September 2011 - Career Point

This minimum speed is called critical speed (v c ). Ifthe speed at A is less than this value, the particle willnot reach up to the highest point. To reach with thisspeed at A, the body should have speed at B given bythe conservation laws viz.Decrease in kinetic energy = increase in potentialenergy1 2 1 mvb – 2 mva = mg.2r2 2v 2 b = v 2 a + 4grfor critical speed v a = v c =gr∴ v 2 b gr + 4gr or v b = 5 grTherefore, the body should have speed at B at least5 gr , so that it can just move in vertical circle.Tension in string at B is given by.T b – mg =mv 2 bror T b = mg +m5vgrr= 6mgThis means that the string should be able to stand to atension, equal to six times the weight of the bodyotherwise the string will break.At any other point P making angle θ with the vertical,from the figure.AT – mg cos θ =CBθmv 2 prTormgv p⎛At point A, θ = 180º; T a = m ⎜⎝ rPQmg cos θ⎛T = m ⎜v 2 r⎝ rv 2 a⎞− g⎟⎠⎞+ g cosθ⎟⎠⎛ ⎞At point B, θ = 0º; T b = m ⎜v 2 b+ g⎟⎝ r ⎠Conical pendulum :A conical pendulum consists of a string AB (fig.)whose upper end is fixed at A and other and B is tiedwith a bob. When the bob is drawn aside and is givena horizontal push. Let it describe a horizontal circlewith constant angular speed ω in such a way that ABmakes a constant angle θ with the vertical. As thestring traces the surface of a cone, it is known asconic pendulum.Let l be the length of string AB. The forces acting onthe bob are (i) weight mg acting downwards,(ii) tension T along the sting (horizontal) componentis T sin θ and vertical component is T cos θ).T cos θ = mgThe horizontal component is equal to the centripetalforce i.e.,Rotational Motion :rAhOTT sin θT cosθBmgCentre of mass of a system of particles :The point at which the whole mass of the body maybe supposed to be concentrated is called the centre ofmass.Consider the case of a body of an arbitrary shape of nXY plane as shown in fig. Let the body consist ofnumber ofY(xP 1(x 1, y 2, y 2)1)(x, y ) P 2P 3(x 3, y 3)XOparticles P 1 , P 2 , P 3 , .... of masses m 1 , m 2 , m 3 , ..... andcoordinates (x 1 , y 1 ), (x 2 , y 2 ), (x 3 , y 3 ), ..... If ( x, y)bethe coordinates of centre of mass, thenm1x1+ m2x2 + m3x3+ .... Σmnxnx ==m + m + m + ..... Σm12m1y1+ m2y2+ m3y3+ ... Σmnynand y ==m1+ m2+ m3+ .... ΣmnWhen there is a continuous distribution of massinstead of being discrete, we treat an infinitesimalelement of the body of mass dm whose position is(x, y, z). In such a case, we replace summation byintegration in above equations. Now we have,x =y =∫∫∫∫x dm=dmy dm=dm∫∫x dmMy dmM3nXtraEdge for IIT-JEE 28 SEPTEMBER 2011