Loading....
Coupon Accepted Successfully!

 

Rigid Body Rotation and Equation of Rotational Motion

The resultant torque acting on the body must be zero.
 

The sum of anticlockwise torques about any axis must be equal to the sum of clockwise torques about the same axis.
 

This means the algebraic sum of all the torques acting on the body about the axis of rotation must be zero.


The general conditions of equilibrium are

  1. The algebraic sum of the resolved components of the forces in any fixed direction must be zero.
  2. The algebraic sum of the resolved components of the forces in a perpendicular direction must be zero.
  3. The algebraic sum of the torques or moments of the forces about any point in their plane must be zero.

Consider a special case of equilibrium of a body acted upon on a number of coplanar parallel forces acting on a body.
 

Using the condition number one, i.e. the algebraic sum of the forces acting on the body is zero,

P + Q - R - S = 0
 

Using condition number three i. e the anticlockwise moment = the clockwise moment,

equating moments about the point A,

P × AD + Q × AF = R× AC + S × AE

P × AD + Q × AF - R × AC - S × AE = 0
 

Thus the algebraic sum of the moments of the forces acting on the body about any point is zero. 

Centre of Gravity

The weight W of an extended body is the vector sum of the gravitational forces acting on the individual elements (the atoms) of the body.
 

Instead of considering all those individual elements, it can be assumed that a single force W effectively acts at a single point called the center of gravity (cg) of the body.
 

If the forces on the individual elements were somehow turned off and force W at the center of gravity turned on, the net force and the net torque (about any point) acting on the body would not change.
 

Up until now, it is assumed that the weight W acts at the center of mass (cm) of the body, which is equivalent to assuming that the center of gravity is at the center of mass. This assumption is valid provided the acceleration g due to the gravitational force is constant over the body.
 

Figure shows an extended body, of mass M and one of it elements, of mass mi.
 

Each such element has weight migi, where gi is the acceleration due to the gravitational force at the location of the element.
 

In Fig, each weight migi produces a torque τ i on the element about the origin O. Using the relation τ = r1F, we can write torque τ i as

τ i = xi mi gi

  1. An element of mass mi in an extended body has weight migi with moment arm xi about the origin O of a coordinate system.
  2. The weight W of a body is said to act at the center of gravity (cg) of the body. The moment arm of W about O is equal to Xcg, Where xi is the moment arm r of force mi gi.

The net torque on all the elements of the body is then

τnet = ∑ τ i = xi mi gi

Figure Shows the body's weight W acting at the body's center of gravity.
 

The torque about O due to W is

τ = xcg W,

where xcg is the moment arm of W.
 

The body's weight W is equal to the sum of the weights migi of its elements. 

So we can substitute mi gi for W,

τ = xcgmi gi

By definition, the torque due to weight W acting at the center of gravity is equal to the net torque due to the weights of all the individual elements of the body.

xcg mi gi = xi mi gi

If the accelerations gi are equal, Substituting the body's mass M for mi on the left side,

xcg = xi mi                 

xcg = xcm

Thus, the body's center of gravity and its center of mass have the same x coordinate.
 

We can generalize this result for three dimensions by using vector notation. The generalized result is: If the gravitational acceleration is constant over a body, the body's center of gravity is at its center of mass.
 

This may not be the case for objects that have non-uniform distribution of mass.
 

we see that the torque due to the weight of a body is zero only if the moment arm xcg is zero. This means that if the body is suspended from some arbitrary point S about which it can rotate, it rotates (due to the torque τ = xcg W about S) until xcg is zero.
 

Its center of gravity is then vertically below the suspension point, as in figure a and b, and the body is in equilibrium. If the body is suspended at its center of gravity, as in figure, then for any orientation of the body, xcg is zero and the body is in equilibrium.

A body free to rotate about a suspension point S will rotate until its center of gravity is vertically below S as in (a) and (b) unless the center of gravity is at S as in (c).




Test Your Skills Now!
Take a Quiz now
Reviewer Name