PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1dN9n7oo0sYE6nlAs_-7YVWxQTMRIeVOY/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true Rotational Motion and Moment of Inertia Centre of mass of a two-particle system, Centre of mass of a rigid body; Basic concepts of rotational motion; moment of a force, torque, angular momentum, conservation of angular momentum and its applications; moment of inertia, radius of gyration, Values of moments of inertia for simple geometrical objects, parallel and perpendicular axes theorems and their applications. Rigid body rotation, equations of rotational motion. C H A P T E R CENTRE OF MASS It is point in a system which moves as if whole mass of the system is concentrated at the point and all external forces are acting on it. Its position is given by ∑mi xi CHAPTER COVERS : Centre of mass Centre of mass of a rigid body xcm = m1x1 + m2 x2 + + mn xn m1 + m2 + + mn m y + m y + ......... + m y = i =1 M ∑mi yi Velocity and acceleration of centre of mass Rotational kinetic y cm = 1 1 2 2 n n = i =1 energy and moment m1 + m2 + .......... .... + mn M n of inertia Theorems for zcm = m1z1 + m2 z2 + + mnzn m1 + m2 + + mn y ∑mi zi = i =1 M moment of inertia Moment of inertia of different objects (x1, y1, z1) (x2, y2, z2) m2 Rolling of a body Angular momentum m1 c.m. (x , y , z ) and its conservation 3 3 3 3 O x CENTRE OF MASS OF A RIGID BODY Mathematically position coordinates of the centre of mass of rigid body are given by xcm = ∫ xdm ∫dm ; y cm = ∫ ydm ∫dm ; zcm = ∫ zdm ∫dm TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 VELOCITY AND ACCELERATION OF CENTRE OF MASS Velocity of Centre of Mass The instantaneous velocity of centre of mass is given by v cm = m1v 1 + m2 v 2 + ....mn v n ; or M v cm = P system M Where P system is the total linear momentum of centre of mass. Acceleration of Centre of Mass Differentiating v cm w.r.t. time we get a cm as a cm = m1a 1 + m2 a 2 + ....mn a n ; or M a cm = ΣF ext M Where ΣFext is the vector sum of forces acting on the particles of system. The centre of mass of a system of particles lies in the region where majority of mass of the system lies. The centre of mass of a body may lie outside the body or on the body. e.g., The centre of mass of a ring lies at its centre not on its body. For a symmetrical body of uniform density, the centre of mass lies at its geometrical centre. For a two particle system, the centre of mass lies on the line joining them and closer to the greater mass. The position co-ordinates of the centre of mass of a system of two masses as shown in the figure. y x = m1(0) + m2d m2d (xcm, 0) m1 (0, 0) COM (d, 0) m2 x cm + m = m1 m2 The centre of mass is a point such that the mass of the system M multiplied by the acceleration (d2Rcm/dt2) of the centre of mass of the system gives the resultant of all forces acting on the system. The centre of mass of a system of two particles lies in between them on the line joining the particles. If we take the centre of mass as the origin, then the sum of the moments of the masses of the system → Σ mi is zero. During translatory motion, the position of the centre of mass changes with time. In the absence of an external force, the velocity of the centre of mass of a body remains constant. The location of the centre of mass depends on the shape and nature of distribution of mass of the body. TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 Applications The position co-ordinates of the centre of mass of a system of three particles of equal masses which are placed on the vertices of an equilateral triangle of side a are calculated as shown. xcm x = m(0) + m(a) + m(a / 2) 3m = 3m a / 2 = a ⎛ a ⎞ cm 3m 2 ⎛ ⎞ ⎜ , a⎟ m(0) + m(0) + m⎜ 3 a ⎟ ⎜ 2 ⎟ a ycm = ⎝ ⎠ = 3m ⎛ a a ⎞ Co-ordinates of the centre of mass of the system are ⎜ 2 , 2 3 ⎟ . m (0, 0) m x A small square is removed from a uniform square plate of side a, as shown in the figure. The position centre of mass of remaining plate relative to the original centre. Can be calculated as shown a/2 The remaining body can be assumed to be a superposition of two masses, one mass is the original mass and the other is the negative mass of the body which has been removed. Original square Removed square + Here, the area of removed square will be taken negative and that of original square positive. By taking the mass (or area) of removed square as negative, the coordinates of centre of mass of the remaining portion are x = A1x1 + A2 x2 and y = A1y1 + A2 y 2 1 + A2 1 + A2 Here, A1 = a2 = area of first body (original square) A2 = –a2/4 = area of second body (removed portion) The remaining body has a line of symmetry, so its centre of mass must lie on this line. Choose the line of symmetry as x-axis with origin at the centre of original square. Now, the x coordinates of the two bodies are as shown in figure. y x Line of y x ⎛ a ⎞ Original square + (0, 0) Body 1 symmetry ⎜ ,0⎟ ⎝ ⎠ TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 2 ⎛ a2 ⎞ ⎛ a ⎞ (a ) × (0) + ⎜− ⎟ × ⎜ ⎟ A x + A x ⎜ 4 ⎟ ⎝ 2 2 ⎠ xcm = 1 1 2 2 = ⎝ ⎠ A1 + A2 (a2 ⎛ 2 ⎞ ) + ⎜− ⎟ ⎝ ⎠ ⇒ x = − a ; the centre of mass of remaining body lies on negative x-axis at a distance of 6 2 from origin. A man (of mass m) stands at the left end of a uniform sled of length L and mass M, which lies on frictionless ice. As the man then walks to the other end of the sled, the sled slides on the ice by a distance x calculated as shown. The net external force on the system (man + sled) is zero. The centre of mass of the system is to be at the same position while their movements. Obviously, the sled will shift towards left. Let the displacement of sled relative to ground = x The displacement of man relative to ground = L – x As the centre of mass remains stationary, − Mx + m(L − x) = 0 m c.m. L Frictionless ice M M + m ⇒ Mx = m (L – x) L – x m x c.m. M L ⇒ x = mL m + M SOME BASIC CONCEPTS OF ROTATIONAL MOTION Rotation of a Rigid Body When a rigid body rotates about a fixed axis, following parameters are needed to describe its motion : Angular velocity ( ω ) : It is defined by the relation ω = dθ . All the particles in the rigid body move in dt circular paths about axis of roation with angular velocity ω. Its direction is along the axis of rotation. ρ dω Angular acceleration ( α ) : It is the rate of change of angular velocity. It is given as α = dt . Tangential acceleration (at) : It is the rate of change of speed of a point on the rigid body. It is related to angular acceleration α as αt = rα. Here, r is the distance of particle or point from axis of roation. Moment of inertia (I) : It is the opposition offered by a rigid body to its state of rest or of uniform rotational motion. Angular momentum ( L ) : It is defined by the mathematical relation L = I ρ . Torque ( τ ) : It is the action of a force which produces a change in its state of rest or of uniform rotational motion. By Newton’s 2nd law, ρ dL or τ = I α τ = dt If a force F acts on the rigid body at a point whose position vector with respect to axis of rotation is ρ ρ ρ r , the torque is given by τ = r × F or τ = F × ⊥ distance from axis of rotation. TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 For uniform angular acceleration (i) (ii) (iii) (iv) ω = ω0 + α t θ = ω t + 1 αt 2 0 2 ω2 = ω2 + 2αθ θ = ⎛ ω + ω0 ⎞ t ⎜ ⎟ ⎝ 2 ⎠ ω = Angular velocity after t time ω0 = Initial angular velocity ROTATIONAL KINETIC ENERGY Consider a rigid body rotating about an axis with angular velocity ‘ω’. Various particles of the body are all rotating on a circle with radius r1, r2 with angular velocity ‘ω’. The total kinetic energy K.E. = 1 Iω2 2 I = m r 2 + m r 2 + .... is called moment of inertia and 1 Iω2 = kinetic energy of rotation. 1 1 2 2 2 MOMENT OF INERTIA A body free to rotate about an axis opposes any change in its state of rest or of uniform rotation, this inherent property of a body is called moment of inertia. Moment of inertia, in rotation motion, is the analogue of mass in linear or translational motion. For System of Particles : I = ∑mi ri2 i =1 mi = mass of i th particle ri = radius of the i th particle For Rigid Bodies : Moment of inertia of a rigid body about any axis of rotation. Mathematically I = ∫dm r 2 TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 THEOREMS FOR MOMENT OF INERTIA Parallel Axis Theorem It states that the moment of inertia of a body about an axis is equal to its moment of inertia about a parallel axis through its centre of mass plus the product of the mass of the body and the square of perpendicular distance between the two axes. dI = I + md 2 Parallel axis Icm = Moment of inertia of the body about its centre of mass I = Moment of inertia of the body about a parallel axis cm m = Total mass of the body d = Perpendicular distance between two parallel axes. Perpendicular Axis Theorem (Applicable only to plane-laminar bodies) The moment of inertia of a plane lamina about an axis perpendicular z to the plane of the lamina is equal to the sum of the moment of inertia of the lamina about two axes perpendicular to each other, in its own plane, and intersecting each other at the point where the perpendicular axis passes through it. y Iz = Ix + Iy Where Ix, Iy and Iz are the respective moment of inertia of the body about x, y and z-axes. Note : The point O need not be the centre of mass of the body. MOMENT OF INERTIA OF DIFFERENT OBJECTS For an axis perpendicular to the plane of the ring A hollow cylinder The axis perpendicular to the plane of the disc. Icm = MR2 2 A disc A solid cylinder MR2 Icm = 2 TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 A thin rod A rectangular plate Ic = ML2 12 Ic = ML2 12 A thin rod about a perpendicular A rectangular plate about one edge axis through its end ω ω ML2 I = 3 A Rectangular Plate z Mb2 (a) Ix = 12 Mass = M l x (b) Ml 2 Iy 12 O (c) Iz = Ix + Iy b (d) Iz = M(l 2 + b2 ) 12 y A Thick Rod The axis is perpendicular to the rod and passing through the centre of mass Icm = ML2 12 MR 2 4 R TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 A Solid Sphere A Hollow Sphere About its diameter About its diameter ω M, R Icm = 2 MR 2 5 I = 2 MR 2 cm 3 RIGID BODY ROTATION In this section, rotation of a body about a stationary fixed axis has been discussed Rotating Disc A tangential force F is applied at the periphery A F τ = F × R [about O] R R I = 1 mR 2 r O 2 C α = τ I = 2F MR Linear acceleration of A is aA = Rα = 2F M (along horizontal) Linear acceleration of B is aB = Rα = 2F M (vertically downwards) Linear acceleration of C is aC = rα = 2Fr MR (along horizontally opposite to A) Hinged Rod The rod is released from rest from horizontal position τ = mg × L 2 (about A) ML2 I = 3 α = τ = 3g I 2L Linear acceleration of COM C is a = L α = 3g cm 2 2L Linear acceleration of point B is aB = Lα = 3g 2 TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 ROLLING OF A BODY Rolling is combination of Translation and Rotation v Rω v + Rω cm vcm R θ v cm θ cm Rω θ v cm vcm O θ vcm vcm ⇒ vcm O θ cm vcm Rω vcm 180 – θ Mass = m vcm Rω vcm Rω Pure translation Case - I : Forward slipping vcm > Rω Pure rotation Rolling Case - II : Backward slipping vcm < Rω friction vcm–Rω Case - III : Pure Rolling vcm = Rω I v = 0 I v = 0 (instantaneous centre of rotation) Kinetic energy of the body during pure rolling (E) E = Translational KE + Rotational KE = ET + ER (instantaneous centre of rotation) = 1 mv 2 + 1 I ω2 2 cm 2 cm 1 ⎛ Icm ⎞ 2 1 2 ⎛ K 2 ⎞ ⎜ m + 2 ⎝ R 2 ⎟vcm = mvcm ⎜1+ ⎟ 2 ⎝ ⎠ (where k is radius of gyration) TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 ⎛ 2 ⎞ E = ET ⎜1+ 2 ⎟ ⎜ ⎟ ⎝ ⎠ ⎛ 2 ⎞ Similarly, E = ER ⎜1+ 2 ⎟ ⎜ ⎟ ⎝ ⎠ Type of body K Fraction of total energy ⎛ K 2 ⎞−1 translational X = ⎜1+ ⎟ ⎜ R 2 ⎟ ⎝ ⎠ Fraction of total energy ⎛ R 2 ⎞−1 rotational Y = ⎜1+ ⎟ ⎜ K 2 ⎟ ⎝ ⎠ Y X 1. Ring or hollow cylinder R 1 = 0.5 = 50% 2 1 = 0.5 = 50% 2 1: 1 2. Spherical Shell 2 R 3 = 0.6 = 60% 2 = 0.4 = 40% 2 : 3 3 5 5 3. Disc or solid cylinder R 2 2 = 0.666 = 66.67% 3 1 = 0.333 = 33.33% 3 1: 2 4. Solid sphere 2 R 5 = 0.714 = 71.4% 2 = 0.286 = 28.6% 2 : 5 5 7 7 Applications A force is applied at the distance h from centre of mass as shown in figure h F fr F ⎜1+ h ⎞ a = ⎝ R ⎠ ⎛ M⎜1+ ⎝ 2 ⎞ ⎟ R 2 ⎟ fr = F(K 2 − hR ) K 2 + R 2 K 2 ≤ μN (must be less than μmg for Rolling) If h < friction is backward R K 2 h > friction become forward R If force is applied at centre of mass then (h = 0) So, a = F ⎛ K 2 ⎞ and fr = FK 2 = K 2 + R 2 ⎛ F R 2 ⎞ M ⎜1+ ⎟ ⎝ ⎠ ⎜1+ ⎝ ⎟ K 2 ⎟ TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 If force is applied at highest point (h = R) a = 2F ⎛ K 2 ⎞ M⎜1+ ⎝ ⎟ R 2 ⎟ ⎛ R 2 − K 2 ⎞ ⎜ K 2 + R 2 ⎟ forward direction ⎝ ⎠ This case is equivalent to case when object is pulled from CM an Rough surface so a = mg m + I mg and T = mg R 2 R 2 1+ K 2 ⎛ ⎞ ⎜ M − M ⎟ 5. a = ⎜ 2 1 ⎟ g ⎜ M + M + I ⎟ ⎜ 2 ⎝ T2 > T1 M2 > M1 1 R 2 ⎟ a a = Mg M + I , α = a , m R R 2 I = mK2 a The rod is released from unstable equilibrium position u = 0 When at B, Mg L (1 + cos θ) = ⎛ 2 ⎞ ⎜ ⎟ω2 ω = at C, θ = 0° ⎜ ⎟ ⎝ ⎠ cos θ 2 ω = (c) at P, θ = 90°, ω = TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 Rolling of a Body on an Inclined Plane By conservation of energy mgh = 1 Iω2 + 1 mv 2 2 2 cm (Total energy) β = 1 + (Rotatory) I (Translatory) K 2 1+ Let MR 2 = (where k is radius of gyration) R 2 R a = mg sinθ = g sin θ f cm vcm = M + I R 2 = K 2 1+ R 2 ω h vcm θ Time = 1 . sinθ Force of friction f = mg sin θ R 2 1+ K 2 i.e., t ∝ Force of friction f = mg sin θ R 2 1+ K 2 Instantaneous power P = (mg sin θ)v ⎛ ⎞ ⎜ ⎟ 6. ⎟ ⎟ ⎠ ANGULAR MOMENTUM AND ITS CONSERVATION L = MvcmR + Iω Case - I : Pure translation | LO | = MVcmh LC = 0 z M h vcm O x TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 Case - II : Rolling Body Lc = Icω LA = ICω + MvcmR LO = Icω + Mvcmb O x Case - III : Pure rotation Put vcm = 0 in the above results so L0 = LA = LC = ICω Case - IV Lc = – Icω LA = – ICω + MvcmR LO = – Icω + Mvcmb Case - V Lc = Icω LA = ICω LO = Icω – Mvcmb vCM x Conservation of Angular Momentum In an isolated system (no external torque) the angular momentum of the system is conserved. ρ dL τext = dt If τext = O ⇒ dL ρ O dt ⇒ L = constant TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 Applications Before After m m Disc (moment of inertia I) L1 = Iω (Picks two particles at diametrically opposite points) L = (I + 2mR2)ω ′ ⇒ ω′ = Iω I + 2mR 2 Before After m θ B Maximum angle of inclination for pure rolling, θmax = tan− 1⎜ ⎜ R 2 ⎜ 1+ ⎝ K 2 Ring : Spherical Shell : Disc : Solid sphere : θmax = tan–1 (2μ), θmax = tan–1(2.5 μ) θmax = tan–1 (3μ) θmax = tan–1(3.5 μ). Moment of inertia of platform I m = mass of man at rest at edge w.r.t. platform v The man starts walking along the edge with speed v in opposite sense w.r.t. the platform then new angular velocity of the platform is ω ′= ω + mvR I + mR A rod is hinged at A and can rotate freely in a vertical plane. A particle of mass m moving with speed u hits and sticks to the rod. A (M) a m u L Only conservation of angular momentum can be applied as the rod is hinged at A. ⎡ ML2 2 ⎤ mua = ⎢ ⎢⎣ 3 + ma ⎥ ω ⎥⎦ ❑ ❑ ❑ TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616