Question

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16sec.? Find
(i) Angular acceleration
(ii) Assuming acceleration to be uniform, how many rotations does the engine make on this bike?
(iii) State law of conservation of linear momentum?

Answer 1

Hint : For solving such questions, we need to have an understanding of what angular momentum and angular acceleration is. Angular momentum is the rotational equivalent of linear momentum and Angular acceleration is equivalent to linear acceleration. The formula which will be used here is $\omega = {\omega _o} + \alpha t$ .

Complete Step By Step Answer:
(i) It is the rate of change of angular velocity with the time of an object in motion. Acceleration is the change in velocity of a moving object with respect to time. If the object moves in a circular direction then its velocity is called angular velocity. The vector direction of the acceleration is perpendicular to the plane where the rotation takes place. Increase in angular velocity clockwise, then the angular acceleration velocity points away from the observer. If the increase in angular velocity is counter clockwise, then the vector of angular acceleration points toward the viewer. The SI unit of angular acceleration is $rad/{s^2}$ .
Here we will be using the equation, $\omega = {\omega _o} + \alpha t$ , where ${\omega _o}$ = initial angular speed in $rad/{s^2}$ .
${\omega _o} = 2\pi \times$ angular speed in rev/s $$
= $\dfrac{{2\pi \times 1200}}{{60}}rad/s$
= $40\pi rad/s$
Similarly we will find $\omega$ = final angular speed in $rad/{s^2}$ .
$\omega = \dfrac{{2\pi \times 3140}}{{60}}rad/s$
$= 104\pi rad/s$ $$
Now, we know that angular acceleration,
$\alpha = \dfrac{{\omega - {\omega _o}}}{t} = \dfrac{{104 - 40}}{{16}}$
$\alpha = 4\pi rad/s$
Thus, the angular acceleration of the engine = $4\pi rad/s$
(ii) The angular displacement in time t is given by
$\theta = {\omega _o}t + \dfrac{1}{2}\alpha {t^2}$
So now we will put all the values in the above equation,
= $(40\pi \times 16 + \dfrac{1}{2} \times 4\pi \times {16^2})rad$
= $(640\pi + 512\pi )rad$
= $1152\pi rad$
So, the number of rotations = $\dfrac{{1152\pi }}{{2\pi }}$ = $576$
(iii) Conservation of linear momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant. Momentum is equal to the mass of an object multiplied by its velocity and is equivalent to the force required to bring the object to a stop in a unit length of time. For any array of several objects, the total momentum is the sum of the individual momenta. There is a peculiarity, however, in that momentum is a vector, involving both the direction and the magnitude of motion, so that the momenta of objects going in opposite directions can cancel to yield an overall sum of zero.
Before launch, the total momentum of a rocket and its fuel is zero. During launch, the downward momentum of the expanding exhaust gases just equals in magnitude the upward momentum of the rising rocket, so that the total momentum of the system remains constant—in this case, at zero value. In a collision of two particles, the sum of the two moments before collision is equal to their sum after collision. What momentum one particle loses, the other gains.

Note :
Remember that if the object moves in a circular direction then its velocity is called angular velocity. Momentum is equal to the mass of an object multiplied by its velocity and is equivalent to the force required to bring the object to a stop in a unit length of time. Calculation mistakes are possible, so try to avoid them and be sure of the final answer.