{"id":38616,"date":"2022-12-03T22:20:22","date_gmt":"2022-12-03T16:50:22","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=38616"},"modified":"2022-12-03T22:20:22","modified_gmt":"2022-12-03T16:50:22","slug":"ts-inter-1st-year-physics-study-material-chapter-7","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-7\/","title":{"rendered":"TS Inter 1st Year Physics Study Material Chapter 7 Systems of Particles and Rotational Motion"},"content":{"rendered":"

Telangana TSBIE\u00a0TS Inter 1st Year Physics Study Material<\/a> 7th Lesson Systems of Particles and Rotational Motion Textbook Questions and Answers.<\/p>\n

TS Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion<\/h2>\n

Very Short Answer Type Questions<\/span><\/p>\n

Question 1.
\nIs it necessary that a mass should be present at the centre of mass of any system? [AP May. ’16; May ’14]
\nAnswer:
\nNo. It is not necessary to present some mass at centre of mass of the system.
\nEx: At the centre of ring (or) bangle, there is no mass present at centre of mass.<\/p>\n

Question 2.
\nWhat is the difference in the positions of a girl carrying a bag in one of her hands and another girl carrying a bag in each of her two hands?
\nAnswer:
\ni) a) When she carries a bag in one hand her centre of mass will shift to the side of the hand that carries the bag.
\nb) When a bag is in one hand some unbalanced force will act on her and it is difficult to carry.<\/p>\n

ii) If she carries two bags in two hands then her centre of mass remains unchanged. Force on two hands are equal i.e. balanced so it is easy to carry the bags.<\/p>\n

Question 3.
\nTwo rigid bodies have same moment of inertia about their axes of symmetry. Of the two, which body will have greater kinetic energy?
\nAnswer:
\nRelation between angular momentum and kinetic energy is, KE = \\(\\frac{L^2}{2I}\\)<\/p>\n

Because moment of inertia is same the body with large angular momentum will have larger kinetic energy.<\/p>\n

Question 4.
\nWhy are spokes provided in a bicycle wheel? [AP May ’14]
\nAnswer:
\nThe spokes of cycle wheel increase its moment of inertia. The greater the moment of inertia, the greater is the opposition to any change in uniform rotational motion. As a result the cycle runs smoother and speeder. If the cycle wheel had no spokes, the cycle would be driven in jerks and hence unsafe.<\/p>\n

Question 5.
\nWe cannot open or close the door by applying force at the hinges. Why? [AP May ’16]
\nAnswer:
\nTo open or close a door, we apply a force normal to the door. If the force is applied at the hinges the perpendicular distance of force is zero. Hence, there will be no turning effect however large force is applied.<\/p>\n

\"TS<\/p>\n

Question 6.
\nWhy do we prefer a spanner of longer arm as compared to the spanner of shorter arm?
\nAnswer:
\nThe turning effect of force, \u03c4 = \\(\\overline{\\mathrm{r}}\\times\\overline{\\mathrm{F}}\\). When arm of the spanner is long, r is larger. Therefore smaller force (F) will produce the same turning effect. Hence, the spanner of longer arm is preferred as compared to the spanner of shorter arm.<\/p>\n

Question 7.
\nBy spinning eggs on a table top, how will you distinguish an hard boiled egg from egg? [AP Mar. ’13]
\nAnswer:
\nTo distinguish between a hard boiled egg and a raw egg, we spin each on a table top. The egg which spins at a slower rate shall be a raw egg. This is because in a raw egg, liquid matter inside tries to get away from the axis of rotation. Therefore, its moment of inertia ‘I’ increases. As \u03c4 = I\u03b1 = constant, therefore, \u03b1 decreases i.e., raw egg will spin with smaller angular acceleration.<\/p>\n

Question 8.
\nWhy should a helicopter necessarily have two propellers?
\nAnswer:
\nIf the helicopter had only one propeller, then due to conservation of angular momen\u00actum, the helicopter itself would turn in the opposite direction. Hence, the helicopter should necessarily have two propellers.<\/p>\n

Question 9.
\nIf the polar ice caps of the earth were to melt, what would the effect of the length of the day be?
\nAnswer:
\nEarth rotates about its polar axis. When ice of polar caps of earth melts, mass concen\u00actrated near the axis of rotation spreads out. Therefore, moment of inertia ‘I’ increases.<\/p>\n

As no external torque acts, L = I\u03c9 = I(\\(\\frac{2 \\pi}{T}\\)) = constant<\/p>\n

with increase of I, T will increase i.e., length of the day will increase.<\/p>\n

\"TS<\/p>\n

Question 10.
\nWhy is it easier to balance a bicycle in motion?
\nAnswer:
\nA bicycle in motion is in rotational equilibrium. From principles of Dynamics of rotational bodies is that the forces that are perpendicular to the axis of rotation will try to turn the axis of rotation but necessary forces will arise it cancel these forces due to inertia of rotation and fixed position of axis is maintained. So it is easy to balance a rotating body.<\/p>\n

Short Answer Questions<\/span><\/p>\n

Question 1.
\nDistinguish between centre of mass and centre of gravity. [AP Mar. 18, 17, 16, 15, 14, 13, May 17; June 15 : TS Mar. 16. 15, May 18, 17]
\nAnswer:<\/p>\n\n\n\n\n\n\n\n\n
Centre of mass<\/td>\nCentre of gravity<\/td>\n<\/tr>\n
1) A point inside a body at which the whole mass is supposed to be concentrated.
\nA force applied at this point produces translatory motion.<\/td>\n		1) A point inside a body through which the weight of the body acts.<\/td>\n<\/tr>\n
2) It pertains (or) contain to mass of the body.<\/td>\n2) It refers to weight acting on all particles of the body.<\/td>\n<\/tr>\n
3) In case of small bodies centre of mass and centre of gravity coincide. (Uniform gravitational field)<\/td>\n3) In case of a huge body centre of mass and centre of gravity may not coincide. (Non uniform gravitational field)<\/td>\n<\/tr>\n
4) Algebraic sum of moments of masses about centre of mass is zero.<\/td>\n4) Algebraic sum of moments of weights about centre of gravity is zero.<\/td>\n<\/tr>\n
5) Centre of mass is used to study translatory motion of a body when it is in complicated motion.<\/td>\n5) Centre of gravity is used to know the stability of the body where it is to be supported.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"TS<\/p>\n

Question 2.
\nShow that a system of particles moving under the influence of an external force, moves as if the force is applied at its centre of mass. [AP May ’18]
\nAnswer:
\nConsider a system of particles of masses m1<\/sub>, m2<\/sub>, ……….. mn moves with velocity
\n\"TS
\nBut Force (F) = ma, so total force on the body is
\nF = MaC.M<\/sub> = m1<\/sub>a1<\/sub> + m2<\/sub>a2<\/sub> + m3<\/sub>a3<\/sub> + ……….. + mn<\/sub> an<\/sub>
\nor Total Force F = MaC.M<\/sub> = F1<\/sub> + F2<\/sub> + F3<\/sub> + ……… + Fn<\/sub><\/p>\n

Hence, total force on the body is the sum of forces on individual particles and it is equals to force on centre of mass of the body.<\/p>\n

Question 3.
\nExplain about the centre of mass of earth-moon system and its rotation around the sun.
\nAnswer:
\nThe interaction of earth and moon does not effect the motion of centre of mass of earth and moon system around the sun. The gravitational force between earth and moon is internal force. Internal forces cannot change the position of centre of mass.<\/p>\n

The external force acting on the centre of mass of earth and moon system is force between the sun and C.M. of earth, moon system. Motion of centre of mass depends on external force. Hence, earth moon system continues to move in an elliptical path around the sun. It is irrespective of rotation of moon around earth.<\/p>\n

Question 4.
\nDefine vector product. Explain the properties of a vector product with two examples. [AP Mar. ’17, ’15 ; TS Mar. ’17, ’16, ’15; APMay ’18. ’17; TS May ’18. ’16]
\nAnswer:
\nVector product (or) cross product :
\nIf the product of two vectors (say \\(\\overline{\\mathrm{A}}\\) and \\(\\overline{\\mathrm{B}}\\)) gives a vector then that multiplication of vectors is called cross product or vector product of vectors.
\n\"TS<\/p>\n

Properties of cross product:
\n1. Cross product is not commutative i.e.
\n\"TS
\n2. Cross product obeys distributive law i.e.,
\n\"TS
\n3. If any vector is represented by the combination of \\(\\overline{\\mathrm{i}},\\overline{\\mathrm{j}}\\) and \\(\\overline{\\mathrm{k}}\\) then cross product will obey right hand screw rule.
\n4. The product of two coplanar perpendicular unit vectors will generate a unit vector perpendicular to that plane
\n\"TS
\n5. Cross product of parallel vectors is zero
\n\"TS<\/p>\n

Examples of cross product:
\n1) Torque (or) moment of force (\\(\\overline{\\mathrm{\\tau}}\\)) :
\nIt is defined as the product of force and perpendicular distance from the point of application.
\n\u2234 Torque \u03c4 = \\(\\overline{\\mathrm{r}}\\times\\overline{\\mathrm{F}}\\)<\/p>\n

2) Angular momentum and angular velocity :
\nFor a rigid body in motion, Angular momentum (\\(\\overline{\\mathrm{L}}\\)) = radius (\\(\\overline{\\mathrm{r}}\\)) x momentum (\\(\\overline{\\mathrm{P}}\\))
\n\u2234 Angular momentum (\\(\\overline{\\mathrm{L}}\\)) = \\(\\overline{\\mathrm{r}}\\) \u00d7 (m\\(\\overline{\\mathrm{v}}\\)) = m(\\(\\overline{\\mathrm{r}}\\times\\overline{\\mathrm{v}}\\))<\/p>\n

Question 5.
\nDefine angular velocity. Derive v = r \u03c9. [TS Mar. 19,’ 17, 16; AP Mar. 19, May. 16; May 14]
\nAnswer:
\nAngular velocity (\u03c9) :
\nRate of change of angular displacement is called angular velocity.<\/p>\n

Relation between linear velocity (v) and angular velocity (\u03c9) :
\nLet a particle P is moving along circumference of a circle of radius r1 with uniform speed v. Let it is initially at the position A, during a small time \u2206t it goes to a new position say C from B. Angle subtended during this small interval is say d\u03b8.
\n\"TS<\/p>\n

By definition angular velocity,
\n\"TS<\/p>\n

Question 6.
\nState the principle of conservation of angular momentum. Give two examples.
\nAnswer:
\nLaw of conservation of angular momentum:
\nWhen no external torque is acting on a body then the angular momentum of that rota-ting body is constant.
\ni.e., I1<\/sub>\u03c91<\/sub> = I2<\/sub>\u03c92<\/sub> (when \u03c4 = 0)<\/p>\n

Example -1:
\nA boy stands over the centre of a horizontal platform which is rotating freely with a speed \u03c91<\/sub> (n1<\/sub>revolutions\/sec.) about a vertical axis passing through the centre of the platform and straight up through the boy. He holds two bricks in each of his hands close to his body. The combined moment of inertia of the system is say I1<\/sub>. Let the boy stretches his arms to hold the masses far away from his body. In this position the moment of inertia increases to I2<\/sub> and let \u03c92<\/sub> is his angular speed.<\/p>\n

Here \u03c92<\/sub> < \u03c91<\/sub> because moment of inertia increases.
\n\"TS<\/p>\n

Example – 2 :
\nAn athlete diving off a high spring board can bring his legs and hands close to the body and performs Somersault about a horizontal axis passing through his body in the air before reaching the water below it. During the fall his angular momentum remains constant.<\/p>\n

\"TS<\/p>\n

Question 7.
\nDefine angular acceleration and torque. Establish the relation between angular acceleration and torque. [TS Mar. ’18, ’17; TS May ’17, June ’15; AP Mar. ’19, June ’15]
\nAnswer:
\nAngular acceleration (\u03b1) :
\nRate of change of angular velocity is called angular acceleration.<\/p>\n

Torque :
\nIt is defined as the product of the force and the perpendicular distance of the point of application of the force from that point.<\/p>\n

Relation between angular acceleration and Torque:
\nWe know that, L = I\u03c9<\/p>\n

On differentiating the above expression with respect to time ‘t’
\n\"TS<\/p>\n

But \\(\\frac{dL}{dt}\\) is the rate of change of angular momentum called ‘Torque (\u03c4)”.<\/p>\n

and \\(\\frac{d \\omega}{dt}\\) is the rate of change of angular velocity called “angular acceleration (\u03b1)”<\/p>\n

\u2234 The relation between Torque and angular acceleration is, \u03c4 = l\u03b1<\/p>\n

Question 8.
\nWrite the equations of motion for a particle rotating about a fixed axis.
\nAnswer:
\nEquations of rotational kinematics :
\nIf ‘\u03b8’ is the angular displacement, Wj is the initial angular velocity, \u03c9f<\/sub> is the final angular velocity after a time ‘t’ seconds and ‘\u03b1’ is the angular acceleration, then the equations of rotational kinematics can be written as,
\n\"TS<\/p>\n

Question 9.
\nDerive expressions for the final velocity and total energy of a body rolling without slipping.
\nAnswer:
\nA rolling body has both translational kinetic energy and rotational kinetic energy. So the total K.E energy of a rolling body is,
\n\"TS<\/p>\n

Long Answer Questions<\/span><\/p>\n

Question 1.
\n(a) State and prove parallel axis theorem.
\n(b) For a thin flat circular disk, the radius of gyration about a diameter as axis is k. If the disk is cut along a diameter AB as shown into two equal pieces, then find the radius of gyration of each piece about AB.
\n\"TS
\nAnswer:
\na) Parallel axis theorem :
\nThe moment of inertia of a rigid body about an axis passing through a point is the sum of moment of inertia about parallel axis passing through centre of mass (IG<\/sub>) and mass of the body multiplied by Square of distance (MR\u00b2) between the axes i.e.,
\nI = IG<\/sub> + MR\u00b2<\/p>\n

Proof :
\nConsider a rigid body of mass M with ‘G’ as its centre of mass. Iq the moment of inertia about an axis passing through centre of mass. I = The moment of inertia about an axis passing through the point \u2019O\u2019 in that plane.<\/p>\n

Let perpendicular distance between the axes is OG = R (say)<\/p>\n

Consider point P in the given plane. Join OP and GP. Extend the line OG and drop a normal from \u2019P\u2019 on to it as shown in figure.
\n\"TS<\/p>\n

The moment of inertia about the axis passing through centre of mass G.
\n(IG<\/sub>) = \u2211mGP\u00b2 ……….. (1)<\/p>\n

M.O.I. of the body about an axis passing through ‘O’ (I) = \u2211mOP\u00b2 ………… (2)
\nFrom triangle OPD
\nOP\u00b2 = OD\u00b2 + DP\u00b2
\n\u21d2 OD = OG + GD
\n\u2234 OD\u00b2 = (OG + GD)\u00b2 = OG\u00b2 + GD\u00b2 + 2OG. GD ………….. (3)
\nFrom Equations (2) and (3)
\nI = \u2211mOP\u00b2 = Em [ (OG\u00b2 + GD\u00b2 + 2OG. GD) + DP\u00b2]
\n\u2234 I = \u2211m {GD\u00b2 + DP\u00b2 + OG\u00b2 + 20G. GN}
\nBut GD\u00b2 + DP\u00b2 = GP\u00b2
\n\u2234 I = \u2211m {GP\u00b2 + OG\u00b2 + 20G. GD}
\n\u2234 I = \u2211m GP\u00b2 + \u2211mOG\u00b2 + 20G \u2211mGD ………. (4)<\/p>\n

But the terms \u2211mGP\u00b2 = IG<\/sub>
\n\u2211mOG\u00b2 = MR\u00b2 (\u2235 \u2211m = M and OG = R)
\nThe term 20G \u2211mGD = 0. Because it represents sum of moment of masses about centre of mass. Hence its value is zero.
\n\u2234 I = IG<\/sub> + MR\u00b2<\/p>\n

Hence parallel axis theorem is proved.<\/p>\n

b) Moment of inertia of a disc of mass ‘M’ and radius ‘R’ about its diameter is,
\nI = \\(\\frac{MR^2}{4}\\)<\/p>\n

If ‘k’ is radius of gyration of disc then, I = Mk\u00b2
\n\u2234 Mk\u00b2 = \\(\\frac{MR^2}{4}\\) \u21d2 k = R\/2<\/p>\n

After cutting along the diameter, mass M of each piece = \\(\\frac{M}{2}\\)
\nMoment of inertia of each piece,
\n\"TS<\/p>\n

Question 2.
\n(a) State and prove perpendicular axis theorem.
\n(b) If a thin circular ring and a thin flat circular disk of same mass have same moment of inertia about their respective diameters as axis. Then find the ratio of their radii.
\nAnswer:
\na) Perpendicular axis theorem :
\nThe moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of the moment of inertia about two perpendicular axis concurrent with perpendicular axis and lying in the plane of the body.
\n\u2234 Iz<\/sub> = Ix<\/sub> + Iy<\/sub><\/p>\n

Proof :
\nConsider a rectangular plane lamina. X and Y are two mutually perpendicular axis in the plane. Choose another perpendicular axis Z passing through the point ‘O\u2019.
\n\"TS<\/p>\n

Consider a particle P in XOY plane.
\nIts co-ordinates are (x, y).
\nMoment of inertia of particle about
\nX-axis is IX<\/sub> = \u2211my\u00b2.
\nM.O.I about Y-axis is IY<\/sub> = \u2211mx\u00b2
\nM.O.I about Z axis is IZ<\/sub> = \u2211 m . OP\u00b2
\nFrom triangle OAP,
\nOP\u00b2 = OA\u00b2 + AP\u00b2 = y\u00b2 + x\u00b2
\n\u2234 Iz = \u2211 mOP\u00b2 = \u2211 m (y\u00b2 + x\u00b2)
\n\u2234 Iz = X my\u00b2 + \u2211 mx\u00b2
\nBut \u2211 my\u00b2 = Ix<\/sub> and \u2211 mx\u00b2 = Iy<\/sub><\/p>\n

\u2234 Moment of Inertia about a perpendicular axis passing through ‘O’ is IZ<\/sub> = IX<\/sub> + IY<\/sub>
\nHence perpendicular axis theorem is proved.<\/p>\n

b) Moment of inertia of a thin circular ring about its diameter is, I1<\/sub> = m1<\/sub> R\u00b21<\/sub><\/p>\n

Moment of inertia of a flat circular disc about its diameter is, I2<\/sub> = \\(\\frac{m_2R^{2}_{2}}{2}\\)
\nGiven that two objects having same moment of inertia i.e., I1<\/sub> = I2<\/sub><\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\nState and prove the principle of conservation of angular momentum. Explain the principle of conservation of angular momentum with examples. [AP Mar. ’16]
\nAnswer:
\nLaw of conservation of angular momentum: When no external torque is acting on a body then the angular momentum of that rotating body is constant.
\ni.e., I1<\/sub>\u03c91<\/sub> = I2<\/sub>\u03c92<\/sub> (when \u03c4 = 0)<\/p>\n

Explanation:
\nHere I1<\/sub> and I2<\/sub> are moment of inertia of rotating bodies and \u03c91<\/sub> and \u03c92<\/sub> are their initial and final angular velocities. If
\n\"TS<\/p>\n

Example -1 :
\nA boy stands over the centre of a horizontal platform which is rotating freely with a speed \u03c91<\/sub> (n1<\/sub> revolutions\/sec.) about a vertical axis passing through the centre of the platform and straight up through the boy. He holds two bricks in each of his hands close to his body. The combined moment of inertia of the system is say I1<\/sub>. Let the boy stretches his arms to hold the masses far away from his body. In this position the moment of inertia increases to I2<\/sub> and let \u03c92<\/sub> is his angular speed.<\/p>\n

Here \u03c92<\/sub> < \u03c91<\/sub> because moment of inertia increases.
\n\"TS<\/p>\n

Example – 2 :
\nAn athlete diving off a high spring board can bring his legs and hands close to the body and performs Somersault about a horizontal axis passing through his body in the air before reaching the water below it. During the fall his angular momentum remains constant.<\/p>\n

Position of centre of mass of some symmetrical bodies :<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
Shape of the body<\/td>\nPosition of centre of mass<\/td>\n<\/tr>\n
1. Hollow sphere (or) solid sphere<\/td>\nAt the centre of sphere<\/td>\n<\/tr>\n
2. Circular ring<\/td>\nAt the centre of the ring<\/td>\n<\/tr>\n
3. Circular disc<\/td>\nAt the centre of disc<\/td>\n<\/tr>\n
4. Triangular plate<\/td>\nAt the centroid<\/td>\n<\/tr>\n
5. Square plate<\/td>\nAt the point of intersection of diagonals<\/td>\n<\/tr>\n
6. Rectangular plate<\/td>\nAt the point of intersection of diagonals<\/td>\n<\/tr>\n
7. Cone<\/td>\nAt \\(\\frac{3h}{4}\\) th of its height from its apex on its own axis<\/td>\n<\/tr>\n
8. Cylinder<\/td>\nAt the midpoint of its own axis.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Comparison of translatory and rotatory motions :
\n\"TS<\/p>\n

Problems<\/span><\/p>\n

Question 1.
\nShow that a \u2022 (b \u00d7 c) is equal in magnitude to the volume of the parallelopiped formed on the three vectors a, b and c. (IMP)
\nSolution:
\nLet a parallelopiped be formed on three
\n\"TS
\nNow \\(\\hat{a}\\) \u2022 (\\(\\hat{b}\\times\\hat{c}\\) x c) = \\(\\hat{a}\\) \u2022 be \\(\\hat{n}\\) = (a) (be) cos 0\u00b0 – abc<\/p>\n

Which is equal in magnitude to the volume of parallelopiped.<\/p>\n

\"TS<\/p>\n

Question 2.
\nA rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope ? Assume that there is no slipping.
\nSoution:
\nHere M = 3 kg ; R = 40 cm = 0.4 m
\nMoment of inertia of the hollow cylinder about its axis, I = MR\u00b2 = 3(0.4)\u00b2 = 0.48 kg m\u00b2
\nForce applied, F = 30 N
\n\u2234 Torque, \u03c4 = F \u00d7 R = 30 \u00d7 0.4 = 12 N – m<\/p>\n

If \u03b1 is angular acceleration produced, then from \u03c4 = I\u03b1
\n\"TS
\nLinear acceleration, a = Ra = 0.4 \u00d7 25
\n= 10 ms-2<\/sup>.<\/p>\n

Question 3.
\nA coin is kept a distance of 10 cm from the centre of a circular turn table. If the coefficient of static friction between the table and the coin is 0.8. Find the frequency or rotation of the disc at which the coin will just begin to slip.
\nSolution:
\nDistance of coin = r = 10 cm = 0.1 m.
\nCoefficient of friction \u00b5 = 0.8.
\nFrequency of rotation = number of rotations per second.
\n\"TS<\/p>\n

Question 4.
\nFind the torque of a force \\(\\mathbf{7} \\overrightarrow{\\mathbf{i}}+\\mathbf{3} \\overrightarrow{\\mathbf{j}}-5 \\overrightarrow{\\mathbf{k}}\\) about the origin. The force acts on a particle whose position vector is \\(\\overrightarrow{\\mathbf{i}}-\\overrightarrow{\\mathbf{j}}+\\overrightarrow{\\mathbf{k}}\\). [AP Mar. ’14, ’13; May ’13]
\nAnswer:
\n\"TS
\n\"TS<\/p>\n

Question 5.
\nParticles of masses 1g, 2g, 3g….., 100g are kept at the marks 1 cm, 2 cm, 3 cm…, 100 cm respectively on a meter scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale.
\nSolution:
\nGiven Masses of 1g, 2g, 3g 100 g are 1, 2, 3 ……….. 100 cm on a scale.
\n\"TS
\ni) Sum of masses 2m = \\(\\sum_{i=1}^n\\)ni
\nSum of n natural numbers S
\n\"TS
\n\u2234 Total mass M = 5051 gr = 5.051 kg \u2192 (1)<\/p>\n

ii) Centre of mass of all these masses is given by
\n\"TS<\/p>\n

iii) M.O.I. = I
\n\"TS
\nSum of cubes of 1st n natural numbers is
\n\"TS<\/p>\n

M.O.I. about C.M. = IG<\/sub> = I – MR\u00b2
\n= 2.550 – 5.05 \u00d7 0.67 \u00d7 0.67 = 2.550 – 2.267 = 0.283 kg.m2<\/p>\n

iv) Perpendicular bisector is at 50 CM.
\nSo shift M.O.I from centre of mass to
\nx1<\/sub> = 50cm point from x = 67 CM
\n\u2234 Distance between the axis
\nR = 67 – 50 = 17cm = 0.17M
\nM.O.I. about this axis I = IG<\/sub> + MR\u00b2
\n= 0.283 + 5.05 \u00d7 0.17 \u00d7 0.17
\n= 0.283 + 0.146 = 0.429 kgm\u00b2
\n\u2234 M.O.I. about perpendicular bisector of scale = 0.429 kg – m\u00b2<\/p>\n

\"TS<\/p>\n

Question 6.
\nCalculate the moment of inertia of a fly wheel, if its angular velocity is increased from 60 r.p.m. to 180 r.p.m. when 100 J of work is done on it. [TS May ’16]
\nSolution:
\nW = 100 J, \u03c91<\/sub> = 60 RPM = 1 R.P.S = 2\u03c0 Rad.
\n\u03c92<\/sub> = 180 R.P.M. = 3 R.P.S = 6\u03c0 Rad.
\n\"TS<\/p>\n

Question 7.
\nThree particles each of mass 100 g are placed at the vertices of an equilateral triangle of side length 10 cm. Find the moment of inertia of the system about an axis passing through the centroid of the triangle and perpendicular to its plane.
\nSolution:
\nMass of each particle m = 100 g; side of equilateral triangle = 10 cm.
\nIn equilateral triangle height of angular bisector CD = \\(\\frac{\\sqrt{3}}{2}\\)l
\n\"TS<\/p>\n

Centroid will divide the angular bisector in a ratio 2 : 1<\/p>\n

So X distance of each mass from vertex to centroid is 2.\\(\\frac{\\sqrt{3}}{2}\\)l = \\(\\frac{\\sqrt{3}}{2}\\)l
\nMoment of Inertia of the system
\n\"TS<\/p>\n

Question 8.
\nFour particles each of mass 100g are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Find also the radius of gyration of the system.
\nSolution:
\nMass of each particle, m = 100 g = 0.1 kg.
\nLength of side’of square, a = 10 cm = 0.1 m
\nIn square distance of corner from centre of square = \\(\\frac{1}{2}\\) diagonal = \\(\\frac{\\sqrt{2}a}{2}=\\frac{a}{\\sqrt{2}}\\)
\n\"TS
\n\u2234 Total moment of Inertia
\n\"TS<\/p>\n

Question 9.
\nTwo uniform circular discs, each of mass 1 kg and radius 20 cm, are kept in contact about the tangent passing through the point of contact. Find the moment of inertia of the system about the tangent passing through the point of contact.
\nSolution:
\nMass of disc = M = 1 kg.
\nRadius of disc = 20 cm = 0.2 m
\n\"TS
\nThey are in contact as shown.
\nM.O.I of a circular disc about a tangent parallel to its plane = \\(\\frac{5}{4}\\) MR\u00b2
\nTotal M.O.I. of the system
\n\"TS<\/p>\n

Question 10.
\nFour spheres each diameter 2a and mass ‘m’ are placed with their centres on the four corners of a square of the side b. Calculate the moment of inertia of the system about any side of the square.
\nSolution:
\nDiameter of sphere = 2a \u21d2 radius = a.
\nSide of square = b.
\nFor spheres 1 and 2 axis of rotation is same and passing through diameters. M.O.I. of solid sphere about any diameter = \\(\\frac{2}{5}\\)MR\u00b2 (put M = m and R = a)
\n\"TS<\/p>\n

Transfer this M.O.I. on to the axis using
\nParallel axis theorem.
\n\"TS
\nTotal M.O.I. of the system
\n\"TS<\/p>\n

Question 11.
\nTo maintain a rotor at a uniform angular speed or 200 rad s-1<\/sup>, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine? (Note : uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
\nSolution:
\nGiven uniform angular speed (\u03c9) = 200 rad s-1<\/sup>
\nTorque, \u03c4 = 180 N – m ; But power p = \u03c4\u03c9
\n\u2234 P =180 \u00d7 200 = 36000 watt = 36 kW<\/p>\n

\"TS<\/p>\n

Question 12.
\nA metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
\nSolution:
\nLet m be the mass of the stick concentrated at C, the 50 cm mark
\n\"TS
\nFor equilibrium
\nabout C’, i.e. at the 45 cm mark,
\n10g (45 – 12) = mg (50 – 45)
\n10g \u00d7 33 = mg \u00d7 5
\nm = \\(\\frac{10\\times33}{5}\\) = 66 grams<\/p>\n

Question 13.
\nDetermine the kinetic energy of a circular disc rotating with a speed of 60 rpm about an axis passing through a point on its circumference and perpendicular to its plane. The circular disc has a mass of 5 kg and radius 1 m.
\nSolution:
\nMass of disc, M = 5 kg; Radius R = 1 m.
\nAngular velocity, co = 60 RPM = \\(\\frac{60\\times 2\\pi}{60}\\) = 2\u03c0Rad\/sec<\/p>\n

M.O.I. of disc about a point passing through circumference and perpendicular to the plane.
\n\"TS<\/p>\n

Question 14.
\nTwo particles, each of mass m and speed u, travel in opposite directions along para\u00acllel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momemtum is taken.
\nSolution:
\nAngular momentum, L = mvr
\nChoose any axis say ‘A’
\nLet at any given time distance between m1<\/sub> & m2<\/sub> = L = L1<\/sub> + L2<\/sub>
\nAbout the axis ‘A’ both will rotate in same direction See fig.
\n\"TS
\n\u2234 Total angular momentum
\nL = L1<\/sub> + L2<\/sub> = muL1<\/sub> + muL2<\/sub> = mu (L1<\/sub> + L2<\/sub>) = muL<\/p>\n

about any new axis say B distance of m1<\/sub> and m2<\/sub>
\nare say L’1<\/sub> and L’2<\/sub>
\nTotal angular momentum,
\nL = mu L’1<\/sub> + mu L’2<\/sub>
\nor L = mu(L’1<\/sub> + L’2<\/sub>) = muL (\u2235 L’1<\/sub> + L’2<\/sub> = L)
\nHence, total angular momentum of the system is always constant.<\/p>\n

Question 15.
\nThe moment of inertia of a fly wheel making 300 revolutions per minute is 0.3 kgm\u00b2. Find the torque required to bring it to rest in 20s.
\nSolution:
\nM.O.I, I = 0.3 kg. ; time, t = 20 sec.,
\n\u03c91<\/sub> = 300 R.P.M. = \\(\\frac{300}{60}\\) = 5. R.P.S.; \u03c92<\/sub> = 0<\/p>\n

Torque, \u03c4 = I\u03b1 = 0.3 \u00d7 \\(\\frac{5\\times 2\\pi}{20}\\) = 0.471 N – m.<\/p>\n

Question 16.
\nWhen 100J of work is done on a fly wheel, its angular velocity is increased from 60 rpm to 180 rpm. What is the moment of inertia of the wheel?
\nSolution:
\nW=100J, \u03c91<\/sub> = 60 RPM = 1 R.PS = 2\u03c0 Rad.
\n\u03c92<\/sub> =180 R.P.M. = 3 R.P.S = 6\u03c0 Rad.
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 17.
\nFind the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are lOOg, 150g and 200g respectively. Each side of the equilateral triangle is 0.5 m long, lOOg mass is at origin and 150g mass is on the X-axis. [TS Mar. 18, June 15; AP Mar. ’18]
\nSolution:
\nMass at A = 100g ; Coordinates = 0, 0
\nMass at B = 150 g; Coordinates = (0.5, 0)
\nMass at C = 200g; Coordinates (0.25,0.25 \u221a3 )
\nCoordinates xcm<\/sub>
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Telangana TSBIE\u00a0TS Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion Textbook Questions and Answers. TS Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion Very Short Answer Type Questions Question 1. Is it necessary that a mass should be present at the centre of … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[27],"tags":[],"yoast_head":"\nTS Inter 1st Year Physics Study Material Chapter 7 Systems of Particles and Rotational Motion - TS Board Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-7\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"TS Inter 1st Year Physics Study Material Chapter 7 Systems of Particles and Rotational Motion - TS Board Solutions\" \/>\n<meta property=\"og:description\" content=\"Telangana TSBIE\u00a0TS Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion Textbook Questions and Answers. 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