SYSTEM OF PARTICLES AND ROTATIONAL MOTION CENTRE OF MASS AND ROTATIONAL MOTION INTRODUCTION- For describing the motion of rigid bodies, we shall introduce the key concept of ‘centre of mass’. This concept enables us to understand how we can apply justifiably the Newton’s laws of motion, in essentially the same form to objects of large size including even the astronomical objects like the planets and the stars. KINDS OF MOTION OF A RIGID BODY- A rigid body may have three kinds of motion- 1) Pure Translation Motion- in such a motion, every particle of the body has the same velocity at a particular instant of time. For e. g. when a rectangular block slides down an inclined plane, any point like P1,P2 of the block, at any instant of time moves with the same velocity. This is because the block is a rigid body and all the particles of the body are moving together, with the same velocity. (2)Pure Rotational Motion- in such a motion, a rigid body rotates about a fixed axis.
Every particle of the body moves in a circle, which lies in a plane perpendicular to the axis, and has its centre on the axis. For e. g. , in an oscillating table fan or pedestal fan, the axis of rotation is horizontal. This axis has an oscillating sideways movement in a horizontal plane about the vertical through the point at which the axis is pivoted. (3) Combination Of Translational And Rotational Motion- for e. g. , when a cylinder rolls down an inclined plane, its motion is a combination of rotation about a fixed axis and translation.
As the cylinder shifts from top to the bottom of inclined plane, the points P1, P2, P3, P4 on the rolling cylinder have different velocities at a particular instant of time. If the cylinder were to roll without slipping, the velocity of the point of contact P3 would be 0, at any instant of time. From the above discussion we conclude that- (i)The motion of a rigid body, which is not provided or fixed in some way is either a pure translation or a combination of translation and rotation. (ii)The motion of a rigid body, which is pivoted or fixed in some way is rotation.
The rotation may be about an axis which is fixed e. g. , in a ceiling fan or about an axis which is moving e. g. , in an oscillating table fan. We shall confine ourselves to rotation about a fixed axis. CONCEPT OF CENTRE OF MASS- The concept of centre of mass of a system enables us to discuss overall motion of the system by replacing the system by an equivalent single point object. We may define centre of mass of a body or a system of bodies as a point at which the entire mass of the body/system of bodies, is supposed to be concentrated.
If all the external forces acting on the body/system of bodies were to be applied at the centre of mass, the state of rest/motion of the body/system of bodies, shall remain unaffected. It is not at all necessary that the total mass of the system be actually present at the centre of mass. CENTRE OF MASS OF TWO PARTICLE SYSTEM- Suppose two particles of masses m1 and m2 are separated by a distance d. If we take the origin at the point mass m1 and the line joining the two particles as x-axis, then the co-ordinates of m1 and m2 are (0,0) and (d,0) respectively. 1 M m2 0 d1 d2 x d so, xcm = (m1*0 + m2d)/(m1+m2) =(m2d)/(m1+m2) ycm=(m1*0+m2*0)/(m1+ m2)=0 zcm=0 Thus the centre of mass lies on the line joining the two particles at a distance of (m2d)/(m1+m2) from the mass m1. i. e. d1=m2d/(m1+m2) and d2=m1d/(m1+m2) or, d1/d2=m2/m1 or, m1d1=m2d2 Obviously, the centre of mass divides the line joining the two particles internally in the inverse ratio of their masses. Hence the centre of mass is closer to the massive particle. NOTE- 1. The centre of mass may lie outside the object.
For e. g. the centre of mass of ring lies outside it. 2. Centre of mass depends on the distribution of mass within the body and is closer to heavy part but for identical particles C. M. lies at the midpoint of the line joining the point masses. 3. If an object is symmetrical and have uniform distribution of mass then its C. M. coincides with geometrical centre. 4. The position of C. M. is independent from the coordinate system. E. g. the centre of mass of a ring of uniform thickness is at its centre whatever be the coordinate system. CENTRE OF MASS OF LAMINAR OBJECTS-
A laminar object means a two dimensional distribution of mass of negligible thickness e. g. paper of your copy. If a large number of laminar type objects are placed in a plane then Xcm=(x1A1+x2A2+…. +xnAn)/(A1+A2+…. +An)=? xiAi/? Ai ycm=(y1A1+y2A2+…. +ynAn)=? yiAi/? Ai where A represents the area of laminar objects. CENTRE OF MASS OF RIGID CONTINUOUS BODIES- A rigid body may be treated as a continuous distribution of matter. The “particle” then becomes differential mass elements “dm” and the summation (? ) becomes integration (? . Hence These integrals are evaluated for all mass elements in the object. SCALAR RELATION BETWEEN v AND ? - When the body rotates through an angle? , the particle P moves a linear distance s along the circle of radius r. Here s = r? …. (? is in radian) or, v = ds/dt = rd? /dt = r? Thus, v=r? …. (scalar relation between v and ? ) Here ? ust be in radian measure. Linear motion Rotational motion (a)If acceleration a=0 and velocity v=constant, then x=vt. If angular acceleration ? =0 and angular velocity ? is constant, then ? =? t. (b)If acceleration a=constant, then (i) v=v0+at (ii)x=v0t+(1/2)at2 (iii)v2=v02+2ax (iv)xnth=v0+[(2n-1)/2]a (v)x=[(v0+v)/2]tIf angular acceleration ? =constant, then (i) ? =? 0+? t (ii)? =? 0t+(1/2) ? t2 (iii)? 2=? 20+2?? (iv)? nth=? 0+[(2n-1)/2]? (v)? =[(? 0+? /2]t (c)If acceleration a is not constant, then (i)(dv/dt)=a >v=v0+? t0adt (ii)(dx/dt)=v >x-x0=? t0vdt (iii)v(dv/dx)=a >? vv0vdv=? x0adxIf angular acceleration ? is not constant, then (i)(d? /dt)=? >? =? 0+? t0? dt (ii)(d? /dt)=? >? -? 0=? t0? dt (iii)?. (d? /d? )=? >??? 0? d? =?? 0 ? d? MOMENT OF INERTIA (I) - Let us consider a rigid body rotating about an axis (AB) with angular speed ?. You know that in rotational motion all particles move in circles having their centres on the axis of rotation.
Suppose the ith particle of the body moves in a circle of radius ri. Its linear speed v=? ri. If mi be its mass, then its kinetic energy is ? mi. vi2=(1/2)mi(? ri)2=(1/2)(mi. ri2)? 2 therefore, the kinetic energy of the whole body is K=? (1/2)(mi. ri2)? 2=(1/2)? (mi. ri2)? 2 or, k = (? )I? 2 where I=? mi. ri2 (summation runs over all the particles) This quantity I=? mi. i2=m1r12+m2r22+……… is called the “moment of inertia (I)” of the body about the axis AB. THEOREMS OF PERPENDICULAR AXES- (i)Theorem Of Perpendicular Axes: the given figure shows a plane body. X-and y-axes are in the plane of the body; and the z-axis is perpendicular to the plane of the body. Three axes are mutually perpendicular. Now this theorem states that Iz=Ix+Iy this theorem is true only for a planar body (lamina) Z N y
P 0 0 x M (ii)Theorem Of Parallel Axes: if I0 be the moment of inertia of a body of mass M about an axis passing through its centre of gravity will be given by I=I0+Md2 where, d is the perpendicular distance between the two axes. MOMENT OF INERTIA OF SOME REGULAR BODIES- Body Axis of rotationFormula ) Thin uniform rod of mass M and length l a) Perpendicular to the length at its mid-point. b) Perpendicular to the length at one of its ends. Ml2/12 Ml2/3 2) Thin circular uniform ring of mass M and radius R a) Perpendicular to its plane at its centre. b) Diameter c) Perpendicular to its plane at the ring. d) Tangent in its plane. MR2 (1/2)MR2 2MR2 (3/2)MR2 3) Uniform circular disc of mass M and radius Ra) Perpendicular to disc at its centre. b) Diameter c) Perpendicular to the plane of the disc at its edge d) tangent in the plane of the disc(? )MR2 1/4)MR2 (3/2)MR2 (5/4)MR2 4) Uniform rectangular plate of mass M, length l and breadth ba) parallel to its length and passing through its centre. b) parallel to its breadth and passing through its centre c) perpendicular to its plane through its centre. Mb2/12 Ml2/12 (M/12)(l2+b2) 5) Hollow cylinder of mass M, radius R and length la) axis of the cylinder b) perpendicular to the axis, at its centreMR2 (Ml2/12)+(MR2/2) 6) Solid cylinder of mass M, radius R and length la) axis of the cylinder b) perpendicular to the axis, at its centre(1/2)MR2 (Ml2/12)+(MR2/4) ) solid sphere of mass M and radius Ra) diameter b) tangent(2/5)MR2 (7/5)MR2 8) hollow sphere of mass M and radius Ra) diameter b) tangent(2/3)MR2 (5/3)MR2 RADIUS OF GYRATION- When a body is rotating about an axis, the total mass (M) of the body may be supposed to be placed at a point at a distance K from the axis of rotation such that the moment of inertia of the body about that axis is equal to I=MK2 Then the distance K is called the radius of gyration. Mathematically K= (vI/M) MOMENT OF FORCE OR TORQUE ( ? ): It is the rotational analogue’ of force.
When a force F acts at a particle P having a position vector r with respect to an origin O the torque ? exerted on the particle by this force with respect to O is a vector quantity defined as ? = r . F………. (definition of torque) Note- 1)Torque is always defined with reference to a given point (or given line). On changing the reference point (or line) the torque may change. 2)As a shortcut the magnitude of torque can be calculated as the product of the force and the length of line perpendicular to the force starting from the point about which the torque is to be calculated. 3)? =0 if r=0, or if F=0, or if ? 0o or 180o. Hence if the line of action of the force passes through the reference point, the torque vanishes. 4)The SI unit of torque is newton-metre. This is also the unit of work and energy. But torque is not work or energy. Usually, the unit of work or energy is written as “joule”. But the torque should not be expressed in “joule”. CENTRE OF GRAVITY- “Centre of Gravity” of a body is a point where entire weight (force of gravity) is assumed to act so that the torques of the entire weight (W) about any point is same as the vector sum of the torques of weights of its constituent particles about the same point.