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Physics NEET MCQ
Centre Of Mass And Rotational Motion. Mcq
Quiz 10
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Q.1
A gramophone record is revolving with an angular velocity ω. A coin is placed at a distance r from the centre of the record. The static coefficient of friction is µ The coin will revolve with the record if: [ CBSE-PMT 2010]
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a) r = µgω 2
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b) r < ω2 / µg
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c) r ≤ µg / ω2
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d) r ≥ µg / ω2
Explanation
For the coin to revolve with the record, centrifugal force ≤ frictional force ⇒ mrω2 ≤ µ mg ∴ r ≤ µg / ω2 Answer: (c)
Q.2
A circular disk of moment of inertia It is rotating in a horizontal plane, its symmetry axis, with a constant angular speed ωi Another disk of moment of inertia Ib is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ωf. The energy lost by the initially rotating disk to friction is: [ CBSE-PMT 2010]
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a)
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b)
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c)
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d)
Explanation
Here Initial Angular momentum= It ω i After placing another disc Final angular momentum is= (It + Ib ) ω f By conservation of angular momentum we get It ω i = (It + Ib ) ω f ∴ ωf = Now loss in K.E ΔK = Initial K.E - Final K.E Now by substituting the value of ωf in above equation we get Answer:(d)
Q.3
A solid sphere, disc and solid cylinder all of the same mass and made of the same material are allowed to roll down (from rest) on the inclined plane, then [1993]
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a) solid sphere reaches the bottom first
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b) solid sphere reaches the bottom last
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c) disc will reach the bottom first
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d) all reach the bottom at the same time
Explanation
For solid sphere For disc and solid cylinder Since acceleration inversely proportional to ratio of K2 / R2 for sphere it is less thus Solid sphere will take minimum time to reach the bottom of the incline Answer: (a)
Q.4
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is [2004]
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a) 1 : √2
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b) 1:3
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c)2:1
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d)√5: √6
Explanation
Moment of Inertia around Axis passing through centre of mass Iy=MR2 /4 Now according parallel axis theorem momentum of Inertia around the tangential axis is I'y= K12 = 5/4 For circular ringMoment of Inertia around Axis passing through centre of mass Iy1=MR2 /2 Now according parallel axis theorem momentum of Inertia around the tangential axis is I'y1= K22 = 3/2 Answer: (d)
Q.5
A boy suddenly comes and sits on a circular rotating table. What will remain conserved?[ CBSE-PMT 2002]
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a) Angular velocity
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b) Angular momentum
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c)Linear momentum
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d) Kinetic energy
Explanation
As net torque applied is zero Hence angular momentum is conservedAnswer: (b)
Q.6
A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom ? [2003]
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a)
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b)
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c)
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d)
Explanation
energy of solid cylinder at the bottom is kinetic energy=Liner K.E + Rolling K.E Total K.E=½ (I ω2 ) + ½ (m v2) Now I=MR2 / 2 and ω2=v2 / R 2 Now Potential energy=K.E mgh=( 3/4) m v2 on solving we get v= Answer:(d)
Q.7
Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side l cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular to A.B and in the plane of ABC, in gram-cm2 units will be [ CBSE-PMT 2004]
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a)
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b)
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c)
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d)
Explanation
Now Moment of Inertia along x-axis be IAX=m(AB)2 + m(OC)2 IAX=ml2 + m(l cosθ)2IAX=ml2 + ml2 / 4 IAX=(5/4) ml2Answer: (d)
Q.8
A weightless ladder 20 ft long rests against a frictionless wall at an angle of 60° from the horizontal. A 150 pound man is 4 ft from the top of the ladder. A horizontal force is needed to keep it from slipping. Choose the correct magnitude of the force from the following [ CBSE-PMT 1998]
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a)175lb
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b) 100lb
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c)120lb
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d)69.21lb
Explanation
Ab is the ladder, F be the horizontal force and W is the weight of man. Let N1 and N2 be normal reactions of ground and wall, respectively. Then for vertical equilibrium W=N1 --eq(1)For horizontal equilibrium, N2=F --eq(2)Since ladder is in rotational equilibriumTaking moments about AN2 ( ABsin60) - W(ACcos60)=0 --eq(3) using (2) and AB=20 ft, BC=4ft we getAnswer: (d)
Q.9
A constant torque of 1000 N-m turns a wheel of moment of inertia 200 kg-m2 about an axis through its centre. Its angular velocity after 3 seconds is [ CBSE-PMT 2001]
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a) 1 rad/s
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b) 5 rad/s
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c) 10 rad/s
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d) 15 rad/s
Explanation
Now τ = I α τ = 1000 N-m, I 200kg-m2 ∴ 1000 = 200 α ∴ α = 5 rad/sec2 ω = ω0 + α t ω = 0 + 3 × 5 = 15 rad/sec Answer: (d)
Q.10
A wheel having moment of inertia 2 kg-m2 about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel's rotation in one minute would be [ CBSE-PMT 2004]
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a) (π/18) N-m
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b) (2π/15) N-m
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c) (π/12) N-m
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d) (π/15) N-m
Explanation
torque = rate of change of momentum Now ωf = 0, and time interval dt = 1 min = 60 Initial angular velocity = 60 × 2 π / 60 sec by substituting we get Answer:(d)
Q.11
Centrifugal force is inertial force when considered by
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a) an observer at the centre of circular motion
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b) an outside observer
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c)an observer who is moving with the particle which experiencing the force
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d)none of the above
Explanation
Answer: (c)
Q.12
Two bodies of mass 1 kg and 3 kg have position vectors i + 2j + k and -3i - 2j + k respectively. The center of mass of this system has a position vector: [ CBSE-PMT 2009]
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a) -2i - j + k
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b) 2i - j - 2k
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c)-j + j + k
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d)-2i + 2k
Explanation
The position vector of center of mass of two particle system is given byAnswer: (a)
Q.13
The instantaneous angular position of a point on a rotating wheel is given by the equation a(t)=2t3 - 6tThe torque on the wheel becomes zero at [ CBSE-PMT 2011]
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a)t=1s
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b) t=0.5s
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c)t=0.25s
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d)t=2s
Explanation
When angular acceleration is zero then torque on the wheel become zero∴ ω=dθ / dt=6t2 - 12 t ∴ α=dω /dt=12t - 12 Now 12t - 12=0 t=1 sec Answer:(a)
Q.14
A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is 90째. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is:[2008]
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a) ML2/24
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b) ML2/12
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c) ML2/6
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d)
Explanation
Let mass of half part be M' and length L' . Moment of inertia around center of mass of half part=M'L'2 / 12 But axis of rotaion for half part is at its end thus from the law of parallel axis theorem Moment of interia about axis shown in figure is M'L'2 / 12 + M'(L'/2)2=M'L'2 (1/3) But M'=M/2 and L'=L/2 thus Total Moment of Inertia=Sum of M.I.s of both parts Total Momentum Of Inertia= Answer: (c)
Q.15
A particle of mass m moves in XY plane with a velocity 'v' along the straight line AB. If the angular momentum of the particle with respect to origin O is LA when it is at A and LB when it is at B then .. [ CBSE -PME 2007]
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a)LA=LB
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b) The relationship between LA and LB depends upon the slope of the line
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c)LA < LB
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d)LA > LB
Explanation
Angular momentum=Linear momentum × perpendicular distance between line of action of liner momentum from originLet d be the perpendicular distance.PA and PB be the momentum at A and B Angular momentum LA=PA × d LB=PB × d As linear momentum are equal LA=LBAnswer: (a)
Q.16
The moment of inertia of a body about a given axis is 1.2 kg mInitially, the body is at rest. In order to produce a rotational kinetic energy of 1500 joule, an angular acceleration of 25 radian/sec2 must be applied about that axis for a duration of [ CBSE- PMT 1990]
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a) 4 seconds
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b) 2 seconds
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c)8 seconds
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d)10 seconds
Explanation
Rotational K.E=Er=(1/2) I ω2 I=1.2 kg/m2Er1500Jα=25 rad/sec2, Answer: (b)
Q.17
rectangle ABCD (BC=2 AB). The moment of inertia is minimum along axis through [ CBSE-PMT 1993]
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a) BC
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b) BD
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c)HF
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d)EG
Explanation
The Moment of Inertia is minimum about EG because mass distribution is at minimum distance from EG. Answer:(d)
Q.18
The centre of mass of a system of particles does not depend upon [ CBSE-PMT 1997]
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a) masses of the particles
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b) forces acting on the particles
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c) position of the particles
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d) relative distances between the particles
Explanation
centre of mass of system depends upon position and masses of particle. Also, it depends upon relative distance between particles. Answer: (b)
Q.19
If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is [ CBSE-PMT 2002]
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a) 2.5m
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b) 1m
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c)1.5m
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d)2m
Explanation
Consider an element of length dx at a distance x from end AHere, mass per unit length λ of rod λ ∝ x ⇒ λ=kx∴ dm=λ dx dm=kx dx position of centre of gravity of rod from end AAnswer: (d)
Q.20
Consider a system of two particles having masses m1 and mIf the particle of mass m1 is pushed towards the centre of mass of particles through a distance d, by what distance would the particle of mass m2 move so as to keep the centre of mass of particles at the original position? [ CBSE-PMT 2004]
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a)
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b)
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c)
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d)d
Explanation
Let d' be the distance by which mass m2 be movedNow according to law of conservation of centre of Massm1 × d=m2 × d'∴ d'=( m1 / m2) d Answer:(c)
Q.21
A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle θ. The fractional force [ CBSE-PMT 2005]
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a) dissipates energy as heat
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b) decreases the rotational motion
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c) decreases the rotational and transnational motion
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d) converts transnational energy to rotational energy
Explanation
Net work done by frictional force when drum rolls down without slipping is zero Wnet=0 Wtrans + Rotational=0 ΔKtrans + ΔKrotational=0 ΔKtrans=- ΔKrotational thus frictional force converts translation energy to rotational energy Answer: (d)
Q.22
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 wheel now rotates with an angular velocity [ IIT 1983]
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a)ωM / ( M + m)
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b) ω(M - 2m) / ( M + 2m)
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c)ωM / ( M + 2m)
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d)ω(M + 2m) / M
Explanation
Objects are placed gently, therefore no external torque acts on the system. Hence angular momentum is conserved. Thus I1ω1=I2ω2Mr2 × ω1=(M + 2m)r2 × ω2 given ω1=ω ∴ ω2=Mω / (M + 2m) Answer: (c)
Q.23
Two point masses of 0.3kg and 0.7kg are fixed at the ends of a rod of length 1.4m and of negligible mass. The rod is set to rotate about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of .. [ AIEEE 1995]
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a) 0.42m from mass of 0.3kg
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b) 0.70 m from mass of 0.7kg
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c)0.98 m from mass of 0.3kg
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d)0.98 m from mass of 0.7kg
Explanation
The moment of inertia of the system about axis of rotation O is I=I1 + I2 I=0.3x2 + 0.7(1.4 - x) 2Thus I=x2 - 1.96x + 1.372 The work done in rotating the rod is converted in to its rotational kinetic energy∴ W=½ I ω2W=½[x2 - 1.96x + 1.372] ω2For work done to be minimumdW/dx=0 ⇒ 2x - 1.96=0 ∴ x=1.96/2=0.98mAnswer: (c)
Q.24
A mass is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin [ IIT 1997]
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a) is zero
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b) remains constant
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c)goes on increasing
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d)goes on decreasing
Explanation
Angular momentum of the mass moving with constant velocity about origin is L=momentum × perpendicular distance of line of action of momentum from origin L=mv × y In the given condition mvy is a constant. Therefore angular momentum is constant Answer:(b)
Q.25
A cubical block of side a is moving with velocity V on a horizontal smooth plane as shown in figure. It hits a ridge at point O. The angular speed of the block after it hits O is [ IIT 1999]
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a) 3V / (4a)
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b) 3V/(2a)
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c) (3V/2a) 1/2
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d) zero
Explanation
r = √2 ( a/2) or r2 = a2/2 Li = MV × (a/2) Final momentum = ( momentum of inertia about CM. + Mr2)ω ( by parallel axis theorem) For cube Moment of inertia about C.M = Ma2 /6 Net torque about O is zero Initial angular momentum = momentum × perpendicular distance of velocity vector from O ∴ angular momentum (L) about O will be conserved or Li = Lf MV (a/2) = ( Icm + Mr2)ω Answer: (a)
Q.26
A cyclist turns around a curve at 15 miles per hour. If turns at double the speed, the tendency to overturn is
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a)doubled
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b) quadrupled
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c)halved
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d)unchanged
Explanation
v ≥ √(rg) r=v2 / gThus if speed is doubled r will be four timesAnswer: (b)
Q.27
A disc of mass M and radius R is rolling with angular speed ω on a horizontal plane as shown in figure. The magnitude of angular momentum of the disc about origin is [ IIT 1999]
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a) ½ MR2ω
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b) MR2ω
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c) (3/2) MR2 ω
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d) 2 MR2ω
Explanation
The disc has two types of motion namely translational and rotational. therefore there are two types of angular momentum and the total angular momentum is the vector sum of these two In the case both the angular momentum have the same direction ( perpendicular to the plane of paper and away from the paper ) L=LT + LR Here LT=angular momentum due to translational motion LR=angular momentum due to rotational motion about C.M L=MV × R + Icmω Here Icm=M.I. about centre of mass C also v=ωR L=M( Rω)R + ½ MR2ω Icm=(3/2)MR2ω Answer: (c)
Q.28
A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity v. It collides elastically and head on with an identical sphere b at rest. Neglect friction everywhere. After collision, their angular speeds are ωA and ωB respectively. Then [ IIT 1999]
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a)ωA < ωB
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b) ωA=ωB
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c)ωA=ω
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d)ωB < ω
Explanation
As spheres are smooth there will be no friction ( no torque) and therefore there will be no transfer of angular momentum. Thus A, after collision will remain with its initial angular momentum ωA=ωAnswer: (c)
Q.29
A cubical block of side L rests on a rough horizontal surface with coefficient of friction µ. A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is [ IIT 2000]
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a) infinitesimal
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b) mg/4
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c)mg/2
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d)mg( 1 - µ)
Explanation
The applied force shifts the normal reaction to one corner as shown. At this situation, the cubical block starts topping about O. Taking torque about O F × L=mg × (L/2) F=mg/2 Answer: (c)
Q.30
A horizontal rod has a bead which can slide along its length and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration α. If the coefficient of friction between the rod and the bead is µ, and gravity is neglected, then the time after which the bead starts slipping is [ IIT 2000]
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a) √ ( µ/α)
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b) µ / √α
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c)1 / √(µα)
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d)infinitesimal
Explanation
When we are given an angular acceleration to the rod, the bead is also having an instantaneous tangential acceleration a= Lα and centrifugal acceleration ω2L this will happen when force is exerted on the bead by the rod. The bead has a tendency to move away from the centre. But due to the friction between the bead and rod, this does not happen to extent to which frictional force is capable of holding the beadThe frictional force here provides the necessary centripetal force. If instantaneous angular velocity is ω then Centripetal force=mLω2 Frictional force=µ(ma) mLω2=µ(ma) ∴ ω2=µα By applying ω=ω0 + αt We get ω=αt ( since ω=0 ) ∴ α2t2=µα ∴ t=√ ( µ / α) Answer:(a)
0 h : 0 m : 1 s
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