{\text{rpm}}$in$2\;\sec $calculate the angle axis rotation of the motor and the number of revolution it makes during this time.
A change in the position of a particle in three-dimensional space can be completely specified by three coordinates. A change in the position of a rigid body is more complicated to describe. It can be regarded as a combination of two distinct types of motion: translational motion and circular motion.
Acceleration of the center of mass is given by
${F_{net}} = M{a_{cm}}$
Where, $M$is the total mass of the system and ${a_{cm}}$is the acceleration of the center of mass.
The initial angular velocity is given as,
\[{\omega _1} = {\text{12}}00\;{\text{rpm}}\]
Now we convert the velocity from revolution per minute into radian per second.
\[{\omega _1} = {\text{4}}0\pi \;{\text{rad}}/{\text{s}}\]
We have given the final angular velocity as,
\[{\omega _2} = {\text{9}}00\;{\text{rpm}}\]
Now we convert the velocity from revolution per minute into radian per second.
\[{\omega _2} = {\text{3}}0\pi \;{\text{rad}}/s\]

As we know that the angular acceleration is the rate of change of the angular speed so we can write,
\[ \Rightarrow \alpha = \dfrac{{{\text{4}}0\pi - {\text{3}}0\pi }}{2}\]
Now we solve the above expression.
\[ \Rightarrow \alpha = {\text{5}}\pi \;{\text{rad}}/{{\text{s}}^2}\]
As we know that the angular displacement is given as,
\[ \Rightarrow \theta = {\omega _1}t + \dfrac{1}{2}\alpha {t^2}\]
Now we substitute the values in the above expression.
\[ \Rightarrow \theta = {\text{4}}0\pi \times {\text{2}} + \dfrac{1}{2} \times {\text{5}}\pi {({\text{2}})^2}\]
Now we solve the above expression.
\[\therefore \theta = {\text{9}}0\pi \;{\text{rad}}\]
As we know that the numbers of revolutions are given as,
\[ \Rightarrow n = \dfrac{{90\pi }}{{2\pi }}\]
Now solve the above expression and we get
\[\therefore n = {\text{45}}\;{\text{rev}}\]

Note: Any displacement of a rigid body may be arrived at by first subjecting the body to a displacement followed by a rotation, or conversely, to a rotation followed by a displacement. We already know that for any collection of particles whether at rest with respect to one another, as in a rigid body, or in relative motion.