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Vol. XXIII No. 10 October 2015 Corporate Office :

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Managing Editor : Mahabir Singh Editor : Anil Ahlawat (be, MbA)

contents

Physics Musing (Problem Set-27) 8

JEE Workouts 11

Core Concept 15

Thought Provoking Problems 20

Ace Your Way CBSE XI 23

Series 2

JEE Accelerated Learning Series 31

Brain Map 46

Bihar CECE 59

Solved Paper 2015

Exam Prep 66

You Ask We Answer 73

Physics Musing (Solution Set-26) 74

Ace Your Way CBSE XII 75

Series 5

Crossword 85

rial

edit

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Editor : Anil Ahlawat

Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited.

cErN discovers new properties of

higgs boson, the fundamental particle

H

iggs boson, the fundamental particle was discovered by cerN,

three years back, in collaboration with the scientists of other laboratories.

the Atlas and cMS collaborators of cerN have obtained the sharpest picture of the Higgs boson for the first time. Now they are on the way to study its properties, production, decay and how its interacts with other particles.

All the measured properties of this particle are in agreement with the standard model. the present studies have the best precision for performing measurements for these particles. A lot of work is yet to be done in the field of more familiar particles.

Krane had been able to measure the radius and the error of determination for half a dozen well-known fundamental particles such as electron, proton, etc. these values were theoretically verified by our scientists.

this story of cerN has proved that with international cooperation, discoveries are faster and their confirmation is also better when more than one lab is attached to the working teams of the bigger laboratories.

Anil Ahlawat Editor

single oPtion correct tyPe

1. In a Young’s double-slit experiment the slit separation is 0.5 mm and the screen is 0.5 m from the slit. For a monochromatic light of wavelength 500 nm the distance of third maxima from the second minima on the other side is

(a) 2.75 mm (b) 2.5 mm

(c) 22.5 mm (d) 2.25 mm

2. A particle of mass m is projected from the surface of earth with a speed v0 (v0 < escape velocity). The speed of particle at height h = R (radius of earth) is

(a) gR (b) v₀2−2gR

(c) v gR₀2_{− (d) v gR}

02+2

3. A uniform magnetic field of induction B is confined to a cylindrical region of radius R. The magnetic field is increasing at a constant rate of

dB

dt (tesla/second). An electron of charge e, placed at

the point P on the periphery of the field experiences an acceleration (a) 1 2 eR m dB dt toward left (b) 1₂eR_m dB_{dt toward right} (c) eR m dB dt toward left (d) zero

4. After one second the velocity of a projectile makes an angle of 45° with the horizontal. After another one second it is travelling horizontally. The magnitude of its initial velocity and angle of projection are (Take g = 10 m s–2₎ (a) 14.62 m s–1_{, 60°} (b) 14.62 m s–1_,tan–1₍₂₎ (c) 22.36 m s–1_{, tan}–1₍₂₎ (d) 22.36 m s–1_{, 60°}

5. A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is T. With what acceleration should the lift be accelerated upwards in order to reduce its period to T

2?(g is the

acceleration due to gravity)

(a) 4g (b) g

(c) 2g (d) 3g

Set 27

P

hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material.

In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You.

The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those who send atleast five correct solutions will be published in the next issue.

We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.

PHYSICS

6. A block released from rest from the top of a smooth inclined plane of angle q1 reaches the bottom in t1. The same block released from rest from the top of another smooth inclined plane of angle q2, reaches the bottom in time t2. If the two inclined planes have the same height, the relation between t1 and t2 is (a) t t2₁ 12 1 2 =    sin sin / q q (b) t t2₁=1 (c) t t2₁ = sinsin 1₂ q q (d) t t2₁ 2 1 2 2 = sin sin q q

7. The magnetic flux f through a stationary loop of wire having a resistance R varies with time as f = at2 _{+ bt (a and b are positive constants). The} average emf and the total charge flowing in the loop in the time interval t = 0 to t = t respectively are (a) a bt a R b t t + , 2+ _{(b) a b} a b R t+ , t2+ t 2 (c) a b at₂+ , t2_R+bt _{(d) 2} 2 2 (a b a), b R t+ t + t

8. Assuming all the surfaces to be frictionless, find the magnitude of net acceleration of smaller block m with respect to ground

(a) ₍₅2 5_{m M+}mg₎ (b) _{(5m M}2mg_{+ )} (c) ₍₅7 5_{m M+}mg₎ (d) none of these

9. VectorsAandBinclude an angle q between them.

If (A B + )and(A B − )respectively subtend angles a

and b withA, then (tana + tanb) is

(a) AB A B sin cos q q 2₊ 2 2 (b) _A22₋AB_B2sin_cosq2_q (c) _A A _B 2 2 2₊ sin2_cosq2_q (d) B A B 2 2 2₋ sin2_cosq2_q

10. Resultant of two vectors having same magnitude forms an angle with any of the vectors. If the magnitude of second vector is reduced to half of initial magnitude without changing the angle between the direction of new resultant vector and first vector is also reduced to half, then the angle between the two vectors is

(a) 120° (b) 60°

(c) 90° (d) 45°

1. A monochromatic light of wave length 500 Å is incident on two identical slits separated by a distance of 5 × 10–4 m. The interference pattern is seen on a screen placed at a distance of 1 m from the plane of slits. A thin glass plate of thickness 1.5 × 10–6 m and refractive index 1.5 is placed between one of the slits and the screen. Find the intensity at the center of the slit now.

2. A totally reflecting, small plane mirror placed horizontally, faces a parallel beam of light as shown in the figure. The mass of the mirror is 20 g. Assume that there is no absorption in the lens and that 30%

of the light emitted by the source goes through the lens. The power of the source needed to support the weight of the mirror is

x × 108 W. Find x. (Take g = 10 m s–2)

3. A certain radioactive material can undergo three different types of decay, each with a different decay constant l, 2l, and 3l. Then, the effective decay constant leff is equal to nl. What is the value of n?

4. A coil is connected to an alternating emf of voltage 24 V and of frequency 50 Hz. The reading on the ammeter connected to the coil in series is 10 mA. If a 1 mF capacitor is connected to the coil in series the ammeter shows 10 mA again. What would be the approximate reading on a dc ammeter (in A) if the coil was connected to a 180 V dc voltage supply? (Take p2 = 10)

5. A metal disc of radius 25 cm rotates with a constant angular velocity 130 rad s–1 about its axis. Find the potential difference in nV between the centre and

rim of the disc if the external magnetic field is absent.

6. Figure shows a potentiometer circuit for determining the internal resistance of a cell. When switch S is open, the balance point is found to be at 76.3 cm of the wire. When switch S is closed and the value of R is 4.0 W, the balance point shift to 60.0 cm. Find the internal resistance of cell C′.

7. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from m25 to m50. The ratio m_m25

50 is 8. An a-particle and a proton are accelerated from rest

by a potential difference of 100 V. After this, their de Broglie wavelengths are la and lp respectively.

The ratio l l_a

p_{, to the nearest integer, is}

9. When two identical batteries of internal resistance 1 W each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J₂. If J₁ = 2.25J₂ then the value of R in W is

10. To determine the half life of a radioactive element, a student plots a graph of ln dN t_dt( ) versus t. Here dN t

dt( ) is the rate of radioactive decay at time t.

If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 years, the value of p is

11. At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4 V?

[Take : ln5 = 1.6, ln3 = 1.1]

12. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from 25

3 m to 50

7 m in 30 seconds. What is the speed of the object in km per hour ?

13. A large glass slab (m = 5/3) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius

R cm. What is the value of R ?

14. Three charges q1 = 3 mC, q2 = –3 mC and q3 are kept at the vertices of a triangle as shown in the figure. If the net force acting on q₁

is F, the charge q 3 is then given as 1_ +1_2

n mC.

Find the value of n.

15. A steady current I goes through a wire loop

PQR having shape of a right angle triangle with

PQ = 3x, PR = 4x and QR = 5x. If the magnitude of

the magnetic field at P due to this loop is k m0_pI_x 48  

 , find the value of k.

solutions

1. (0) : Take a point at a distance y from the centre of the slit. Path difference between waves reaching this point,

Dx dy= _D − −(m 1)t

At the centre of slit y = 0 \ Dx = –(m – 1)t Phase difference φ=2_lpDx= −2_lp(m−1)t Intensity, I=4I _{  =} I _− − t_ 2 4 1 0cos2 φ 0cos2 p_l(m ) = − − × ×       − − 4 1 5 1 1 5 10 5000 10 0 2 6 10 I cos p( . ) . =4I cos0 232p =0

2 (1) : Let n photons (each of frequency f) per second are emitted from source. Then power of source is

P = nhu

But only 30% of the photons go towards mirrors. Then force exerted on mirror is

F=  n h nh_c P_c

  = =

2 30

100 l 35 u 35

This force should be equal to weight of mirror, so 3 5Pc =20 10× −3×g or P = × ×5 3 10 20 10× × − ×10= × W 3 1 10 8 3 8 \ x = 1

3. (6) : Effective decay constant will be the sum of all different decay constants

So, leff = l l + 2l + 3l = 6l, hence n = 6. 4. (1) : Z= L +R = × − ( )w 2 2 24 ₃ 10 10 R L R L C 2₊ 2 ₌ 2₊ ₋ 1 2   ( )w w w ( )w w w L L C = − + 1 L C = = × × × − 1 2 1 2 100 100 10 2 6 w p p

= 5 H;

(2400)2 = (500p)2 + R2

R = (2400) (2− 5p×100)2 =10 24( )2−25p2 = ×_{10 326 180}≅ _{, I}=_1A

5. (3) : Centripetal force required for circular motion of electron is generated by radial electric field caused by the redistribution of the electron in the disc.

F = eE = mrw2 _{or E} mr e = w2 From dV = –E dr, or dV m= − w_e2 r dr or dV m e r dr R V V = −

_∫

∫

₁ w2 0 2 or V V m R e 1 2 2 2 2 − = w = × × − − ( . )( ) ( . ) ( )( . ) 9 1 10 130 0 25 2 1 6 10 31 2 2 19 = 3.0 × 10–9 V = 3.0 nV 6. (1) :

Let e be the emf of the cell C′ and r its internal resistance. Let l = AJ be the balance length when switch S is open. When a resistance R is introduced by closing the switch a current begins of flow through the cell C′ and resistance R. The potential difference between the terminals of the cells falls and the balance length decreases to l′ = AJ′. The terminal resistance of the cell is given by

r E V I

= −

where V is the terminal voltage of C′ and I is the current in the circuit involving C′ and R. Also I V= ._R \ r=_E− _ V 1 R But E V =ll′. \ r R l l= −_l ′    ′ =4 0×_76 3 60 0− _ 60 0 1 . . . .  W 7. (6) : Magnification, m f f u = +

According to cartesian sign conventions

m₂₅ 20 20 25 4 = − = − m₅₀ 20 20 50 32 = − = − \ = = m m25₅₀ 122 6 8. (3) : de Broglie wavelength l = h mK 2 l = h = mqV K qV 2 [ kinetic energy, ] l_p p p h m q V = 2 , la= _{a a} h m q V 2 \ l = × l_a a a p p p h m q V m q V h 2 2 = m q = m q m q m q p p p p p p a a (4 )(2 ) _{= 8 2 2 3= ≈} 9. (4) : In series,

Rate of heat produced in R is J_{1 R}2 2R

2 = +     e In parallel

Rate of heat produced in R is

J R R 2 2 1 2 = +         e ₌ +     2 2 1 2 e R R \ = +     × +    J J R R 1 2 2 2 2 2 2 1 2 e e = + +     2 1 2 2 R R

According to given problem

\ = + +     2 25 2 1 2 2 . R R or 1 5 2 1 2 . = + + R R or 1.5R + 3 = 2R + 1 ⇒ 0.5R = 2 \ R = = 2 0 5. 4W

10. (8) : According to radioactive decay

N = N₀e–lt dN dt = −lN e0 −lt dN dt = −lN e0 −lt dN dt =lN e0 −lt

Taking natural logarithms of both sides of above equation, we get

ln dN ln( )

dt = lN0 −lt

ln dN ln( )

dt = −lt+ lN0

Comparing the above equation with equation of a straight line i.e. y = mx + c, we get

From graph, slope = − = − −

l 3 4

6 4 or l =1year−

2 1

Half lifeT_{1 2/} =0 693 2 0 693. = × . years

l =1.386 years

4.16 years is approximately 3 half lives Nuclei will decay by a factor of 23 = 8 \ p = 8

11. (2) :The equivalent resistance of the two parallel resistors is R = + = ( )( ) ( ) ( ) 2 2 2 2 1 M M M M M W W W W W

The equivalent capacitance of the two parallel capacitors is

C = 2 mF + 2 mF = 4 mF

The corresponding equivalent diagram is as shown in the figure.

Time constant of the circuit, t = RC = (1 MW)(4 mF) = 4 s Since V(t) = V0(1 – e–t/t) Here, V(t) = 4 V, V0 = 10 V \ 4 = 10(1 – e–t/t) 10e–t/t = 6 or e−t/t₌ 6 ₌ 10 3 5 −t_{= −} t ln3 ln5 or t t=ln5−ln3 1 6 1 1 0 5= . − . = . or t = (0.5)(4 s) = 2 s

12. (3) : Focal length of a convex mirror,

f R= =

2 20

2 m = 10 m For first object

v₁ 25 f

3 10

= + m, = + m

Using mirror formula 1 1 1_{v u f}+ =

\ 1 25 3 1 1 10 1 ( / )+u = or 1 1 10 3 25 1 u = − u1 = – 50 m For second object

v₂ 50 f 7 10 = + m, = + m \ 1 + 1 =1 2 2 v u f 1 50 7 1 1 10 2 ( / )+u = or 1 1 10 7 50 2 u = − u2 = – 25 m

Speed of the object=25 − 30 m s 1 = × − 25 30 18 5 km h = 3 km h1 –1 13. (6) : sinθ m c= =1 3₅ ...(i) From figure, sinθ_c R R = + 2 ₈2 ...(ii)

The centre of mass (COM) for any given system of masses represents that particular point for the entire system whose motion is not dependent upon the point of application of force but rather what matters is how much force has been applied and in which direction the force has been applied, and not where it has been applied.

To understand this better, let us consider three cases of a uniform rod being hit at three different points.

  F F F C C vc vc vc C

The diagram clearly shows, that the motion of different points of the rod is different in all the three cases but there, at point C (in this case the geometrical centre) whose motion is identical in all the three cases. This point C is the COM.

Hence, one more definition of COM is also prevalent amongst most books which says it is one such point for the entire system of given masses where the entire mass may be assumed to be concentrated.

Mathematically, the location of COM is calculated as shown :

1. For discrete point masses

Let m_{ } 1, m_2, .... mn, be point masses with position vectors

r r_{1 2}, , .... then the position vector of COM is,r_n

    r m r m r_{m m m}m rn n n COM= 1 1+₊ 2 2₊+_......₊+ 1 2 =Σ Σ m r mi i_i  \ r_COM=x_COMi y^+ _COMj z^+ _COMk^

where x_COM m x_{m y}i i _COM m y_{m zCOM} m z_m i i i i i i i =Σ_Σ , =Σ_Σ , =Σ_Σ

where (xi, yi, zi) are the x, y and z co-ordinates of the

ith _mass.

Note : They are co-ordinates and hence + or – has to be taken care of.

2. For continuous mass distribution

Choose an elemental mass dm, the position vector of whose COM is r, then the COM of the entire system becomes   r dm r dm COM=

∫

∫

Now, with the formula being known, let us calculate the location of COM for some standard configuration.

Centre of Mass

COM of a two point mass system

Let us take COM to be somewhere in between as shown and from there we have taken x and y axis for simplification of calculation. \ = = + + x m x m x m m COM 0 1 1 2 2 1 2 = − + + + m r m r m m 1 1 2 2 1 2 ( ) ( ) ⇒ m1r1 = m2r2

Hence greater the mass, lesser its distance from COM. Thus COM is shifted more towards heavier mass. also, r1 + r2 = r r m m m r r m m m r 1 1 1 2 2 1 1 2 = +     = +     , ...(i)

Clearly, if the masses been equal, the COM would be located midway on the Line of Symmetry (LOS). Hence if we can find any LOS for the given system then draw it and the COM will lie on this LOS.

If there are two or more LOS, then their point of interaction gives COM.

Q1. Locate the COM of the system shown in the figure.

Soln.: Seeing this as three separate uniform rods, their respective COM will be at their geometric centre and mass being proportional to length, we can see the same figure as shown below

Q2. From a uniform disc of radius 2R, a disc of radius R is cut out touching the periphery as shown. Locate the COM.

Soln.: The LOS is drawn as below

Let us fill in the portion which was cut-out, now the COM of 2 is known as well as combined 1 + 2 is known.

Let us assume the COM of the given figure (part 1 in our diagram) at a distance x from the combined COM.

For 1 + 2 combined system

x m x m x

m m

COM = 1 1+₊ 2 2 =

1 2 0

⇒ m1x1 + m2x2 = 0

But since the disc is uniform, m ∝ A (area) \ A1x1 + A2x2 = 0

⇒ p[(2R)2 _{– R}2_{] (–x) + (pR}2_{)(R) = 0} ⇒ x = R/3

COM of non-uniform rod with linear mass density linearly dependent on distance measured from one end

dx x

A B

( ) = (x ax + b)kg m–1

Given : The linear mass density of rod AB of length l is proportional to distance measured from one end A. To locate COM, we choose an elemental mass dm where,

dm = ldx

x xdm dm ax bx dx ax b dx l l COM = = + +

∫

∫

∫

∫

( ) ( ) 2 0 0 = + + = + + al bl al _bl al bl al b 3 2 2 2 3 2 2 2 3 3( 2 ) \ = + + x al bl al b COM 2_{3 2}3 2 ( ) ...(ii)

Note : If the linear mass density is given to be proportional to distance being measured from one end, l(x) = ax i.e., b = 0

\ x_COM=2l 3

COM of a thin uniform semicircular ring

R Rsin = y d  m R dm = ·Rd = md_ COM LOS R \ = = ⋅

∫

∫

∫

y dmy dm m d R m COM p θ θ p sin 0 =R

_∫

d = R p θ θ p p sin 0 2 ...(iii) Q3. Locate the COM of a thin uniform quarter ring.

45°

C _{OC = ?}

LOSO 45° COM

Soln.:Let us add the other quarter so that it becomes a half ring as shown in the figure below

\ y_COM=2R p

A quarter ring is identical from y-axis as well as x-axis hence, x_COM =y_COM = 2R p \ =    OC 2 2R p COM of a uniform semicircular disc

Let us divide the entire disc, into small triangular strips as below :

Since, each thin triangular sheet is uniform, their COM will lie at centroid, i.e., at 2

3R from common vertex.

Joining all the point masses, we have a ring of radius

r= 2R 3 \ y_COM=2r_{= }2 2R_{ =} R 3 4 3 p p _p ...(iv)

COM of a thin uniform hollow hemisphere

We divide the hemisphere into thin rings.

\ =    dm m R R Rd 2p 2 (2p sin )(θ θ) \ =

∫

∫

y dmy dm COM = ⋅

∫

m d R m sin cos / θ θ θ p 0 2 =R

∫

d =R 2 ₀ 2 2 2 sin / θ θ p ...(v) COM of a uniform solid hemisphere

We divide the hemisphere, into thin hemispherical shells.

COM y = r2 r dr R dm M R r dr =         2 3 2 3 2 p ( p ) = 33 2 m R r dr

The COM of this element is at r_2.

\ =

∫

∫

y dmy dm COM = ⋅ ⋅

∫

33 2 ₂ 0 m R r dr r m R = 3 8 R _...(vi)

There are certain standard results which needs to be memorized. Motion of COM   r m r mi i_i COM =Σ_Σ \ Velocity of COM,     v dr dt m dri dt m m v m i i i i i COM= COM = = Σ Σ Σ Σ \ = + + + + + +     v m v m v m v m m m_nn n COM 1 1 2 2 1 2 ... ... \ Linear momentum of COM is,_{ }

   p_COM =MV_COM =m v m v_{1 1}+ _{2 2}+...+m v_{n n} Where M = m1 + m2 + ... + mn \ dp = + + + dt m dv dt m dv dt m dv dt n n     COM 1 1 2 2 ... =m a m a_{1 1} + _{2 2} +...+m a_{n n} = +F F₁ ₂+...+F_n

which is the vector sum of all the forces acting on the combined system which is the rate of change of momentum of COM. This proves that for COM, the net force is important and not the point of application of force.

This gives us one important conclusion, if there is no net external force in any direction for combined system, the linear momentum of the system in this direction cannot be changed. And specifically speaking, if the COM was initially at rest and due to internal forces, objects start moving, then they move such that COM does not move.

\ = =

⇒ =

+ +

If _COM initially and _ext COM      v F r m r m r 0 0 0 1 1 2 2 D D D ... +m r_{n n}D 0=

Q4. A man of mass m stands on a rough horizontal plank of mass M whose lower surface is smooth.

If the man walks a distance l over plank, find the distance moved by plank.

Soln.: Even though we understand that the plank would recoil towards left due to friction force of man’s foot, we assume the plank to be shifting towards right for simplification of calculation.

On the combined system,

Fhorizontal = 0 ⇒ DxCOM = 0 ⇒ m1Dx1 + m2Dx2 = 0 ⇒ m(x + l) + Mx = 0 ⇒ x ml m M = − +    

\ Negative sign signifies that plank moves opposite to assumed direction by a distance | |x ml .

m M

= +

This result gives us the answer without calculations in few questions as shown here.

m m x _R M R Smooth m m l   M M Smooth x lsin  | |x = mR_{m + M}  | | =x _{m + M}mlsin M

Acceleration of COM     a dv dt m dv dt m m a m i i i i i i COM= COM = = Σ Σ Σ Σ \ = + + + + + +     a m a m a m a m m m_n n n COM 1 1 2 2 1 2 ... ... = vector sum of all the forces

total mass

Q5. m₂ k m₁ F

Smooth

The spring is in its natural length and a constant force starts acting on m1 as shown. In a time t, m1 displaces by x1 then find the displacement of m2?

Soln.: a F m m COM = ₊ =constant 1 2 \ = = + + S a t m x m x m m COM 1₂ COM 2 1 1 2 2 1 2 ⇒ 1 − = 2Ft2 m x m x 1 1 2 2 ⇒ = −     x F m t m m x 2 2 2 1 2 1 1 2 Q6. _m1 k _m2 F2 Smooth F₁

Find the maximum elongation in spring if the spring is initially in its natural length.

(Assume F2 > F1)

Soln.: aCOM =m mF F towards right.

− +

2 1

1 2

If we analyse the motion from COM, we need to include a pseudo force for both, as shown below

\ = + − +     F F m F F m m 1 1 1 2 1 1 2 eff =m F m F2 1_{m m+}+ 1 2 1 2 F F m F F m m 2 2 2 2 1 1 2 eff = − _{ +}− _ =m F m F2 1_{m m+}+ 1 2 1 2

F1eff = F2eff = Feff

With respect to COM, equal and opposite forces act on the two objects.

m₂ m₁ m₁ m₂ x₁ x2 F_eff F_eff At maximum elongation v1 = v2 = 0 \ Using work energy theorem,

F x F x_{eff 1 2} PE 1k x x_{1 2} 2 2 + _eff =D = ( + ) ⇒ F_eff(x x₁+ ₂)=1k x x( ₁+ ₂)2 2 ⇒ = + = = + + x x x F k m F m F m m k max 1 2 (₍ 2 1 ₎1 2) 1 2 2 _eff 2 Q7. For (m2 > m1), find : (i) aCOM

(ii) Net force acting on the combined system. Soln.: Clearly, a a m m m m g a 2 1 2 1 1 2 = = − +     = a m a m a m m COM= 1 − 1₊+ 2 2 1 2 ( ) ( ) = − +     m m m m₁2 1₂ g 2 F m m a m m g m m net=( 1+ 2) COM=(₍2−₊ 1)₎ 2 1 2 nn

1. Determine the moment of inertia about the x-axis of the solid spherical segment of mass M.

O

R/2 R/2

X

2. A body consisting of a cone and hemisphere of radius r fixed on the same base rests on a table, the hemisphere being in contact with the table. Find the maximum height of the cone, so that the combined body may stand upright.

3. Consider the motion of the cylinder of mass m and radius r moving up a plane inclined at angle q to the horizontal. To begin with, cylinder rotates with angular velocity wo about its axis when gently placed

on the plane so that its initial translational velocity is zero. Find the distance moved by cylinder before sliding stops.

 _o

4. As shown in figure, a homogeneous slider bar with a mass of 4 kg and a length of 500 mm being pushed

along the smooth horizontal surface by a horizontal force F = 60 N. Determine the angle q for translation. What is the accompanying acceleration?

 F 500 mm

5. The mass of block A is twice the mass of block B. Find the acceleration of A in terms of the gravitational acceleration. Neglect the mass of the pulleys shown in figure.

6. A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as

shown in figure. Ground

The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m s–2_. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and coefficient of friction between the stick and the ring is x/10. Find the value of x?

ROTATIONAL MOTION

By : Prof. Rajinder Singh Randhawa*

SOLUTIONS

1. The solid sphere is generated by revolving circle about the diameter.

Mass of spherical segment,

M = Density × Volume of segment

Consider a small disc of radius y and thickness dx as shown in figure.

Mass of small disc, dm = r·py2_{·dx ...(i)} r p p = = = −

∫

∫

M V M y dx M R x dx R R R R 2 2 2 2 2 / / ( ) r p p = M = R M R 5 24 24 5 3_/ 3 ...(ii) O x A B C R x dx x2_{+ =}y2 R2 y y

Moment of inertia of disc about x-axis

dI=1dm y⋅ = y dx 2 1 2 2 _rp 4 (using (i)) \ I=

_∫

dI=

_∫

y dx=

_∫

R −x dx R R R R 1 2rp ₂ 4 2 ₂ 2 2 2 rp / / ( ) I R R x x dx R R =rp

_∫

− + 2 _/2( 4 2 2 2 4) =  − +        24 10 2 3 5 3 4 2 3 5 2 p p M R R x R x x R R / (using (ii)) Solving, we get, I= 53 MR 200 2

2. The composite body is symmetrical about y axis therefore its centre of gravity will lie on this axis. Now consider two parts of the body, hemisphere and cone. Let the bottom of the hemisphere be the axis of reference as shown in figure.

Weight of hemisphere, W₁ g2 r3

3 = r p .

Distance of centre of gravity of hemisphere from bottom y₁ 5r 8 = h r O y x Weight of cone, W₂ g r h2 3 = r p .

Distance of centre of gravity of cone from bottom,

y₂ = r + h/4.

Centre of gravity of combined body

y W y W y W W = + + 1 1 2 2 1 2 y g r r _{g r h r h} g r g r h = × + _ + _ + r p r p r p r p 2 3 5 8 1 3 4 2 3 1 3 3 2 3 2

For stable equilibrium, y r≤

r p r p r p r p g r r g r h r h g r g r h r 2 3 5 8 3 4 2 3 1 3 3 2 3 2 × + _ + _ + ≤ \ h2 _{≤ 3r}2 _{⇒ h ≤ 1.732 r}

3. The cylinder initially slides on the plane. Surface of the cylinder slides downward (backward) with respect to the plane; hence, the frictional force acts upwards (forward) as shown in figure. The equation of translational motion is mmgcosq – mgsinq = ma

or a = g(mcosq – sinq) ...(i)

Velocity of cylinder after time t is

v = at = g (mcosq – sinq)t ...(ii)

Frictional force produces a torque. Hence, t = (mmgcosq)r = Ia or a=mmg qr= m q mr g r cos . cos 1 2 2 2 ...(iii)

Angular velocity decreases with angular retardation a with time t,

At the moment sliding stops and pure rolling begins, we have v = wr, i.e. rolling begins at time t = T.

w= =v w −a r o T ⇒ g T r o T ( cosm q−sin )q _{=w −a} ⇒ w_o mg qT m q q r g T r

−2 cos = ( cos −sin )

or T=_g or

− w

m q q

( cos3 sin )

The distance moved by the cylinder before sliding stops is S₁ 1aT2 2 = = − × −     \ = − 1 2 3 2 1 2 2 g r g S r o o ( cos sin ) ( cos sin ) ( cos s m q q w m q q w m q iin ) ( cos sin ) q m q q 2 3g − 2

4. Free body diagram of slider bar is shown in figure.

Angular acceleration, a = 0 and linear acceleration = a

Applying the equation SFH = ma ⇒ 60 = 4a

\ a = 15 m s–2 _...(i)

SFV = 0 ⇒ RA = 4g = 39.24 N ...(ii)

Taking moment about G, we get, 60 × 0.25sinq – RA × 0.25cosq = 0

⇒ tanq 39 24= . = .

60 0 654

q = tan–1 _{(0.654) ⇒ q = 33.18°}

5. The free body diagram of two masses are shown in figure. A B y_{1 y} 2 T a2 a1 mg 2mg 2m l T T T TT 2T

Equation of motion for body A is

2mg – T = 2ma2 ...(i)

Equation of motion for body B is

2T – mg = ma1 ...(ii)

Length of the rope L = y1 + (y1 – l) + (y2 – l) + 2 × (half of circumference) or 2y1 + y2 = constant

Differentiating twice w.r.t. t, we get,

2d y2₂1 2₂2 0 2 _{1 2} 0

dt

d y

dt a a

+ = ⇒ + =

Acceleration of body A = 2 × Acceleration of body B From equation (i),

2mg – T = 2ma2 = 4ma1 ...(iii)

Solving (ii) and (iii), we get

a1 = g/3, a2 = 2a1 = 2g/3

\ Acceleration of body A= 2g 3

6. There is no slipping between ring and ground. Hence f2 is not maximum.

f₁ N₁ f₂ a mg N₂ 

But there is slipping between ring and stick. Therefore, f1 is maximum. Now, I mR= 2=2 0 5( . )2=1kg m2 2 N1 – f2 = ma or N1 – f2 = 2 × 0.3 = 0.6 N ...(i) a R R I R f f RI R f f I = a= t= ( ₂− ₁) = 2( ₂− ₁)/ or 0.3 = (0.5)2 _(f 2 – f1)/I or 0 3 0 5 1 2 2 2 1 . ( . ) ( ) / = f − f or f2 – f1 = 0.6 N ...(ii) N1 = 2 N ...(iii) Further, f₁ N₁ x N₁ 10 =m = _ _ ...(iv)

Solving these four equations, we get, x = 4

CHAPTERWISE UNIT TEST : Laws of Motion|

Work, Energy and Power| System of Particles and Rotational Motion

1. Why are porcelain objects wrapped in paper or straw before packing for transportation?

2. A light body and a heavy body have the same kinetic energy. Which one will have the greater momentum?

3. Find the radius of gyration of a hollow sphere rotating about its diameter.

4. Why are rockets given a conical shape?

5. About 4 × 1010 kg of matter is converted into energy in the Sun each second. What is the power output of the Sun?

6. A rigid spherical body is spinning around an axis without any external torque. Due to change in temperature, the volume increases by 1%. What will be the percentage change in angular velocity? 7. What is the tension in rod of length L and mass M

at a distance y from F1 when the rod is acted on by two unequal forces F1 and F2 (F2 < F1) at its ends? 8. A sphere of mass m moving with a velocity u hits

another stationary sphere of same mass. If e is the coefficient of restitution, what is the ratio of the velocities of two spheres after the collision?

9. A particle is projected making an angle of 45° with horizontal having kinetic energy K. What is the kinetic energy at highest point ?

10. A sphere of radius r is rolling without sliding. What is the ratio of rotational kinetic energy and total kinetic energy associated with the sphere?

OR

An automobile engine develops 100 hp when rotating at a speed of 1800 rpm. Find the torque acting.

11. An object of weight W hangs from a rope that is tied to other ropes that are fastened to the ceiling as shown in figure. The upper ropes make angles q and φ with the horizontal. Find the tensions T1, T2 and

T3 in the three ropes.  T1 T3 T2 W   

12. A ball A moving with a velocity of 9 m s–1 strikes an identical stationary ball B such that after collision the direction of each ball makes an angle of 30° with the original line of motion. Find the speeds of the two balls after the collision. Is the kinetic energy conserved in the collision process?

13. A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane, the centre of mass of the cylinder has a speed of 5 m s–1. (a) How far will the cylinder go up the plane? (b) How long will it take to return to the bottom ?

Series 2

Series 2

GENERAL INSTRUCTIONS (i) All questions are compulsory.

(ii) Q. no. 1 to 5 are very short answer questions and carry 1 mark each. (iii) Q. no. 6 to 10 are short answer questions and carry 2 marks each. (iv) Q. no. 11 to 17 are also short answer questions and carry 3 marks each. (v) Q. no. 18 is a value based question and carries 4 marks.

(vi) Q. no. 19 and 20 are long answer questions and carry 5 marks each. (vii) Use log tables if necessary, use of calculators is not allowed.

14. Explain why

(a) a horse cannot pull a cart and run in empty space.

(b) passengers are thrown forward from their seats when a speeding bus stops suddenly.

(c) It is easier to pull a lawn mower than to push it. 15. A student holds two dumbbells without stretched

arms while standing on a turntable. He is given a push until he is rotating at a rate of 0.5 rps. Then the student pulls the dumbbells towards his chest. What is the new rate of rotation? Assume the dumbbells are originally 60 cm from his axis of rotation and are pulled into 10 cm from the axis of rotation. The mass of the dumbbells is such that the student and dumbbells have equal angular momentum when at 60 cm distance.

16. A locomotive of mass m starts moving so that its velocity varies according to the law v k s= , where

k is constant, and s is the distance covered. Find the

total work performed by all the forces which are acting on the locomotive during the first t second after the beginning of motion.

OR

Two discs of moment of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre) and rotating with angular speeds w1 and w2 are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? (Take w1 ≠ w2).

17. If 28 × 1023 molecules of a gas strike a surface of area 14 cm2 normally per second with velocity of 500 m s–1 and rebound in the opposite direction with the same speed, find the pressure exerted by the gas on the surface if mass of each molecule is 5 × 10–23 g.

18. Sam went to shopping mall to purchase certain goods .There he noticed an old lady struggling with her shopping. Immediately he showed her the lift and explained to her how it carries the load from one floor to the next. Even then the old lady was not convinced. Then Sam took her in the lift and showed her how to operate it. That old lady was very happy.

(a) What values does Sam possess?

(b) An elevator can carry a maximum load of 1800 kg in moving up with a constant speed of 2 m s–1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.

19. A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v0 at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C as shown in figure. Obtain an expression for (a) v0 ; (b) the speeds at points B and C; (c) the ratio of the kinetic energies at B and C (KB/KC).

Comment on the nature of the trajectory of the bob after it reaches the point C.

OR

Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by v gh k R 2 2 2 2 1 = + ( / )

using dynamical consideration (i.e., by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.

20. A particle of mass m slides from the top of the surface of sphere of radius r. It loses contact and strikes the ground. At what distance below the top, the particle will lose contact with the surface? At what distance from the initial position the particle strikes the ground?

OR

A small mass slides down an inclined plane of inclination q with the horizontal. The coefficient of friction, µ = µ₀x, where x is the distance through

which the mass slides down and µ₀ is a constant. Find the distance covered by the mass before it stops. What is the maximum speed over this distance?

solutions

1. When porcelain objects are wrapped in paper or straw, the time of impact between themselves is very much increased during jerk while transportation.

Since F p

t

=D

D , as Dt increases, F decreases.

Hence, force on the porcelains is reduced during transportation, and saves them from breakage.

2. Kinetic energy, K p

m p mK

= 2 \ =

2 , 2 .

Since K is same for both bodies, p∝ m, i.e., the heavier body has more momentum than the lighter body.

3. The moment of inertia of a hollow sphere about its diameter = 2

3MR 2 ..(i)

If K is the radius of gyration, then moment of inertia of a body = MK2 ...(ii) From (i) and (ii)

2

3MR2 =MK2 \ K R= 2 3

4. Rockets are given a conical shape because conical shape of rockets minimize the atmospheric friction. It also helps to maintain its direction.

5. Energy liberated per second by the conversion of 4 × 1010 kg of matter, i.e.,

E = mc2 =[(4 × 1010) (3 × 108)2] J = 3.6 × 1027 J Power output of the Sun = energy (work) liberated per second

= 3.6 × 1027 J s–1 = 3.6 × 1027 W 6. Angular momentum of spherical body,

L I= w=2 w 5MR Q I2 = MR   25 2 Also, volumeV = R R =  V    4 3 3 4 3 2 2 3 π π or / \ L= M V    2 5 3 4 2 3 π w /

Q L and M do not change with temperature \ w ∝V−2 3/ d dV V w w ×100 = − × 2 3 100 % %

So, the angular velocity decreases by 0.67%. 7. L y F1 A B F2 T

Refer to figure, acceleration of the rod along F1,

a F F

M

=( 1− 2)

Mass of the part (say AB) of length y,

m M

L y

= _ _

If T is the tension in AB, then

F T ma M L y F F M F F yL 1− = = _{ }( 1− 2)=( 1− 2) or T F F F y L = −₁ ( ₁− ₂) =F _ −y_+ _ _ L F yL 1 1 2 8. Here, u₁ = u, u₂ = 0 \ e=_{u u}v v− v v_u − = − − 2 1 1 2 2 1 0 or v₂ – v₁ = e u ...(i)

By the law of conservation of momentum,

mu + m × 0 = mv1 + mv2

or v1 + v2 = u ...(ii)

Adding (i) and (ii), 2v2 = u + eu = u (1 + e) or v₂ u1 e

2

= ( + ) ...(iii)

Again, from (ii),

v u v_{1 2} u u1 e u e

2

1 2

= − = − ( + = −) ( ) ...(iv)

Divide eqn (iii) by eqn (iv)

v v e e 2 1 1 1 = + − .

9. Initial kinetic energy, K= 1mu

2 2

Velocity at the highest point, u'

= horizontal component of u =ucos 45° = u 2 Hence KE at the highest point,

′ = ′ K 1mu 2 2 ′ =     =  = K 1m u mu K 2 2 1 2 1 2 2 2 2 _.

KE at highest point will be half of the initial kinetic energy.

10. Translational kinetic energy, E_T = 1mv

2 2

and Rotational kinetic enegy E_R= 1I

Total energy, E E= _T+E_R=1mv + I 2 1 2 2 _w2 =1 + × × 2 1 2 2 5 2 2 2 2 mv mr v r =1 + = 2 1 5 7 10 2 2 2 mv mv mv \E = = E mv mv R 1 5 7 10 2 7 2 2 . OR Here, N = 1800 rpm, P = 100 hp = 100 × 746 W Angular velocity, w=2π =2π×1800= π − 60 60 1 N rads Now power = twor 100 746× = ×t 60π ⇒ t=746 100×_π 60 = 395 9. Nm

11. The free body diagram for the forces acting at O, is shown here.

From F∑ x =0,

T₂cosφ−T₁cosq=0 ...(i)

From F∑ y =0,

T₁sinq+T₂sinφ−W= 0 ...(ii) From eqns. (i) and (ii),

T_1sin T1cos W cos sin q q φ φ +    =

or T₁ (sin q + cos q tan φ) =W or T1=_{(sinq+cos tan )}W _{q φ}

Similarly, T₂= W

+

(sinφ cos tan )φ q and T₃ = W

12.

Initial momentum of A and B along X-axis

= m × 9 + m × 0 = 9m ...(i)

Final momentum of A and B after collision along

X-axis

= mv₁ cos 30° + mv₂ cos 30° =m 3 v v+

2 ( 1 2) ...(ii)

From eqns. (i) and (ii),

m 3 _{v v m}

2 ( 1+ 2)=9

or (v v₁+ ₂)=6 3 ...(iii)

Applying the law of conservation of momentum along Y-axis,

mv₁ sin 30° = mv₂ sin 30°

or v1 = v2 ...(iv)

From eqns. (iii) and (iv),

v v₁= =₂ 3 3m s−1 Initial KE =1 = = 2 9 81 2 40 5 2 m( ) m . m Final KE =1 + = + 2 1 2 1 2 12 22 12 22 mv mv m v( v ) =1 + = 2m[(3 3) (2 3 3) ]2 27m

KE lost in the collision 40.5 m – 27 m = 13.5 m So, kinetic energy is not conserved.

13. Here q = 30°, v = 5 m s–1, I= 1mr

2 2

(a) As the cylinder goes up the plane, it acquires potential energy at the expense of its KE of translational and rotational motion. Let the cylinder go up the plane upto height h. Then according to principle of conservation of energy, we have,

1 2 1 2 2 2 mv + Iw =mgh or 1 2 1 2 1 2 2 2 2 2 mv mr v r mgh + _ _ = [... v =rw]

or 3 4mv2 =mgh \ h v g = = × × = 3 4 3 5 4 9 8 1 913 2 _{( )}2 . . m

Suppose that the cylinder covers a distance S along the incline in reaching height h on the plane. Then,

sin sin . sin . q q = = = °= h S S h or 1 913 m 30 3 826

(b) For a cylinder rolling down an inclined plane, linear acceleration. a=2g = × × ° = − 3 2 3 9 8 30 3 27 2 sinq . sin . ms Now S v t= ₀ +1at2= + at2 2 0 12 {Q v0 = 0} \ t S a = 2 =1 53. s

14. (a) While trying to pull a cart, a horse pushes the ground backwards with a certain force at an angle. The ground offers an equal reaction in the opposite direction, on the feet of the horse. The forward component of this reaction is responsible for motion of the cart. In empty space, there is no reaction and hence a horse cannot pull the cart and run.

(b) This is due to inertia of motion. When the speeding bus stops suddenly, lower part of the bodies in contact with the seats stop. The upper part of the bodies of the passengers tend to maintain the uniform motion. Hence the passengers are thrown forward.

(c) While pulling a lawn mower, force is applied upwards along the handle. The vertical component of this force is upwards and reduces the effective weight of the mower, as shown in figure (i). While pushing a lawn mower, force is applied downward along the handle. The vertical component of this force is downwards and increases the effective weight of the mower, as shown in figure (ii). As the effective weight is lesser in case of pulling than in case of pushing, therefore, pulling is easier than pushing.

15. As L = mvr = mr2w,

initial angular momentum of both the dumbbells of total mass m, Ld₁= mr12w.

If L_s1is the initial angular momentum of the student, then initial angular momentum of the system,

L L₁= _s1+L_d1=L_s1+mr₁2_{w 1 ...(i)}

Final angular momentum of the system,

L₂ L_s L_d L_s mr₂2

2

2 2 2

= + = + w ...(ii)

From the law of conservation of angular momentum

L1 = L2 ...(iii)

From eqns. (i), (ii) and (iii),

Ls₂+mr22w2=Ls₁+mr12w 1 ...(iv) Since the angular momentum of the student is proportional to his rate of spin,

Ls₂ 2Ls₁

1 =w

w ...(v)

From eqns. (iv) and (v), w w2₁ Ls1 mr22w2 Ls1 mr12w1    + = + or w w₁2 12w1 22w2 12w1 12w1    mr +mr =mr +mr = 2mr w12 1 (Q L_s1 L_d1 mr₁2 ) 1 = = w or w₂ 12 w 12 22 1 2 2 2 2 2 0 6 0 6 0 1 0 5 = +   _r r_r _ =_{( . ) ( . )}( . )₊ _( . ) = 0.97 rps 16. Given: v k s= dv dt k s ds dt k sv = = 2 2 = × = k s k s k 2 1 2 2 Force on the locomotive,

F m dv dt mk = = 1 2 2 Again, dv dt k = 2 2 or dv k dt= 2 2 Integrating, v k t c= 2 + 2

where c is the constant of integration. Suppose v = 0 at t = 0. Then, c = 0 \ v k t= 2 ds_dt =k t2

2 or 2

or ds k t dt= 2 2

Integrating, s k t= 2 2 + ′c 4 Suppose s = 0 at t = 0. Then, c′ = 0 \ s k t= 2 2 4 Work done = Fs =1 × = 2 2 4 8 2 2 4 2 mk k t mk t OR

(a) Total initial angular momentum of the discs,

L₁ = I₁w₁ + I₂w₂

Moment of inertia of the two disc system = (I1 + I2) If w is the angular speed of the combined system, then, final angular momentum of the system,

L2 = (I1 + I2)w

As no external torque is acting, therefore, according to principle of conservation of angular momentum, we have, (I₁ + I₂)w = I₁w₁ + I₂w₂ or w= w + w + I I I I 1 1 2 2 1 2

(b) Initial KE of two discs, E₁ I_{1 1}2 I

2 22 1 2 1 2 = w + w

Final KE of the system, E₂ 1 I I_{1 2} 2 2 = ( + )w \ E E_{1 2} I_{1 1}2 I I I 2 22 1 2 2 1 2 1 2 1 2 − = w + w − ( + )w

Putting the value of w from part (a) and solving, we get, E E I I I I 1 2 1 2 1 2 2 1 2 2 − = − + ( ) ( )

w w _{which is a positive quantity}

Hence the rotational KE of the combined system is less than the sum of the initial rotational KE of the two discs. Note that there is loss of KE in the process. This loss of energy is due to dissipation of energy due to frictional contact of the two discs. It may be noted that angular momentum is conserved in the process as torque due to friction is only an internal torque.

17. Let the direction in which the molecules rebound after striking the surface be taken as positive. Momentum of each molecule after striking the surface =mv₂ = 5 × 10–26 kg × 500 m s–1

Momentum of each molecule before striking the surface = mv1 = 5 × 10–26 kg × (– 500 m s–1) 28 × 1023 molecules strike the surface per second. Change in momentum of the molecules striking the surface in 1 second

= 28 × 1023 [(5 × 10–26 × 500) – 5 × 10–26 (– 500)] kg m s–1 = 28 × 1023 × 5 × 10–26 × 1000 kg m s–1

= 140 kg m s–1

\ Rate of change of momentum =140 kg ms

1s kg ms

-1

-2 =140

But rate of change of momentum is equal to the applied force = 140 kg m s–2 = 140 N

By Newton's third law of motion this must also be the magnitude of the force exerted by the molecules on the surface.

\ Force exerted by molecules on surface = 140 N Area of surface = 14 cm2 = 14 × 10–4 m2 Pressure on surface = = × − force area 140 14 10 4 = 105 N m–2

18. (a) Sam is sympathetic and also has the attitude of helping others. He has patience.

(b) The downward force on the elevator is F= mg + f = (1800 × 10) + 4000 = 22000 N. The motor must supply enough power to balance this force.

Hence P = F.v = 22000 × 2 = 44000 W = 59 hp (1hp = 746 W)

19. (a) Two external forces act on the bob are gravity and tension (T) in the string. At the lowest point

A, the potential energy of the system can be taken

zero. So at point A,

Total mechanical energy = kinetic energy

E= 1mv

2 02 ...(i)

If TA is the tension in the string at point A, then

T mg mv

L

A− = 0

2

...(ii) At the highest point C, the string slackens, so the tension T_C becomes zero. If v_C is the speed at point

C, then by conservation of energy,

E K U= + or E=1mv_C+ mgL 2 2 2 ...(iii) Also, mg mv LC = 2 ...(iv) or mv_C2 =mgL ...(v)

Using (v) in (iii),

E=1mgL mgL+ = mgL

2 2

5

2 ...(vi)

From equations (i) and (vi), we get 5

2mgL m v= 2 02 or v0= 5gL ...(vii)

(b) From equation (iv), we have

v_C = gL

The total energy at B is

E=1mv_B+mgL

2 2 ...(viii)

From equations (i) and (viii), we get 1 2 1 2 2 02 mvB+mgL= mv 1 2 1 2 5 2 mv_B+mgL= m× gL [using (vii)] v_B= 3gL

(c) The ratio of kinetic energies at B and C is

K K mv mv B C B C = = 1 2 1 2 3 1 2 2 OR

Consider a body of mass M and radius R rolling down a plane inclined at an angle q with the horizontal, as shown in figure. It is only due to friction at the line of contact that body can roll without slipping. The centre of mass of the body moves in a straight line parallel to the inclined plane.

The external forces on the body are

(i) The weight Mg acting vertically downwards. (ii) The normal reaction N of the inclined plane. (iii) The force of friction acting up the inclined plane.

Let a be the downward acceleration of the body. The equation of motion for the body can be written as

N – mg cos q = 0 F = ma = mg sin q – f

As the force of friction f provides the necessary torque for rolling, so

t = f × R = I a = mk2 _a_R_

or f m k

R a

= 2₂

where k is the radius of gyration of the body about its axis of rotation. Clearly

ma = mg sin q – m k R 2 2 a or a g k R = + sin ( / ) q 1 2 2

Let h be height of the inclined plane and s the distance travelled by the body down the plane. The velocity v attained by the body at the bottom of the inclined plane can be obtained as follow:

v2 – u2 = 2as or v2 – 02 = 2 g k R s sin ( / ) q 1₊ 2 2 or v gh k R 2 2 2 2 1 = + / Qhs =     sinq or v gh k R = + 2 1 2 2 ( / )

20. The particle of mass m is initially at A. At the point

P where the particle loses contact with the surface

of the sphere, normal reaction is zero and the only force acting on the particle is its weight mg acting vertically downwards, as shown in figure.

The radial component (mg cos q) of the weight provides the necessary centripetal force,

mg mv r cosq= 2 or v rg2 _{= cosq ...(i)} As cosq =OQ = −( ) , OP r h r v2 = (r – h)g ...(ii)

Since the velocity v has been acquired by the particle after falling through a height h,

v= 2 gh ...(iii) From eqns. (ii) and (iii),

2gh = (r – h)g \ h r=

3 ...(iv)

From eqns. (iii) and (iv),

v= 2gr

3 ...(v)

If x is the horizontal and y is the vertical distance covered by the particle as it hits the ground at C after time t,

x = (v cos q)t ...(vi)

and y v=( sin )qt+1gt

2 2 ...(vii)

From eqns. (vi) and (vii),

y x gx v = tan + cos q q 2 2 2 2 ...(viii) It is clear that, y= − = − =2r h 2r r r 3 5 3 , cosq = − = −r h / = , r r r r 3 2 3

and tanq= sec2q− =₁ ( / )_{3 2}2− =₁ 5 2 From eqn. (viii),

5 3 5 2 2 2 3 2 3 2 2 r x gx gr =    + ( / )( / ) or 27x2+(8 5r x) −( / )80 3r2=0 i.e., x= − r+ r + r × 8 5 320 2880 2 27 2 2

(neglecting negative root) or x= −17 9 56 6r+ r= r= r

54

38 7

54 0 72

. . . _.

If s is the distance from the initial position where the particle strikes the ground at C,

s = BC = BD + DC = rsin q + x =    + = + r 5 r r r 3 0 72. 0 75 0 72. . ≈ 1.5r OR

Let us consider the mass after it has slide down a distance x [as shown in figure].

If a is the acceleration at this instant, then

ma = mg sin q – f

= mg sin q– µ₀x mg cos q

[Q f = µR = (µ0x)mg cos q]

or a = g(sin q – µ₀x cos q) ...(i)

As a dv dt dv dx dx dt dv dx v = = _ __ _= _ _{ ,} v dv = a dx = g (sin q – µ0 xcos q) dx or _∫v_{v dv=( sin )g q ∫}x_{dx−(µ gcos )q∫}x_{x dx} 0 0 0 0 or v2 g x ₀g x2 2 = − 2     ( sin )q µ cosq

or v2 = (2g sin q)x – (µ₀g cos q)x2 ...(ii) If the mass comes to rest after covering a distance s, then if x = s, v = 0.

From eqn. (ii),

(2g sin q)s – (µ₀g cos q)s2 = 0 or s[2g sin q – (µ₀g cos q)s] = 0 As s ≠ 0, 2g sin q – (µ0g cos q)s = 0 or s g g = 2 = 2 0 0 sin cos tan q µ q µ q ...(iii) When v = vmax , a dv dt d dt v = = ( _max)=0

If x = x0, for a = 0, from eqn. (i)

g(sin q – µ x₀ cos q) = 0 or µ x₀ cos q = sin q or x₀ 0 = tanq µ ...(iv)

Putting v = v_max and x = x₀ = tanq µ₀  



 in eqn. (ii),

v_max2 ( sin ) tang ( gcos ) tan

0 0 0 2 2 =    −     q q µ µ q q µ or v_max2 gsin tan g sin tan

0 0 2 = − µ q q µ q q = g µ₀sin tan q q or vmax = _µg sin tanq q

0