Mass Moment of Inertia, $I_G$

$I_G$ for a body depends on the body’s mass and the location of the mass. The greater the distance the mass is from the axis of rotation, the larger $I_G$ will be. For example, flywheels have a heavy outer flange that locates as much mass as possible at a greater distance from the hub. $I_G$ is the “mass moment of inertia” for a body about an axis passing through the body’s mass center, G.

$I_G$ is defined as: $I_G=\int r^2dm$

Units: kg-$m^2$ or slug$ft^2$ $I_G$ is used for several kinds of rigid body rotation problems, including: (a)$F=ma$ analysis moment equation ($\sum M_G=I_G{\alpha}$ ). (b) Rotational kinetic energy ($T=\frac{1}{2}I_Gw^2$ ) (c) Angular momentum ($H_G = I_{Gω}$ )

$I_G$ is the resistance of the body to angular acceleration. That is, for a given net moment or torque on a body, the larger a body’s $I_G$, the lower will be its angular acceleration, α. $I_G$ also affects a body’s angular momentum, and how a body stores kinetic energy in rotation.

Radius of Gyration, $k_G$, for Complex Shapes Some problems with a fairly complex shape, such as a drum or multi-flanged pulley, will give the body’s mass m and a radius of gyration, $k_G$, that you use to calculate $I_G$. If given these, calculate $I_G$ from: $I_G = mk_G^2$ As illustrated below, using $k_G$ in this way is effectively modeling the complex shape as a thin ring.

D’Alembert’s Principle

Consider rigid body acted upon by a system of forces. The system may be reduced to a single resultant force acting on the body whose magnitude is given by the product of the mass of the body and the linear acceleration of the center of mass of the body.

According to Newton's second law of motion. $F=m.a$........... (i) Where, F= Resultant forces acting on the body,

A little consideration will show, that if the quantity($–m.a$) be treated as a force, equal, opposite and with the same line of action as the resultant force F and include this force with the system of the forces of which F is the resultant, then the complete system of forces will be in equilibrium. This principle is known as D-Alembert's principle. The equal and opposite force (–$m.a$) is known as reversed effective force or the inertia force (briefly written as $F_I$).

The equation (ii) may be written as $F+F_I=0$

Thus, D-Alembert's principle states that the resultant force acting on a body together with the reversed effective force are in equilibrium. This principle is used to reduce a dynamic problem into an equivalent static problem.

D’Alembert’s principle makes the venerable Atwood’s machine, illustrated in figure -1, trivial to analyze. The state of the machine is determined by the positions of the two masses along the U-shaped coordinate s looping over the frictionless pulley with the string. The work done by gravity on the left-hand mass under the displacement $δ-s$ is $δW_L = −mgδ_s$, while the gravity acting on the right-hand mass produces work $δW_R = +M_gδ_s$. The inertial work on the two masses gives $δW_I = −(M + m)(d^2s/dt^2 )δ_s$, resulting in

$δ_W = (M − m)g − (M + m)( d^2s/dt^2) δ_s = 0$

$d^2s/dt^2 =\frac{g(M – m)}{( M + m)}$ .

Note that the force of constraint by the pulley (which is assumed to be free to rotate but held rigidly in place and with negligible moment of inertia) does not enter the problem at all.

This technique is useful for solving statics problems, with static forces of constraint. A static Force of constraint is one that does no work on the system of interest, but merely holds a certain part of the system in place. In a statics problem there are no accelerations. We can extend the principle of virtual work to dynamics problems, i.e., ones in which real motions and accelerations occur, by introducing the concept of inertial forces. For each parcel of matter in the system with mass m, Newton’s second law states that F = ma

Kinetic Energy of Rotation considers a rigid body rotating with a constant angular speed w about a fixed axis AOB as show on in the figure-2. As the body is made up of a number of particles of masses $m_1 , m_2 , m_3 , ......$ at distances $r_1 , r_2 , r_3 , ......$ from the axis of rotation, all these particles describe circular paths of radii $r_1 , r_2 , r_3 , ......$ with the same angular speed ω. As the linear velocity of the particles $V = rω$, V is different for different particles. Let $V_1 , V_2 , V_3 , ......$ be the linear velocities of the particles. Then the total K.E. of rotation of the body,

$E=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{1}{2}m_3v_3^2+......$

$E=\frac{1}{2}m_1r_1^2w_1^2+\frac{1}{2}m_2r_2^2w_2^2+\frac{1}{2}m_3r_3^2w_3^2+......$

$E=\frac{1}{2}w^2(m_1r_1^2+m_2r_2^2+..........)$

$E=\frac{1}{2}w^2I$ (where $I=MI$ of the body about the axis rotation)

$E=\frac{1}{2}Iw^2$

Fig.- 2 Rotation of Rigid body

If we define kinetic energy as $KE= \frac{1}{2}mv^2$ the result is the work-energy principle, the work done in changing the state of motion of an object is equal to the change in the kinetic energy.

$W_net=KE_2−KE_1=∆KE$

This relationship is not anything which is not contained in Newton's laws, it is a restatement in terms of energy, and makes many problems much easier to solve.

Potential energy

While Kinetic energy is associated with motion potential energy is associated with position. Whereas kinetic energy has an obvious zero, when the velocity is zero, potential energy is only definedas a change. Potential energy itself has come in different kinds.

From our kinetic relationships we have $v_f^2 =v_i^2+2gh$

Since we drop the object the initial velocity is zero so we have $v_f^2 =2gh$

To see how this relates to energy we multiply both sides of the above equation by $\frac{1}{2}(m)$

we get

$\frac{1}{2}mv_f^2 =\frac{1}{2}m(2gh)$

$\frac{1}{2}mv_f^2=mg h$

This means that the object has acquired a kinetic energy equal to $m g h$. Because we can object to a height h, and then drop is and have is acquire a kinetic energy equal to $mgh$ we call the quantity (mgh) gravitational potential energy.

It is potential energy, because it can be turned into actual kinetic energy. Potential energy is only meaningfully defined as a change when an object moves from one position to another. $∆PE_{Grav}=mg(y_2− y_1)=mgh$ Work and potential energy.

$W_{ext}=mg (y_2− y_1)$ so, combining this with the equation above it gives $W_{ext}=∆PE$

The work-energy principle extended to include non-conservative forces Earlier we stated that the network, done on an object is equal to the change in kinetic energy. $W_{net}=KE_2−KE_1=∆KE$. If non conservative forces are involved then we need to divide the work done into a part done by the conservative forces and the work done by non-conservative forces.

$W_C+ ∆W_{NC}=∆KE$ If we rearrange this we get $W_{NC}=∆KE−W_C$ The work done by a conservative force such as gravity we have seen above as $W_{Grav}=−∆PE$.

If we put this into the formula above it we get $W_{NC}=∆KE+∆ PE$

Mechanical Energy and its Conservation If we consider the above equation in the case in which there is no non-conservative work then we get $∆KE+∆PE=0$ If we define mechanical energy as $∆KE+∆PE$, then we see that, in that absence of any nonconservative force, the change in mechanical energy is zero, so mechanical energy is conserved.