Question
Question: If motor revolving at \(1200\;{\text{rpm}}\) slows down uniformly \(900\;{\text{rpm}}\) in \(2\;\sec...
If motor revolving at 1200rpm slows down uniformly 900rpm in 2sec calculate the angle axis rotation of the motor and the number of revolution it makes during this time.
Solution
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 at1200rpmslows down uniformly 900rpmin2seccalculate 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
Fnet=Macm
Where, Mis the total mass of the system and acmis the acceleration of the center of mass.
The initial angular velocity is given as,
ω1=1200rpm
Now we convert the velocity from revolution per minute into radian per second.
ω1=40πrad/s
We have given the final angular velocity as,
ω2=900rpm
Now we convert the velocity from revolution per minute into radian per second.
ω2=30πrad/s
As we know that the angular acceleration is the rate of change of the angular speed so we can write,
⇒α=240π−30π
Now we solve the above expression.
⇒α=5πrad/s2
As we know that the angular displacement is given as,
⇒θ=ω1t+21αt2
Now we substitute the values in the above expression.
⇒θ=40π×2+21×5π(2)2
Now we solve the above expression.
∴θ=90πrad
As we know that the numbers of revolutions are given as,
⇒n=2π90π
Now solve the above expression and we get
∴n=45rev
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.