{"id":47356,"date":"2024-02-17T08:05:08","date_gmt":"2024-02-17T02:35:08","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=47356"},"modified":"2024-02-17T16:15:47","modified_gmt":"2024-02-17T10:45:47","slug":"plus-one-physics-chapter-wise-questions-answers-chapter-7","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/plus-one-physics-chapter-wise-questions-answers-chapter-7\/","title":{"rendered":"Plus One Physics Chapter Wise Questions and Answers Chapter 7 Systems of Particles and Rotational Motion"},"content":{"rendered":"
Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 1. Question 2. Answer: Question 3. Answer: 2. When a cat falls, it stretches its body. So that the moment of inertia becomes large. As I\u03c9 = constant, the value of angular speed will be decreased due to the increased value of moment of inertia. So cat lands on its feet gently.<\/p>\n Question 4. The accumulation of water in equatorial line will increase the moment of inertia I of earth. In order to keep the angular momentum as a constant, \u03c9 will decrease. The decrease in \u2018\u03c9\u2019will increase the length of day.<\/p>\n Question 5. Question 6. Question 7. Question 8. Question 1. Answer: Question 2. Answer: 2. The direction of torque is always along the axis of rotation of door.<\/p>\n 3. If the door handle is fixed at middle, more force must be applied to get maximum torque that is required to open the door.<\/p>\n Question 3. Answer: 2. A raw egg has more monemt of inertia than boiled egg. Hence raw egg spins more time than boiled egg.<\/p>\n Question 4. Question 1. 2. we know if torque acting on the body is zero, its angular momentum will be conserved Question 2. Answer: 2. Consider a rigid body rotating about an axis passing through the point O. Let co be the uniform angular velocity of the body. 3. Moment of inertia will be changed.<\/p>\n Question 3. Answer: 2. Angular velocity decreases.<\/p>\n 3. Initial kinetic energy K.E1<\/sub> = \\(\\frac{L^{2}}{2 I_{1}}\\) ____(1) Question 4. Answer: 2. Question 5. Answer: 2. Statement of conservation of angular momentum.<\/p>\n 3. I1<\/sub>\u03c91<\/sub> = I2<\/sub>\u03c92<\/sub> Question 1. Answer: 2. Moment of inertia of ring, 3. I = I0<\/sub> + ma2<\/sup> Question 2. Answer: 2. This statement is correct. 3. New volume of earth = 1\/4 original volume of earth Question 3. 2. Using theorem of perpendicular axis, the moment of inertia of the disc about an axis passing through the centre of the disc Question 4. Question 5. Question 6. Answer: Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 7 Systems of Particles and Rotational Motion Plus One Physics Systems of Particles and Rotational Motion One Mark Questions and Answers Question 1. The dimension of angular momentum is (a) M\u00b0L1T-1 (b) M1L2T2 (c) M1L2T-1 (d) M2L1T-2 Answer: (c) M1L2T-1 Angular momentum = Moment of […]<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[42728],"tags":[],"yoast_head":"\n
\nThe dimension of angular momentum is
\n(a) M\u00b0L1<\/sup>T-1<\/sup>
\n(b) M1<\/sup>L2<\/sup>T2<\/sup>
\n(c) M1<\/sup>L2<\/sup>T-1<\/sup>
\n(d) M2<\/sup>L1<\/sup>T-2<\/sup>
\nAnswer:
\n(c) M1<\/sup>L2<\/sup>T-1<\/sup>
\nAngular momentum = Moment of inertia x angular velocity
\n(Angular momentum) = [M1<\/sup>L2<\/sup>][T-1<\/sup>] = [M1<\/sup>L2<\/sup>T-1<\/sup>].<\/p>\n
\nIf the density of material of a square plate and a circular plate shown in figure is same, the centre of mass of the composite system will be
\n
\n(a) inside the square plate
\n(b) inside the circular plate
\n(c) at the point of contact
\n(d) outside the system
\nAnswer:
\n(a) inside the square plate<\/p>\n
\nWhy spokes are provided in by cycle wheel?
\nAnswer:
\nThis increases moment of inertia even when the mass is small. This ensures uniform speed.<\/p>\n
\nA ballet dancer, an acrobat and an ice skater make use of an important principle in physics. Which is that principle?
\nAnswer:
\nConservation of angular momentum.<\/p>\n
\nA cat is able to land on her feet after a fall. Which principle of physics is being used by her?
\nAnswer:
\nPrinciple of conservation of angular momentum.<\/p>\n
\nA body is rotating in steady rate. What is torque acting on the body?
\nAnswer:
\nZero. Torque is required only for producing angular acceleration.<\/p>\n
\nA flywheel is revolving with constant angular velocity. A chip of its rim breaks and flies away. What will be the effect on the angular velocity?
\nAnswer:
\nThe reduction in mass will decrease moments of inertia. Hence angular velocity will be increased in order to conserve angular momentum.<\/p>\n
\nIs radius of gyration of a body constant quantity?
\nAnswer:
\nNo. It changes with change in position of the axis of rotation.<\/p>\n
\nWhat is another name for angular momentum?
\nAnswer:
\nMoment of momentum.<\/p>\nPlus One Physics Systems of Particles and Rotational Motion Two Mark Questions and Answers<\/h3>\n
\nMoments of inertia of some bodies with axis are given in the table below. Fill in the blanks
\n
\nAnswer:
\n<\/p>\n
\nMatch the following:<\/p>\n\n\n
\n (a) Moment of force<\/td>\n \u03a4 \u2206 \u03b8<\/td>\n<\/tr>\n \n (b) F\u2206r<\/td>\n Linear motion<\/td>\n<\/tr>\n \n (c) Couple<\/td>\n Torque<\/td>\n<\/tr>\n \n (d) 1\/2 I\u03c92<\/sup><\/td>\n \u03a4 \u2206 r<\/td>\n<\/tr>\n \n <\/td>\n Rotational motion<\/td>\n<\/tr>\n \n <\/td>\n 1\/2 MR2<\/sup><\/td>\n<\/tr>\n \n <\/td>\n L2<\/sup>\/2I<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\n(a) Torque
\n(b) \u03c4 \u2206 \u03b8
\n(c) Rotational motion
\n(d) \\(\\frac{1}{2} \\frac{L^{2}}{1}\\).<\/p>\n
\nA cat is able to function its feet after a fall, taking the advantage of principle of conservation of angular momentum.<\/p>\n\n
\n1. When there is no external torque, the total angular momentum of a body or a system of bodies are a constant.
\n\u03c4 = \\(\\frac{d L}{d t}\\) (when \u03c4 = 0 , we get \\(\\frac{d L}{d t}\\) = 0).
\nie L = constant.
\nBut L = I\u03c9
\n\u2234 I\u03c9 = a constant.<\/p>\n
\nIf the polar ice cap melts what will happen to the length of the day?
\nAnswer:
\nFor earth, angular momentum is a constant (L\u03c9 = constant, ie no torque acts on the earth). When the polar ice cap melts, the water thus formed will flow down to the equatorial region.<\/p>\n
\nA girl has to lean towards right when carrying a bag in her left hand. Why?
\nAnswer:
\nWhen a girl carries her bag in her left hand, the centre of gravity of system will shift towards left. In order to bring it in the middle, the girl has to lean towards right.<\/p>\n
\nIf the earth loses the atmosphere what will happen to the length of the day?
\nAnswer:
\nFor earth, the angular momentum (L = I\u03c9) is a constant, because there is no torque acting on it. When earth loses the atmosphere, I decreases and \u03c9 increases to keep L as constant. Hence length of the day decreases.<\/p>\n
\nA girl standing on a turn table. What happens to the rotation speed, if she stretches her hand?
\nAnswer:
\nlf a girl rotating with a uniform speed on turn table, it\u2019s angular momentum (L = I\u03c9) will be a constant. When she suddenly stretches her hand, I Increases and \u03c9 decreases to keep L as constant.<\/p>\n
\nHow does a circus acrobat and a diver take advantage of conservation of angular momentum? Answer:
\nThe diver while leaving the spring board, is throwing himself in a rotating motion. When he brings his hands and legs close, I decrease and \u03c9 increases. But before reaching water he will stretch his hands and legs. Hence I increases and \u03c9 decreases. So, that he gets a smooth entry into the water.<\/p>\nPlus One Physics Systems of Particles and Rotational Motion Three Mark Questions and Answers<\/h3>\n
\nA rigid body consists of \u2018n\u2019 particles of mass m1, m2, m3,……The body rotates about an axis with
\nan angular velocity \u03c91, \u03c92, \u03c93……..<\/p>\n\n
\n1. Consider a body rotating about an axis passing through some point O with uniform angular velocity \u2018\u03c9\u2019. The body can be considered to be made up of a number of particles of masses m1, m2, m3……etc at distances r1, r2, r3……etc. All the particles will have same angular velocity \u03c9 But their linear velocities will be different say v1, v2, v3…….etc.
\n
\nK.E of 1st<\/sup> particle = \\(\\frac{1}{2}\\)m1<\/sub>v1<\/sub>2<\/sup>
\n\\(\\frac{1}{2}\\)m1<\/sub>(r1<\/sub>\u03c9)2<\/sup>
\n(\u2235 v = r\u03c9)
\nK.E of IInd<\/sup> particle = \\(\\frac{1}{2}\\)m2<\/sub>(r2<\/sub>\u03c9)2<\/sup>
\n\u2234 K.E of whole body =
\n
\nBut we know moment of inertia,
\n
\n\u2234 KE = \\(\\frac{1}{2}\\)I\u03c92<\/sup>
\n2. Rotation inertia is measured in terms of moment of inertia. Hence moment of inertia is also called rotational inertia.<\/p>\n
\nThe handle of a door is always found at one edge of the door which is located at a maximum possible distance away from hinges.<\/p>\n\n
\n1. Torque \u03c4 = r F sin \u03b8
\nFrom the above equation it is clear that, we get maximum torque when the handle of a door is located at a maximum possible distance (r) away from hinge.<\/p>\n
\nMoment of inertia depends on the mass, axis of rotation and distribution of mass of the body.<\/p>\n\n
\n1. Moment of inertia I = mr2<\/sup>
\nRadius of gyration K = \\(\\sqrt{\\frac{I}{m}}\\).<\/p>\n
\nTable below given analogy between translational and rotational motions. Match the following.
\n
\nAnswer:
\n<\/p>\nPlus One Physics Systems of Particles and Rotational Motion Four Mark Questions and Answers<\/h3>\n
\n1. Show that the total angular momentum of a rotating system remains constant if no torque acts on the system
\n
\n2. A disc of moment of inertia I1<\/sub> is rotating freely with angular speed \u03c91 and a second non rotating disc with moment of inertia I2<\/sub> is dropped on it as shown in the figure. The two then rotate as one unit. Find the angular speed of rotation of the system.
\nAnswer:
\n1. we know torque \u03c4 = \\( \\frac{d L}{d t}\\)
\nif \u03c4 = 0, we get \\( \\frac{d L}{d t}\\) = 0
\nie. L = constant.<\/p>\n
\nie. I1<\/sub>\u03c91<\/sub> = I2<\/sub>\u03c92<\/sub>
\nangular momentum of system, \u03c92<\/sub> = \\(\\frac{I_{1} \\omega_{1}}{\\left(I_{1}+\\mathrm{I}_{2}\\right)}\\).<\/p>\n
\nA rigid body can rotate an axis with a constant angular velocity and angular momentum L.<\/p>\n\n
\n1. L = I\u03c9
\nie. I = L\/\u03c9<\/p>\n
\n
\nThe body is imagined to be made up of large number of particles. Consider one such particle of mass \u2018m\u2019 at a distance \u2018r\u2019 from the axis of rotation.
\nLinear Velocity of the particle v = r\u03c9
\nK.E of the particle = 1\/2mv2<\/sup> = 1\/2 mr2<\/sup>\u03c92<\/sup>
\nK.E of whole body = \u03a31\/2mr2<\/sup>\u03c92<\/sup> = 1\/2\u03c92<\/sup>\u03a3mr2<\/sup>
\nK.E = 1\/2I\u03c92<\/sup>
\nWhere \u03a3mr2<\/sup> = I, moment of inertia of the body.<\/p>\n
\nA platform diver holds his hands and legs straight and makes loops in air before entering into water.<\/p>\n\n
\n1. Conservation of angular momentum
\nStatement<\/span>
\nConservation of angular momentum states that, if the total torque acting on a system is zero, its angular momentum will be conserved.<\/p>\n
\nwhere I1<\/sub> is the moment of inertia of diver when he makes loops in air
\nfinal kinetic energy K.E2<\/sub> = \\(\\frac{L^{2}}{2 I_{2}}\\) _____(2)
\nwhere I2<\/sub> is the moment of inertia of diver when he stretches his hands
\nBut I1<\/sub> < I2<\/sub>
\nHence from eq(1) and eq(2), we get
\nKE1<\/sub> > KE2<\/sub>
\nwhich means that rotational kinetic is not conserved.<\/p>\n
\nMoment of inertia of a thin ring of radius R about an axis passing through any diameter is 1\/2MR2<\/sup><\/p>\n\n
\n1.
\n<\/p>\n
\n<\/p>\n
\n
\nFigures show the two different spinning poses of a ballet dancer.<\/p>\n\n
\n1. The pose shown in figure (B).<\/p>\n
\n1\/2 I1<\/sub>2<\/sup>\u03c91<\/sub>2<\/sup> = 1\/2 I2<\/sub>2<\/sup>\u03c92<\/sub>2<\/sup>
\nI1<\/sub>(1\/2 I1<\/sub>\u03c91<\/sub>2<\/sup>) = (1\/2 I2<\/sub>\u03c92<\/sub>2<\/sup>)I2<\/sub>, I1<\/sub> > I2<\/sub>
\n(1\/2 I2<\/sub>\u03c92<\/sub>2<\/sup>) > 1\/2I1<\/sub>\u03c91<\/sub>2<\/sup><\/p>\nPlus One Physics Systems of Particles and Rotational Motion Five Mark Questions and Answers<\/h3>\n
\nThe moment of inertia of a thin ring of radius R about an axis passing through any diameter is \\(\\frac{1}{2}\\)MR2<\/sup><\/p>\n\n
\n
\n1. perpendicular axis theorem.<\/p>\n
\nI = mr2<\/sup> ____(1)
\nMoment of inertia of ring in terms of radius of gyration,
\nI = mk2<\/sup> ____(2)
\nFrom eq(1) and eq(2), we get
\nmk2<\/sup> = mr2<\/sup>
\nradius of gyration, k = r.<\/p>\n
\n<\/p>\n
\nThe earth is moving around the sun in an elliptical orbit and this brings out the seasons.<\/p>\n\n
\n1. Conservation of angular momentum.<\/p>\n
\nWhen earth approaches near the sun, its moment of inertia decreases. To maintain angular momentum as constant, the angular velocity increases.<\/p>\n
\n
\nTotal angular momentum before shrinking = Total momentum after shrinking.
\nie. I1<\/sub>\u03c91<\/sub> = I1<\/sub>\u03c92<\/sub>
\nmr1<\/sub>2<\/sup>\u03c91<\/sub> = m2<\/sub>2<\/sup>\u03c92<\/sub> ______(2)
\nSub eq(1) in eq(2)
\n
\nwe know period
\n
\nT2<\/sub> = 3 hr
\nDuration of day T2<\/sub> = 3 hr.<\/p>\n
\n1. Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR2<\/sup>\/5, where M is the mass of the sphere and R is the radius of the sphere.
\n2. Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR2<\/sup>\/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
\nAnswer:
\n1. Applying theorem of parallel axes, moment of inertia of sphere about a tangent
\n<\/p>\n
\n
\nUsing theorem of parallel axes, moment of inertia of the disc passing through a point on its edge and the normal to the disc
\n<\/p>\n
\nA solid cylinder of mass 20kg rotates about its axis with angular speed 100rad s-1<\/sup>. The radius of the cylinder is 0.25m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of the angular momentum of the cylinder about its axis?
\nAnswer:
\nM = 20kg
\nangularspeed, co – 100 rad s-1<\/sup>; R = 0.25m
\nMoment of inertia of the cylinder about its axis
\n= \\(\\frac{1}{2}\\)MR2<\/sup> = \\(\\frac{1}{2}\\) \u00d7 20 \u00d7 (0.25)2<\/sup>kgm2<\/sup> = 0.625kgm2<\/sup>
\nRotational kinetic energy,
\nEr = \\(\\frac{1}{2}\\)I\u03c92<\/sup> = \\(\\frac{1}{2}\\) \u00d7 0.625 \u00d7 (100)2<\/sup>J = 3125 J
\nAngular momentum,
\nL = I\u03c9 = 0.625 \u00d7 100 Js = 62.5 Js.<\/p>\n
\nA rope of negligible mass is wound round a hollow cylinder of mass 3kg and radius 40cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N? What is the linear acceleration of the rope? Assume that there is no slipping.
\nAnswer:
\nM = 3kg, R = 40cm = 0.4m
\nF = 30N, \u03b1 = ?, a = ?
\nThe torque acting on the cylinder,
\n\u03c4 = force \u00d7 radius of the cylinder = 30N \u00d7 0.4m = 12Nm
\nThe moment of inertia of the hollow cylinder about its axis,
\nI = MR2<\/sup> = 3kg \u00d7 (0.4)2<\/sup>m2<\/sup> = 0.48kgm2<\/sup>
\nNow, \u03c4 = I\u03b1, where \u03b1 is angular acceleration
\n\u2234 The angular acceleration of the cylinder, r 12
\n\u03b1 = \\(\\frac{\\tau}{I}=\\frac{12}{0.48}\\)rad s-2<\/sup> = 25 rad s-2<\/sup>
\nThe linear acceleration of the rope,
\na = R\u03b1 = 0.4 \u00d7 25m s-2<\/sup> = 10 m s-2<\/sup>.<\/p>\n
\nExplain why friction is necessary to make the disc in roll in the direction indicated.<\/p>\n\n
\nFriction is necessary for rolling.<\/p>\n\n
Plus One Physics Chapter Wise Questions and Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"