Unformatted text preview: rotation as in positive for counter-clockwise and negative for
By applying the Pythagoras
Theorem with P as the origin:
IP = IX + IY .
x 194 11
Example 1 1: Consider a thin rod AB of mass M of uniform density and length L
along the x-axis. The rod is free to rotate about the y-axis passing through the center
of mass and perpendicular to its length.
4321 − L/ x 2 dx B
2 Since the mass M is of uniform density we have: mass per unit length = M/ L .
So an element of the rod has mass dM = M/ L . dx
The moment of iner tia of this element of the rod about the y-axis is dependent
on the distance x from the y-axis.
Therefore: dIY = (M/ Ldx) . x2 = M/ Lx2dx .
The moment of iner tia of the entire rod about the y-axis is got by continuously
summing up or integrating over the range of x from − L/ to +L/ .
IY = ∫ dIY +L/ = 2 ∫ M/ Lx2dx
−L = /2 +L/
2 M/ L [x /3 ]
3 − L/ 2 = M/ 3L[(+L/ )3 − (−L/ )3]
IY = ML/12 195 If we now shift the rod so that the axis of rotation (y-axis) passes thru one end of the
rod, what is the moment of iner tia IY ?
y y 4321
4321 0 dx B
+L x A
4321 A dx B
In both cases IY = ML/3 A position vector has both magnitude (distance) and direction. The direction may be
positive or negative depending on which side of the axis the rigid body is located. It is
impor tant to note that the measur e off the tendency to rotate is defined
independent of the side of the axis the mass is located. The orientation is
the distance of the mass from the axis or point of rotation.
We observe that the greater the distance of the mass from the axis or point of
rotation, the greater the moment of iner tia . Hence the easier it is to rotate
or turn the rigid body and so lesser energy is required.
If the rigid body happens to be in rotational motion about...
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- Fall '09
- Limit, Δx