SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

∑ m i i r = . Remember that the position vectors ( r i ) are taken with respect to the CG. Now, in accordance with the reasoning given below Eq. ( .4a) in Sec.

. , if the sum is zero, the origin must be the centre of mass of the body. Thus, the centre of gravity of the body coincides with the centre of mass in uniform gravity or gravity- Fig. .

Determining the centre of gravity of a body of irregular shape. The centre of gravity G lies on the vertical AA through the point of suspension of the body A. u u free space. We note that this is true because the body being small, g does not vary from one point of the body to the other.

If the body is so extended that g varies from part to part of the body, then the centre of gravity and centre of mass will not coincide. Basically, the two are different concepts. The centre of mass has nothing to do with gravity. It depends only on the distribution of mass of the body.

In Sec. . we found out the position of the centre of mass of several regular, homogeneous objects. Obviously the method used there gives us also the centre of gravity of these bodies, if they are small enough.

Figure . illustrates another way of determining the CG of an irregular shaped body like a cardboard. If you suspend the body from some point like A, the vertical line through A passes through the CG. We mark the vertical AA .

We then suspend the body through other points like B and C. The intersection of the verticals gives the CG. Explain why the method works. Since the body is small enough, the method allows us to determine also its centre of mass.

Example . A metal bar cm long and . kg in mass supported on two knife- edges placed cm from each end. A .

kg load is suspended at cm from one end. Find the reactions at the knife-edges. (Assume the bar to be of uniform cross section and homogeneous.) Answer Fig. .

Figure . shows the rod AB, the positions of the knife edges K and K , the centre of gravity of the rod at G and the suspended load at P. Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homogeneous; hence G is at the centre of the rod; AB = cm.

AG = cm, AP = cm, PG = cm, AK = BK = cm and K G = K G = cm. Also, W = weight of the rod = . kg and W = suspended load = . kg; R and R are the normal reactions of the support at the knife edges.

For translational equilibrium of the rod, R +R –W –W = (i) Note W and W act vertically down and R and R act vertically up. For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments about is G. The moments of R and W are anticlockwise (+ve), whereas the moment of R is clockwise (-ve).

For rotational equilibrium, –R (K G) + W (PG) + R (K G) = (ii) It is given that W = . g N and W = . g N, where g = acceleration due to gravity. We take g = .

m/s . With numerical values inserted, from (i) R + R – . g – . g = or R + R = .

g N (iii) = . N From (ii), – . R + . W + .

R = or R – R = . g N = . N (iv) From (iii) and (iv), R = . N, R = .

N Thus the reactions of the support are about N at K and N at K . ⊳ Example . A 3m long ladder weighing kg leans on a frictionless wall. Its feet rest on the floor m from the wall as shown in Fig.

. . Find the reaction forces of the wall and the floor. Answer Fig.

. The ladder AB is m long, its foot A is at distance AC = m from the wall. From Pythagoras theorem, BC = m. The forces on the ladder are its weight W acting at its centre of gravity D, reaction forces F and F of the wall and the floor respectively.

Force F is perpendicular to the wall, since the wall is frictionless. Force F is resolved into two components, the normal reaction N and the force of friction F. Note that F prevents the ladder from sliding away from the wall and is therefore directed toward the wall. For translational equilibrium, taking the forces in the vertical direction, N – W = (i) Taking the forces in the horizontal direction, F – F = (ii) For rotational equilibrium, taking the moments of the forces about A, F − ( / ) W = (iii) Now W = g = × .

N = . N From (i) N = . N From (iii) . / .

N W From (ii) . N . N N The force F makes an angle α with the horizontal, tan , tan ( ) N F ≈ ⊳ . MOMENT OF INERTIA We have already mentioned that we are developing the study of rotational motion parallel to the study of translational motion with which we are familiar.

We have yet to answer one major question in this connection. What is the analogue of mass in rotational motion? We shall attempt to answer this question in the present section. To keep the discussion simple, we shall consider rotation about a fixed axis only.

Let us try to get an expression for the kinetic energy of a rotating body . We know that for a body rotating about a fixed axis, each particle of the body moves in a circle with linear velocity given by Eq. ( . ).

(Refer to Fig. . ). For a particle at a distance from the axis, the linear velocity is i r υ .

The kinetic energy of motion of this particle is i i m r υ where m i is the mass of the particle. The total kinetic energy K of the body is then given by the sum of the kinetic energies of individual particles, i i K m r ω Here n is the number of particles in the body. Note ω is the same for all particles. Hence, taking ω out of the sum, i i K m r We define a new parameter characterising the rigid body, called the moment of inertia I , given by i i m r