Q. No. The moment of inertia of a body does not depend on : Option The mass of the body Option The angular velocity of the body Option The axis of rotation of the body Option The distribution of the mass in the body Correct Answer I= miri I is independent of angular velocity of the body Q. No. Three point masses m, m and m are located at the vertices of an equilateral triangle of side a. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through m? Option Option Option Option Correct Answer ( m +m ) a ( m +m ) a ( m +m ) a ( m +m +m ) a a a I=m ( 0 ) +m +m = ( m +m ) a Q. No. A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius R is made from an iron plate of thickness t. Then the relation between the moment of inertia I X and I Y is : Option I Y = I X Option I Y = 6 I X Option I Y = I X Option I Y = 6 I X Correct Answer Moment of inertia of disc. I = MR I x = { T R t} R t I y = T ( R ) ( HR) I Y = 6 I X
Q. No. The ratio of the squares of radii of gyration of a circular disc and a circular ring of the same radius about a tangential axis is : Option : Option : 6 Option : Option : Correct Answer Moment of inertia I t I c + mr (about tangential axis) I Ring= MR =MK I Disc= M R =M K K = = K 6 {By perp. And parallel axis theorem} Q. No. Moment of inertia of a uniform annular disc of internal radius r and external radius R and mass M about an axis through its centre and perpendicular to its plane is : Option M ( R -r ) Option M ( R +r ) Option M( R +r ) ( R +r ) Option M ( R +r ) ( R -r ) Correct Answer I of given disc = I of disc of radius R -I of disc of radius r I= T( R ) R - T( r ) r Also, M=T( R - r ) I= T { R -r } = ( M ) ( R -r ) R -r = M( R +r )
Q. No. 6 For the same total mass which of the following will have the largest moment of inertia about an axis passing through the centre of gravity and perpendicular to the plane of the body? Option A disc of radius a Option A ring of radius a Option A square lamina of side a Option Four roads forming square of side a Correct Answer I Disc = Ma I Ring = Ma ( ) M a I Plate = = Ma 6 M( a) 6 I rods = +Ma = Ma Moment of inertia of rods of side a is largest. Q. No. If the radius of a solid sphere is cm, calculate the radius of gyration when the axis is along a tangent : Option 0cm Option cm Option cm Option cm Correct Answer I tangent=i c +MR = Ma +MR = MR MR =Mk K= ( ) = cm Q. No. 8 The moment of inertia of a straight thin rod of mass M, length L about an axis perpendicular to its length and passing through its one end is : Option ML Option ML Option ML Option ML Correct Answer ML I end = I c +Md = +M ML =
Q. No. 9 A closed tube partly filled with water lies in a horizontal plane. If the tube is rotated about perpendicular bisector, the moment of inertia of the system : Option Increases Option Decreases Option Remains constant Option Depends on sense of rotation Correct Answer Since, the tube is partly filled, on rotation the particles are thrown away from the axis due to centrifugal force. Since R increases for particles, moment of inertia of system increases. Q. No. 0 Two rings of same and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass of the ring = m, radius = r) Option mr Option mr Option mr Option mr I total = I Ring + I Ring In the given situation, the rings are perp. To each other and the axis is perp to ring. Thus the axis is in plane of the other ring. I =MR Ring I Ring = MR (by perp. Axis theorum) I total = mr Q. No. What is the moment of inertia I of a uniform solid sphere of mass M and radius R, privoted about an axis that is tangent to the surface on the sphere? Option MR Option MR Option 6 MR Option MR Correct Answer I tangent =I c+md =I c +MR
= MR +MR = MR Q. No. The moment of inertia of a solid cylinder of mass M, radius R and Length L about its axis Option ML Option MR Option MR L Option MR Correct Answer Consider the cylinder made up of stack of discs of radius R one over other. Let mass of an element (disc) is dm di Disc = dmr MR I= di Disc = dmr = Q. No. The moment of inertia of a meter stick of mass 00 gm, about an axis at right angles to the stick and located at 0 cm mark, is : Option 8. 0 g-cm Option Option 8g-cm.. 0 g-cm Option None of these Since, it s a meter stick, the mark is 0 cm from the center. I Mark = I c + Md ( ) 00 00 = +00 0 = 0 + 0 = 0 gm-cm Or. 0 gm-cm ( ) Q. No. The moment of inertia of a solid sphere about an axis passing through centre of gravity Option Option Option is MR ; then its radius gyration about a parallel axis at a distance R from first axis is : R R R
Option R Correct Answer By parallel axis the. I A = I C + Md ( ) = MR +M R = MR =MK K= R Q. No. Four sphere of diameter a and mass M are placed with their centres on the four corners of a square of side b. Then the moment of inertia of the system about an axis along one of the side of the square is : Option Ma +Mb Option 8 Ma +Mb Option 8 Ma Option Ma +Mb Correct Answer I = I + I + I + I = Ma + Ma Mb 8 = Ma +Mb Q. No. 6 For the given uniform square lamina ABCD, whose centre is O Option I AC =IEF Option I AD = IEF Option I AC =IEF Option I AC = IEF ML I ef =
ML I 0=I Ef+I E'f' = {By perp axis theorum due to symm. I Ef = I E f } 6 I AC + I BD = I O {By perp axis th. And. I AC = I BD (symm)} L I AD=I Ef +M ML I AD = I AC = I Ef { parallelaxistheroum } Q. No. Moment of inertia of a circular wire of mass M and radius R about its diameter is : Option MR Option MR Option MR Option MR Correct Answer I O = I x + I y {Perp axis th} MR = I x {I x = I y, symm} MR I x = Q. No. 8 One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moments of inertia about their diameters are respectively I A and I B such that Option I A =IB Option I > I Option I < I Option IA d = I d A A B I A= MR and I B = MR I < I B B A B A B
Q. No. 9 Three point masses, each of mass m, are placed at the corners of an equilateral triangle of side l. Then the moment of inertia of this system about an axis along one side of the triangle is : Option ml Option ml Option m l Option m l l I=M +m 0 +m 0 = ML ( ) ( ) Q. No. 0 Moment of inertia of a uniform circular disc about a diameter is I. Its moment of inertia about an axis perpendicular to its plane and passing a point on its rim will be : Option I Option I Option 6 I Option I MOI about diameter. MR I=I dia = {Perp axis th and symm} I A = I C + MR {Parallel axis th.} I A = MR I =6I A Q. No. The moment of inertia of a circular ring of radius R and mass M about a tangent in its plane is : Option MR Option MR Option MR Option MR
MR I dia = Perp axisth symm {. } I t=i dia +MR = MR Q. No. A wheel comprises of a ring of radius R and mass M and three spoke of mass m each. The moment of inertia of the wheel about its axis is : Option m M+ R Option (M + m) R Option (M + m) R Option M+m R Correct Answer I total = I Ring + I Spoke MR =MR + = (M + m) R Q. No. Four identical rods are joined end to end to form a square. The mass of each rod is M. The mass of each rod is M. The moment of inertia of the square about the median line is : Option Ml Option Ml Option Ml 6 Option Ml Correct Answer I total = I + I + I + I
L ML ML L =M + + +M ML ML = + = ML 6 ( I & I are for roods, axis passing through center and for < each particle is at dish L Q. No. When a steady torque or couple acts on a body, the body : Option Continues in a state of rest or of uniform motion by Newton s st law Option Gets linear acceleration by Newton s nd law Option Gets an angular acceleration Option Continues to rotate at a steady rate Torque rotates a body, thus provides an angular acceleration. Q. No. A uniform rod is kept on a frictionless horizontal table and two forces F and F are acted as shown in figure. The line of action of force F R (which produces same torque) is at a perpendicular distance C from O. Now F and F are interchanged and F is reversed. The new forces F R (which produces torque of same magnitude in the present case) has its line of action at a distance C from O. If the F R : FR in the ratio :, then a : b is ( assume F a Fb) > : Option F -F F -F Option F +F F -F Option F -F F +F Option F +F F +F Correct Answer Torque by f R = Net Torque by f and f fr ( C ) =fa+f b ()
f and f are interchange d and f is reversed. Torque by f R = Net torque by f and f. C f R =f a-f b () On dividing () and () fr fa+fb ( ) = = ( fr : f R = : ) f f a-fb R fa+fb=fa-fb (f - f )a = (f + f b) a f +f = b f -f Q. No. 6 Option Option Option Option What is the torque of force F =i -j +k acting at a point r =i +j +k about the origin? 6i -6j +k -6i +6j -k i -6j -k -i +6j +k F=r f i j k ( ) ( ) ( ) = =i 8+9 - j -6 +k -9- - =i-6 j-k Q. No. A cubical blocks of mass M and edge a slides down a rough inclined plane of inclination θ with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude : Option Zero Option Mga Option Mga sinθ Option Mga sinθ Correct Answer On a slope, N does not pass through the COM, it shifts in order to balance the torque due to friction about COM. Γ = Γ N J a Γ Γ N =magsin θ { J = magsin θ,asthevelisconst a=0}
Q. No. 8 A T-shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C. Option l Option l Option l Option l Correct Answer If an object is free to rotate, it rotates about its COM. If the force only causes translational motion, it acts on COM of body. m( l ) +m( l) l COM= = m Q. No. 9 An equilateral prism of mass m rests on rough horizontal surface with coefficient of friction µ. A horizontal force F is applied on the prism as shown in the figure. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is : Option Option Option mg mg µ mg Option µ mg Correct Answer The block will topple about A. and will only topple if the torque by f is more than
torque by mg. Γ Γ f f mg mg Q. No. 0 The driving side belt has a tension of 600 N and the slack side has 00 N tension. The belt turns a pulley 0 cm in radius at a rate of 00 rpm. This pulley drives a dynamo having 90% efficiency. How many kilowatts are being delivered by the dynamo? Option. Option 6. Option.8 Option. Correct Answer Work done on pulley = net work done by torque of 600N. and 00N. net ( ) w= Γ d θ = Γ θ dw dθ Power= = Γ net = Γ net ( w) dt dt ( ) ( 00 =00 0. ) =.0 0 6 The efficiency is 90% Power delivered = 0.9 (.8) kw =. kw. Q. No. The angular velocity of a wheel increases from 00 rps to 00 rps in 0 s. Then number of revolutions made during that time is : Option 600 Option 00 Option 000 Option 000 Correct Answer w 0 = 00, w f. = 00, t = 0s. w f =w 0 + θ w f =w 0 + t 00 = 0 =0 ( ) ( 00 ) = ( 00 ) +( 0)( θ) 8x 0 =0θ θ =000 Q. No. When a ceiling fan is switched off, its angular velocity falls to half while it makes 6 rotations. How many more rotations will it make before coming to rest? Option Option 6 Option 8 Option
Correct Answer Assume, a constant avg. angular retardation w 0 = w w w = ( ) θ =6 0 0 w =w + θ w - = 6 Now, forom this point to the moment, the fan stops. w w = w f = 0 ( )( ) w 0= + θ w θ θ = = 8 No. of rotati ons= = Q. No. A rigid body rotates about a fixed axis with variable angular velocity equal to α - βt β at time t where α and β are constants. The angle through which it rotates before it comes to rest is : Option α β Option α - β α Option α - β β Option α ( α - β ) Correct Answer w= - β t At t = 0 w= it starts with angular velocity for stopping, final angular velocity w t = 0. dw Angular acc n = =- β dt w -w = θ t 0 ( ) - = - β θ θ= β
Q. No. A wheel is subjected to uniform angular acceleration about its axis. Initially, its angular velocity is zero. In the first s, it rotates through an angle θ, in the next s, it rotates through an angle θ. The ration of θ θ Option Option Option Option Angle rotated in sec. = ( ) = = θ Angle rotated in sec = ( ) =8 Angle rotated between to sec =6 = θ θ =. θ is : Q. No. The linear velocity of a particle on the equator is nearly (radius of the earth is 000 miles) : Option Zero Option 0 mile/hr Option 00 mile/hr Option 000 mile/hr Correct Answer Linear velocity, ν =wr ( ) 000 = 000miles/ hr Q. No. 6 A stone of mass m is tied to a string of length L and rotated in a circle with a constant speed v ; if the string is released the stone files : Option Radially outward Option Radially inward Option Tangentially Option With an acceleration mv L For a particle, moving in a circle, is velocity is always tangential. Therefore, when the string is cut, it will fly off tangentially. Q. No. A disc of radius r rolls on a horizontal ground with linear acceleration a and angular acceleration α as shown in figure. The magnitude of acceleration of point P shown in figure at an instant when its linear velocity is v and angular velocity is ω will be : Option Option ( a+r α) + ( r ω ) ar R α ω
Option r α +r ω Option r α Correct Answer The particle has accelerations, tangential and radial. Net acceleration = a+r ( ) + ( rw ) Q. No. 8 An electric fan has blades of length 0 cm as measured from the axis of rotation. If the fan is rotating at 00 rpm, the acceleration of a point on the tip of a blade is about : Option 0 m/sec Option 0 m/sec Option 600 m/sec Option 0 m/sec Correct Answer =0, a=w R 00 = 0. 0m/sec 60. ( ) ( ) Q. No. 9 A flywheel of mass 0 kg and radius of gyration about its axis of rotation of 0. m is acted upon by a constant torque of. Nm. Its angular velocity at t = sec is : Option. rad/sec Option rad/sec Option. rad/sec Option 0 rad/sec Correct Answer Moment of inertia, I = MK = 0 (0.) =. Torque=Iα. =. ( α ) After sec. w=0+ α t =rad/sec. α=rad/sec Q. No. 0 A uniform meter stick of mass M is hinged at one end and supported in a horizontal direction by a string attached to the other end. What should be the initial acceleration (in rad/sec ) of the stick of the string is cut? Option g Option g Option g Option g Correct Answer When the string is cut, torque is provide by its weight. Γ =I
L ML mg = = g L ( ) Q. No. A thin hollow cylinder is free to rotate about its geometrical axis. It has a mass of 8 kg and a radius of 0 cm. A rope is wrapped around the cylinder. What force must be exerted along the rope to produce an an angular acceleration of rad/sec? Option 8. N Option.8 N Option.8 N Option None of these MOI for follow cylinder, I = MR = f( R ) =I f ( 0. ) =0.( ) =.8 N. = 8(0.) 0. Q. No. In the pulley system shown, if radii of the bigger and smaller pulley are m and m respectively and the acceleration of block A is m/s in the downward direction, then the acceleration of block B will be : Option 0 m/s Option m/s Option 0 m/s Option m/s Correct Answer In the given problem, both the pulley will rotate by the same angle in any interval of time. a a A= B A = r r A rb a B=a A = = ms r A B
Q. No. Figure shows a uniform rod of length l and mass M which is pivoted at end A such that it can rotate in a vertical plane. The free end of the rod B is initially vertically above the pivot and then released. As the rod rotates about A, its angular θ is Option Option Option g cos θ l g tanθ l g sin θ l Option g sinθ l Correct Answer Torque is only provided by Mg. Γ Mg =I l Mg cosθ = Ml = gcosθ l Q. No. In the above question, the end B of the rod will hit the ground with a linear speed : Option gl Option Option gl gl Option g l Taking the rod + earth as a system. The only external force is hinge force, but it does no work, as hinge has 0 displacement. W = K.E+ P.E ext l 0= Iw -0+0-Mg ML l w =Mg
g w= l Linear velocity of B, V B = wl = gl Q. No. A uniform rod of mass M and length L lies radially on a disc rotating with angular speed ω in a horizontal plane about its axis. The rod does not slip on the disc and the centre of the rod is at a distance R from the centre of the disc. Then the kinetic energy of the rod is : Option L mω R + Option m R ω Option m ω L Option None of these Correct Answer The rod is rotating with angular speed w, w.r.t the center of the disc K.Eofrod= IC w ML I c = +MR (By parallel axis theorum) ML K.E= +MR w Q. No. 6 A uniform rod of length L is free to rotate in a vertical plane about a fixed horizontal axis through B. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle θ its average angular velocity ω is given as : Option Option Option Option 6g sin θ L 6g θ sin L 6g θ cos L 6g cos θ L
Correct Answer Taking rod + earth as a system. w ext = K.E+ P.E l l 0= Iw -0+Mg cosθ-mg ML l w =Mg ( -cosθ) g θ -cos θ w= ( -cos θ) = g sin l l =sin θ Q. No. A string of negligible thickness is wrapped several times around a cylinder kept on a rough horizontal surface. A man standing at a distance from the cylinder holds one end of the string and pulls the cylinder towards him. There is no slipping anywhere. The ratio of length of the string passed through the hand of the distance moved by centre of mass of cylinder is Option Option Option Option Correct Answer Since, there is no slipping anywhere if the vel. Of center of cylinder is ν, then vel of top point (and thread) is ν. Dist travelled by center = ν t=d Dist travelled by center =ν t=l l = d Q. No. 8 A solid sphere of mass M and radius R is placed on a rough horizontal surface. It is pulled by a horizontal force F acting through its centre of mass as a result of which it begins to roll without slipping. Angular acceleration of the sphere can be expressed as: Option F MR Option Option Option Correct Answer F MR F MR F MR For linear motion F - f = Ma
For rotational Motion Γ =I and a f( R ) = MR a R = R f= Ma F F= Ma a= M a F = = R MR Q. No. 9 A sphere cannot roll on : Option A smooth horizontal surface Option A rough horizontal surface Option A smooth inclined surface Option A rough inclined surface For a body to roll,a= R. In option (c), the body will have a but there is no force to provide torque and thus = 0. This situation cannot satisfya= R. The body cannot roll in this case. Q. No. 0 A hoop rolls on a horizontal ground without slipping with linear speed v. Speed of a particle P on the circumference of the hoop at angle θ is : Option Option Option vsin θ vsinθ vcos θ Option vcosθ Correct Answer The particle has two velocities, one due to linear motion of COM and the other due to rotation. ( ) ( ) ( ) ν = V + WR +V WR cos - θ net = V +V (-cosθ) { V=WR } =V ( -cosθ ) θ -cos =V sin =sin θ θ
=vsin θ Q. No. A sphere of mass m rolls without slipping on an inclined plane of inclination θ. The linear acceleration of the sphere is : Option gsin θ Option gsin θ Option gsin θ Option gsin θ Correct Answer For linear motion. F = Ma. Mgsinθ -f=ma For rotational motion Γ =I a a fr= MR = R R Mgsin θ = Ma a= gsin θ Q. No. In the above question, the force of friction on the sphere is : Option Mgsin θ Option Mgsin θ Option Mgsin θ Option Mgsin θ Correct Answer f= Ma= Mgsin θ Q. No. In the above question, the minimum value of coefficient of friction so that sphere may roll without slipping is : Option sin θ Option cos θ
Option tan θ Option cot θ f = µ N Mgsin θ = µ Mgcos θ µ µ = tan θ Q. No. A hoop rolls without slipping down an incline of slope 0 0. Linear acceleration of its centre of mass is Option g Option g Option g Option g 6 0 θ=0 Mgsin θ -f=ma a fr=mr R Mgsin θ =Ma gsinθ g a= = Q. No. A 6 kg ball starts from rest and rolls down a rough gradual slope until it reaches a point 80 cm lower than its starting point. Then the speed of the ball is : Option.9 ms - Option. ms - Option. ms - Option.8 ms - By taking ball + earth as a system. w = K.E+ P.E ext 0= MV + IW + (-Mgh) v MV w=,i = MR =Mgh R and,k.eistranslational + rotational V =gh 0
0( 0.8) 0 V= =.m/s Q. No. 6 A uniforms solid sphere rolls on a horizontal surface at 0 ms -. It then rolls up an incline having an angle of inclination at 0 o with the horizontal. If the friction losses are negligible, the value of height h above the ground where the ball stop is : Option. m Option 8.6 m Option. m Option 9.8 m Correct Answer Again by taking ball + earth as a system. w = K.E+ P.E ext v =mgh 0 ( ) ( ) 0 h= =8.6m 0 98 {By above question} Q. No. A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity v m/s. If it is to climb the inclined surface then v should be : Option Option Option gh 0 gh gh Option 0 gh Correct Answer Since, the surface is smooth, only translational K.E cost in climbing up the height. k.e P.E MV Mgh v gh Q. No. 8 A disc is rolling on an inclined plane. What is the ratio of its rotational K.E. to the total K.E.? Option : Option : Option : Option : Correct Answer Total kinetic energy, K.E T = K.E rotational + K.E translational V = MR + MV R
= MV Rotational K.E= ( Iw ) = MR w = MV R K.E = = K.E T Q. No. 9 A spherical a ball of mass 0 kg is stationary at the top of hill of height 00 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 0 m and finally rolls down to a horizontal base at a height of 0 m above the ground. The velocity by the ball is : Option 0 m/s Option 0 m/s Option 0 m/s Option 0 0m/s Correct Answer Since the surface is smooth, only translational K.E is gained by the ball. K.E = P.E MV = Mg h f -h i ( ) v= ( 0)( 80 ) =0m/s Q. No. 60 Figure shows a hemisphere of radius R. A ball of radius R is released from position P. It rolls without slipping along the inner of the hemisphere. Linear speed of its centre of mass when the ball is at position Q is : Option Option Option Option Correct Answer 0gR gr 0gR 9 6gR Again, taking ball + hemisphere + earth as system. K.E = P.E ( w =0) MV + MV =Mg R by ( ) ( θ ) ext
v= 0gR Q. No. 6 If a spherical ball rolls on a table without slipping, the fraction of its total energy associated with rotation is Option Option Option Option Correct Answer K.E T = MV 0 Rotational K.E= MV K.ER = = K.E T 0 Q. No. 6 A particle of mass m is projected with a velocity v making an angle of 0 with the horizontal. The magnitude of angular momentum of the projectile about an axis of projection when the particle is maximum height h is : Option Zero Option mv Option Option Correct Answer g mv g m gh At max height, the particle has horizontal velocity u sin θ v And is at height H= = g g Angular momentum, L=m r ν =mh ν mv = g v vcos θ =
Q. No. 6 A particle of mass m = units is moving with a uniform speed v= m in the XOY plane along the line Y = X +. The magnitude of the angular momentum of the particle about the origin is : Option Zero Option 60 unit Option. unit Option 0 unit Correct Answer y=x+ Slope of line = (by y = mx + c) And intercept = θ= 0 0 v x =vcos and v y = vsin 0 v x = and v y = Angular momentum, L=m R ν = m (xv y ) + m y v x ) = (0)() + ()() = 60 units Q. No. 6 A particle is moving along a straight line parallel to x- axis with constant velocity. Its angular momentum about the origin: Option Decreases with time Option Increases with time Option Remains constant Option Is zero v x = v, v y = 0 L =m R ν =mxv y+myvx =my ν = constant ( Particle travels parallel to x axis. Y does not change, v k m) Q. No. 6 If a particle moves in the X-Y plane, the resultant angular momentum has : Option Only x-component Option Only y-component Option Both x and y component Option Only z-component Correct Answer Angular momentum, L =m r ν L Is perp to both rand ν. Now, since motion is in x - y plane L is in z direction. Q. No. 66 A constant torque acting on a uniform circular wheel changes its angular momentum from A 0 to A 0 in seconds. The magnitude of this torque is :
Option A0 Option A 0 Option A 0 Option A 0 Correct Answer L i = A 0 L j = A 0. L=Lj -L i =A0 dl L A Γ= = = dt t 0 Q. No. 6 A particle moves in a force field by : F =rf(r), where r is a unit vector along the position vector r, then which is true? Option The torque acting on the particle is not zero Option The torque acting on the particle produces an angular acceleration in it Option The angular momentum of the particle is conserved Option The angular momentum of the particle increases f =fr Torque, Γ = r f =0 r &f areparallel dl Since, Γ =0 =0 L is constant dt Angular momentum is conserved. Q. No. 68 A rigid body rotates with an angular momentum L. If its rotational kinetic energy is made times, its angular momentum will become : Option L Option 6L Option L Option L Correct Answer L=Iw and k.e= Iw L k.e= I L= ( I)( k.e ) If k.e is mode times, angular momentum is doubled. L j =L Q. No. 69 The position of a particle is given by : r = i +j -k and its linear momentum is given by : P =i +j -k. Then its angular momentum, about the origin in perpendicular to: Option YZ plane Option z-axis Option y-axis Option x-axis Correct Answer
Angular momentum, L = r P I J K = - - ( ) ( ) ( ) =I -+ -J -+ +K -6 =0I - J-K L is perpendicular to x-axis direction. Q. No. 0 Option Option If the radius of earth contracts of its present day value, the length of the day will be n approximately : h n h n Option n h Option n h Correct Answer If the radius of earth changes, its moment of inertia changes and since there is no external torque the angular momentum of earth remains conserved. I w =I w R MR = M n T T= h n w= T Q. No. A disc of moment of inertia I is rotating freely with angular velocity ω when a second, non-rotating disc with moment of inertia I is dropped on it gently the two then rotate as a unit. Then the total angular speed is : Option I ω I Option Option I ω I I ω I +I Option ( I +I ) I ω Taking both the disc as a system, there is no external torque and thus angular momentum remains conserved. I ω =i ω i i f f ( ) 0+I ω = I +I I w = I +I ω ω
Q. No. A thin circular ring of mass M and radius R is rotating about its axis with a constant angular velocity ω. Two objects, each of mass m, are attached gently to the opposite ends of a diameter of the ring. The ring rotates now with an angular velocity : Option ωμ M+m Option ω( Μ-m) Option M+m ωμ M+m Option ω( Μ+m) M Initial moment of interia, I i = MR After attaching the masses, MOI, I f = MR + MR (Ring + masses) Taking ring and both the masses as a system. External torque = 0, thus L i = L f MR ω = M+m R ω ω new ( ) ωm = M+m new Q. No. If a gymnast, sitting on a rotating stool with his arms outstretched, suddenly lowers his arms : Option The angular velocity increases Option His moment of inertia increases Option The angular velocity remains same Option The angular momentum increases Correct Answer When the gymnast lowers his stretched arms, the mass are brought nearer to the axis and thus his MOI decreases further, as there is no external torque, angular momentum must conserved. Iω = K Thus, if I decrease, w increases. Q. No. A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to the plane with angular velocity ω. Another disc of same mass but half the radius is gently placed over it coaxially. The angular speed of the composite disc will be : Option ω Option ω Option ω Option ω Correct Answer Moment of inertia of rotating disc, I = MR R MR Moment of inertia of second disc, I = M = 8
Γ ext Taking both the disc as a sys. =0 i L =L ( ) I ω = I +I j ω i i new MR MR ω = { ω new } + ω new = ω