# Two unique concepts in Turning Moment Diagram

I was discussing the doubt of one of my students when I realized that these concepts were not explicitly mentioned and explained in any of the books, though few were using them without any explanation. So, I thought of explaining them here.

Concept A:

In solving the questions of Turning Moment diagram of flywheel, we do not realize that there are some equations of dynamics which we cannot use here. As we already know, these equations are:

1. T = I . alpha
2. ω = ω(0) + alpha.t
3. ω^2 – ω(0)^2 = 2. alpha . θ
4. θ = ω(0).t + ½ . alpha . t^2

where, alpha = angular acceleration.

These equations are valid only when there is a case of constant angular acceleration/deceleration and constant torque. However, in case of flywheels, both angular acceleration/deceleration and torque vary. Hence, we cannot just apply them for any two points on turning moment diagram. (Though we can apply 1st one for mean torque since that is constant)

For example, in the Turning Moment diagram below, we cannot use it between A and B since actual torque is varying between A and B and so angular acceleration is also varying with torque. Concept B:

This concept will require you to know the basics of Turning Moment diagram for flywheel. (All these concepts are covered in detail in Online Classes of Mechanical Academy)

So,
T = I . alpha is not applicable for any two selected points.
But we can modify this equation to make it usable. How? See the actual problem is that both T and alpha are varying but integrating them consider the variations in torque and reflect it on alpha.

Considering a  partial change in torque,
dT = I. d(alpha)

Integrating torque from T1 to T2 and angular acceleration from alpha(1) to alpha(2).
(T2 – T1) = I . [ alpha(2) – alpha(1) ] ————–(i)

Now, in the image below, A is the point of minimum energy and B is the point of maximum energy. Hence, angular velocity is minimum at A and maximum at B. Variation of angular velocity can be seen in the graph below. So what will be angular acceleration alpha at 1? It will be zero. You can see from the curve of angular speed that slope of the curve is zero at 1.

Slope dω/dt = alpha. Hence, alpha = 0.

Putting (ii) in (i)

(T2 – T1) = I . alpha(2)
or,

∆T = I . alpha

You can practice this concept on the question below. This question is not for those who do not know the basics of Turning moment diagram for flywheel.

The turning moment diagram of an engine is shown below. The speed of the wheel varies between 300 rpm and 360 rpm. What is the crank angle measured from TDC, when the angular acceleration of the wheel is 92.15 rad/s^2?

(Ans: 30°) ### EXERGIC

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