We know that Newton’s laws of motion define the equations that govern motion of bodies in macro scale. As far as translation is concerned,
• First Law: At its crux defines Inertia or the resistance offered by body to change in state(rest or motion). For translation, inertia is same as mass of body & is the resistance offered to change of translation state of the body.
• Second Law: This defines the relation between inertia and force required to induce an momentum change in the body as
( )
• Third Law: This proposes the action and reaction pairs of Forces.
Now, if we extend this to rotation, the concept analogous to these are generated as follows.
• First Law: At its crux defines Moment of Inertia or the resistance offered by body to
change in state(rest or motion). For rotation, moment of inertia is the resistance offered to change of rotation state of the body.
• Second Law: This defines the relation between moment of inertia and the torque required to induce a change in angular momentum in the body as
( )
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• Third Law: This proposes the action and reaction pairs of Torques.
The moment of the inertia force on a particle around an axis multiplies the mass of the particle by the square of its distance to the axis and forms a parameter called the moment of inertia (denoted by I). The moment of inertia of an object is defined by the distribution of mass around an axis. It depends not only on the total mass of the object but also on the square of the perpendicular distance from the axis to each element of mass. This means the moment of inertia increases rapidly as masses are distributed more distant from the axis. For example, consider two wheels that have the same mass, one that is the size of a bicycle wheel and one that is half the size.
The larger wheel has four times the moment of inertia even though it is only twice the diameter.
• For system of particles, moments of inertia of individual particles sum to define the moment of inertia of a body rotating about an axis ie ∑
• For rigid bodies moving in a plane, such as a compound pendulum, the moment of inertia is a scalar got by integrating over the body mass ie I ∫ r dm ∫ r dV
• For rigid bodies moving in three dimensions, such as a spinning top, the moment of inertia becomes a matrix, also called a tensor with 9 components (6 independent components).
For easiness, moment of inertia is written in lumped form as where k is called radius of gyration.
Theorems for computing Moment of Inertia for compound bodies
The main theorems which will help in computing Moment of Inertias are as follows.
• Superposition: The moment of inertia of the body is additive. That is, if a body can be decomposed (either physically or conceptually) into several constituent parts, then the moment of inertia of the whole body about a given axis is equal to the sum of moments of inertia of each part around the same axis ie ∑
• Perpendicular Axis: If Ix, Iy, Iz are moments of inertia around three perpendicular axis passing through the bodies center of mass, then each of them cannot be greater than the sum of two others: For example . Here the equality holds only if the body is flat/planar body (negligible z values ie z ≈ 0)). When body is planar, ∫ ∫(
) ∫ ∫ ∫( ) ∫( ) ( ≈ ) ⇒ ,thereby proving the above identity.
• Parallel Axis: If the objects moment of inertia around a certain axis passing through the center of mass is known (ICM), then the parallel axis theorem or Huygens Steiner theorem provides a convenient formula to compute the moment of inertia of the same body around
Figure 1: Computation of Moment of Inertia
a different axis (Id), which is parallel to the original and located at a distance d from it. The formula is only suitable when the initial and final axis are parallel & is given by Id = ICM + Md2
Examples 25
1. Given 2 masses - m@2r & 2m@r from axis of rotation, to find the moment of inertia, we see that this is a particle system and hence have Inet = £\ Ij = 2m(r)2 + m(2r)2 = 6mr2. 2. Given a disk of radius R & mass M, we can divide the continuum into discrete rigid bodies
and take integral as follows. I= L,r2dm = f„ 7rnfR nr2(rd6dr^L) = ±MR2. However, note that in case of a loop/hoop it has same mass and radius,
The Moment of Inertia / = MR2because in that case the whole of the mass is distributed at a larger distance from the rotation/central axis.
3. For a given right circular cone with radius R, mass M and height H about the vertical axis, we have as follows. We divide the cone into stacked discs of diminishing radius as shown and do a summation of individual I’s for getting the total moment of inertia of the
compound body. ∫ ( ) ∫ (
) ⇒ ( ) On
Simplifying, we get
Further, for any axis passing through the apex of cone, parallel to base (3rd figure in above diagram), we proceed as follows. Here, a common mistake people make is to employ the perpendicular axis theorem but the point to be noted is that it can be used only for planar bodies. So, we again break up the bodies into smaller bodies and do integration as follows.
We see that the moment of inertia about an axis parallel to base but passing through the elemental disc shown is given by (obtained by applying perpendicular axis
Figure: Computation of Moment of Inertia for a cone Figure: Computation of Moment of Inertia for a disc
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theorem for the disc which is a planar body). Now, applying the perpendicular axis theorem for getting the moment of inertia about the axis parallel to base passing through the apex, we get IdApex = Id + dm(H - h). Finally, integrating this, we get as follows. 5 Thin rectangular plate of height h and of width w and mass m
( ) 6 Ellipsoid (solid) of semi-axis a, b, and c with axis of rotation a and
mass m ( )
Example 26
The drum shown in Fig. (a) has a radius of gyration of 30 cm and weighs 1.8 kN. It is supported by means of small hubs which rest in bearings. A weight of 1 kN is attached to one end of a rope, the other end being wrapped around the drum. Neglecting friction in the bearings, determine the acceleration of the weight, the angular acceleration of the drum, and the tension in the rope.
Solution
Moment of inertia of drum,
( ) 6
The free body diagram is shown in Fig. (b) Let T be the tension in the rope Using D’ Alembert’s principle, we have for the 1kN weight,
T
∑ ( )
1000 – T- (a) And for the drum
∑ ( )
∴ 6 (b) Now a = r = 0.375 .. (c) Substituting Eq. (c) in (a), we get 1000 – T –
∴ 6 Substituting in Eq. (b), we get 0.375 (1000 – 38.226 )- 16.5 375 – 14.335 - 16.5 = 0 6 /
6 6 /
6 6 Example 27
Two weights and of 8 kN and 5 kN are attached at the ends of a flexible cable. The cable passes over a pulley of 100 cm diameter. The weight of the pulley is 600 N with a radius of gyration of 50 cm about its axis of rotation. Find the torque which must be applied to the pulley to arise the 8 kN weight with an acceleration of 1.5 / . Neglect friction in the pulley bearing.
Solution
Let T and T be the tensions in the cable on the weights W and W sides respectively and be the torque applied to the pulley to accelerate the 8000 N weight with 1.5 / as shown in Fig.
Using D’ Alembert’s principle, we have for , T = (a) and for
(b) For the pulley, we have
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∑
Or M ( ) (c ) Also a = r (d) From Eq. (d), we get
/
∴
∴ ( ) 6
( )
a
5kN a
8kN
W2
W1
α
50 cm
600N
Example 28
A step pulley and mass system is shown in the figure along side. Consider pulley is mass less &
frictionless, string is non extendable
Find the m2>m1 and m2 moving downward Solution
F.B.D is,
From the F.B.D, we can write,..
S1-m1a1-m1g
&
S2-m2a2-m2g 2 m1g
S1
a1
m1a1
m2g S2
a1
m2a2
m
1m
2r
1r
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For the rotation of ‘O’ of the step pulley, we can write
x
1=r
10 x
2= r20
∴
we can write,Considering equilibrium at that instant S1r1=S2r2
From eg 2,3,4, we can write
From eq 1 & 5 we can write
( )
( )