Circular Motion & WPE Solution

IIT-ian’s PACE Edu.Pvt.Ltd.
Circular Motion & WPE Rg-C&WPE-11
CIRCULAR MOTION
EXERCISE – 1
1. d
 = rate of change of angle
1
  1 as they both complete 2 angle in same time.
2
2. c
mv 2
mg  N 
r
mv 2
 N  mg 
r
Since rA  rB
 NA  NB
3. a
Force is always perpendicular to displacement
 work done = 0
4. b
a resul tan t  a c2  a tan
2
g.
2
 30 2  2 181
   2   2.7 m / s 2
 500  5
5. a
f = mg
  N  mg
  mr2  mg
g
 
r
6. c
If the coin just slips at a distance of 4r from centre
  mg = m4r2 … (1)
If angular velocity is doubled
 mg = mR (2)2 … (2)
From (1) and (2)
 R=r
7. d
T = mr 02 … (1)
2
2T = mr  … (2)
   20  2  5 rpm
8. c
Since force is always perpendicular to velocity particle moves in circle. It’s speed is constant
and velocity variable.
9. c
1
IIT-ian’s PACE Edu.Pvt.Ltd. Circular Motion & WPE Rg-C&WPE-11
mv 2max
mg 
r
v max  gr  0.3  10  300
 30m / s  108 km / hr
10. c
mv 2
Fnet ma rad. 
r
11. b
T = mr2 = 0.2  0.5  42 = 1.6N
12. c
Centripetal force is provided by friction
13. a
mv 2
N A  mg 
r
mv 2 mv 2
N A  mg  , N B  mg 
rA rB
mv 2
N C  mg 
rc
14. a
Real forces are mg and T only
15. a
Bead starts slipping, when
N = mL2
mL = mL2
 = (0 + t)2

 t

EXERCISE – 2
1. (a, c)
2h 25
Time to fall =   1s
g 10
Distance covered in y direction = v  t
=31
= 3m
Since there is no velocity along x-direction x is always 2m.
2. (a, b)
For no wear and tear friction is zero
v2
  tan15o
Rg
 v  Rg tan15o  28.1 m / s
2
Rg(  tan )
v max   38.1m / s
1   tan 
3. (b, c)

 dr  
v ;  v || dr (i.e. along the tangent)
dt

 v
a avg  0
t
4. (a, b)
dS
vK S
dt
S t
dS
 0 S 0 Kdt

2 S  Kt
K 2t 2
S
4
dS K 2 t
v 
dt 2
5. (a, b, c)
T m2 , v   , Fvert = 0
6. (a, b, d)
v
   constant,  = t
R
Fy = –F sin  = –F sin t
mv 2
 sin t
R
Vr = –vsin = –v sin t
x-coordinate = R cos t
7. (b, d)
If  = 0.1, fmax = 0.1  0.5  10 = 0.5N
Req. centripetal force = mr2 = 0.5  1  0.52 = .125N
1
 f  N, Tension  zero
8
1 1
If   , f max   0.5  10  0.25 N
20 20
1
 f  N, Tension  zero
8
1 1
If   , f max   0.5  10  0.125 N
40 40
3
1
 f  N, Tension  0
8
EXERCISE – 3
1. (a)
v2
 K 2 rt 2 (given)
r
(a) Centripetal force = mK2rt2
dv
(b) Tangential force = m  mKr
dt
 
(c) Power of centripetal force  Fcentripetal . v  0
 
(d) Power of tangential force  Ft . v  Ft v
 m.K 2 r 2 t
2. (b)
Assuming no friction between m1 and m2
T
a1 = R2 -
m1
T
a2 = R2 -
m2
 a1  a2
 Friction on upper block acts towards left and on lower block towards right.
3. (a)
Let the required angular velocity be 
Then
T + m1g = m1R2 … (1)
T = m1g + m2R2 … (2)
 2 m1g = (m1 – m2)R2
2m1g
   6.3 rad / s
(m1  m 2 )R
4. (b)
T = m1g + m2R2
= 0.5  2  10 + 1  0.5  40 = 30N
5. (b)
v 2design 1
tan    ;
gR 2
 v2 
f  m  g sin   cos  
 R 
 300 5 m / s
6. (a)
4
 v2 
f  m  cos   g sin  
R 
 500 5 m / s
7. (a)
1
  tan 1
2
8. (c)
v2
mg sin  = m … (1)
L
1 1
m( 3gL) 2  mv 2  mgL(1  sin ) … (2)
2 2
1 1
   sin 1   ; Also v 2  gL
3 3
9. (c)
0 2  v 2 cos 2 
h max  L(1  sin ) 
2g
40L

27
10. (b)
1
m( 3gL)2  mg h max
2
3L
 h max 
2
EXERCISE – 4
1. Here frictional force will provide the required tangential and  v2 

2
2
centripetal force for the circular motion of car. a net  a 
R 
Force of friction will act along the direction of net acceleration.  
2
 v2 
2
f  m a  
R
2 a
 v2  2 2
Car will skid when m a    v
R R
a4  t4
 m a2 
R2
1/4
 ( 2g 2  a 2 )R 2 
 t 
 a4 
5
1 2 R 2g2  a 2
Till this time distance traveled by car is D  at . Putting value of t we get D  .
2 2a
2. The insect will slide when mg sin  becomes equal to f max  N
limiting friction. At every instant the insect is in N
equilibrium. 
So, N = mg cos
mg cos 
N = mg sin  mg sin 
–1
  mg cos  = mg sin    = tan    = tan  = mg
tan–1(1/3)
3. Applying Newton’s law towards the centre of circle we get 

N = m2R
B
Let  be the minimum angular speed for which man is not falling.
At this instant its weight will be balanced by limiting friction acting /6
upwards.
i.e, N = mg  0.15  70 2  3 = 70  10   3
  = 4.7 rad/sec. 2
A 3 T

T P
Mg
4. Applying Newton’s law along vertical we get Q

cos  cos 
T T  mg … (1)
6 3
2mg
 T P R
O
3 1
Applying Newton’s law along horizontal.
   3
We get T cos  T cos  m2 R  m2
3 6 2
2g
 2 
 3

5. Particle P and Q will be at same angular position whenever 5t = 2t +  2n .
2
1 2n
 t  (n = integer) … (1)
6 3
Similarly, particle P and R will be at same angular position.
Whenever 5t = 3t =  + 2m (m = integer)
1
 t  m … (2)
2
6
All three particles will be at same angular position when (1) = (2)
1 1 2n
 m  
2 6 3
 2n = 1 + 3m
Smallest integral value of m & n satisfying the above equation is m = 1, n = 2.
Putting these values in (1) & (2) we get t = 1.5 sec.
So, they will meet for the first time at t = 1.5 sec.
v2
6. According to question, a c  a t 
R
2 v t
dv v dv dt
    2 
dt R v0
v 0
R
1 1 t Rv0
    V(t) 
v v0 R R  tv0
2 R T
dt  Rv 0
  dx  Rv0  R  tv  2R  (n(R  tv0 ) 0T
0 0 0 v0
R
 T (1  e 2  )
v0
7. Lets assume that the particles meet at time t. Distance traveled by
A = distance traveled + R by B
v
1 72v 2 t 2 B at
 vt    vt  R A
2 25R
5R v
 t
6v
dis tan ce travelled 11
Angle traced by A  
R 6
v  at 17v
Angular velocity  
R 5R
EXERCISE – 5
1. (c, d)
  mv 2
F  v  P = 0  kinetic energy = constant; F is constant (given)   constant
R
 R = constant.
2. (a)
Radius of curvature in (a) is minimum
3. (a)
1
mg sin  = mg cos   cot  = 3
3
4. (c)
   
a  a tan gential  a normal and a tan gential is downward.
7
R mg 5 3
5. Kx cos 30° + mg cos 30° = mat. As x  and K  , at  g;
4 R 8
3mg
N + Kx cos 60° = mg cos 60°  N 
8
8
WORK POWER & ENERGY

EXERCISE – 1
1. D
x 5
x  x2 x 5
2 2 3
W  Fdx   (7  2x  3x )dx  (7x  x  x ) y
x  x1 x0 x 0
= 135 J
2. A
This is the statement of Work – Kinetic Energy Theorem
3. B
Since the force acting on the particle is perpendicular to the displacement everywhere, the work
done iz zero
4. B
 
Instantaneous power, P = F . v  (10iˆ  10ˆj  20k)
ˆ . (5iˆ  3jˆ  6k)
ˆ
= 140 W
5. D
M
Mass of the hanging part = ; when the hanging part of the chain is paralled on the table, its
3
L
centre of mass is raised by .
6
 M   L  MgL
The work done = rise in potential energy =    g   
 3  6 18
6. B
1
mv 2  mgR  v  2gR
2
7. B
Klonger llonger = Koriginal loriginal
k 3
 K longer   k
(2 / 3) 2
8. C
WF = increase in potential energy = mgL (1 – cos )
9. D
work done by gravity
Mean power of gravity = 0
time elapsed
10. B
9
Acceleration, a = –kx
 F = –Kx  loss of KE : gain of potential energy  x2.
11. C
t3
x  v  t2
3
1
Now, W = KE = m(vf2  v 22 ) = 16J
2
12. C
KEmax = Maximum loss of KE = Mgl (1 – cos )
13. A
h
Centre of mass of the rope is lifted by and the back by h. Therefore,
2
h  m
W = Mgh + mg   M   gh
2  2
14. D
x v
dv
x mg = m v    xg dx   mv dv
dx x 0 v 0
 E  x2
15. D
 3mg 
W  d
 4 
16. C
B
  FR
W   F . dr 
A
2
17. C
dW 3t 2
P   W  4J  v  2m / s
dt 2
18. C
W1 : W2 : W3 = Ratio of corresponding displacements = 12 : (22 – 11) : (32 – 22)
=1:3:5
19. C
1/6
dv 12a 6h  2a  a b b2
 13  7  0  x     U min    
dx x x  b  (2a / b) 2 (1a / b) 4a
b2
 Minimum energy required =
4a
10
20. D
P
 mg vmax = P  vmax =
 mg
21. C
mg
W  163.3 J
1 
22. C
The KE intercepted  v3
23. D
v2  5g 
T – mg = m  T  mg    6mg
   
24. A
1 2
mg(h + x) = kx  980x2 – 2  9.8 (0.4 + x) = 0
2
2
 50x – x – 0.4 = 0
 (10x – 1) (5x + 0.4) = 0
 x = 0.1 m = 10 cm
25. D
y a
W  (kx ˆj) . (ai)

ˆ 
 k(yiˆ  aj)ˆ . dy ˆj  ka
2
y 0
EXERCISE – 2
1. c, d
 v2
Since no work is done by the force speed is constant not velocity. a  along the centre of
r
circle.
2. b, c
W.d by all forces = K.E.
 Wg + WN = K.Ef – K.Ei
1
 mgh + 0 = mv 2  0
2
 v  2gh  v P vQ
where h is the initial height of both blocks from ground.
3. b, c
4. b, c
 
w.d = F . d
 (î  2ˆj  3k)
ˆ . 3jˆ
=6J
11
 
w.d. = F . d
 (î  2ˆj  3k)
ˆ . (3jˆ  4k)
ˆ  18J
5. a, b, c
W.d = Area enclosed by the F-x graph
6. a, d
Fnet 10  0.2  2  10
a   3 m / s2
m 2
v(t = 4s) = 0 + 3  4 = 12 m/s
1
s (t = 4s) =  3  42  24 m
2
w.d by net force = K.E
1 1
 mv 2   2  122
2 2
= 144 J
w.d. by applied force = 10  24 = 240 J
 
w.d by friction = F . d = – 4  24 = –96 J
7. b, c
8. b, c, d
9. a, b, d
w.d by all forces = K.E.
 Fx – mgx = 40 J
 20x = 40 J
x=2m
w.d.gravity = – mgx = –2  10  2 = – 40 J
w.d.tension = Fx = 40  2 = 80 J
10. a, d
 
Power = F .v  Fv  F  at or F  2ax
Since ‘a’ and ‘F’ are constants
Power varies linearly with time and parabolically with displacement
11. a, c
Hint: Direction of spring force and displacement are same in (a) & (c)
12. b, c
 
Pmg  F . v
 mg(ˆj) .[u cos î  (u sin   gt)ˆj]
= – mg (u sin  – gt)
u sin  2u sin  u sin 
 P  0 for t  and P  0 for t
g g g
12
13. a, c, d
 F U ˆ U ˆ
a  i j   3iˆ  4jˆ
m mx my
v(at x  0)  u 2  2as
 02  2  5  10
= 10 m/s
   1
x  x 0  ut  at 2
2
1
 6iˆ  4ˆj  0  (3iˆ  4ˆj)12
2
ˆ
 4.5i  2 j ˆ
14. c, d
w.d by all force = increase in spring energy
2
1 mg 2 1  mg 
 Fx0 + mgx0 = k(x 0  )  k 
2 k 2  k 
2F
 x0 
k
15. b, c, d
 
w.d. by F2 15   6  45 J
2

w.d. by F3  30  6  180 J
  
w.d. by F1   F1 cos  90   rd
 2

F1 is conservative in nature as it is always directed towards P.
16. b, d
mv2
At highest point Fnet 

2
mv
 2mg  mg 

 v  3g
Conserving energy velocity at lowest point
v  7g
17. b, d
 
w.d. = F . d  (6iˆ  6ˆj) . ( 3iˆ  4ˆj)
= –18 – 24 = – 42 J
Had there be no initial velocity particle must have moved along straight line making an angle of
45° with x-axis.
13
18. a, c, d
 
P  F . v   ive angle is acute
= – ive angle is obtuse
Area under graph = K.E.
= 20J
19. (a, c) v2
m v1 M
mv1 = Mv2, where v1 and v2 are speeds of mass m and M,
l
as seen from ground. The velocity of m relative to M is v12
= v1 – (– v2).
1 1
Hence, or t   or v1 + v2 = l/t.
V12 V1  V2
20. (a, b)
dU 2A B 2A
F    3  2 at equilibrium F = or, r 
dr r r B
2 2
2A B B
At infinity U = 0 r  ,U U  .
B 4A 4A
21. (a, c)
From conservation of linear momentum (1 + 2)v = (6 x 1) + (2 – 3) v = 4m/s (of both the blocks)
1
From work energy therom i.e., Wtotal = KE on 1kg block, Wf   1 (42  6 2 )  10J
2
1
on 2kg block Wf   2(42  32 )  7J .  Net work done by friction is –3J.
2
22. (b, d)
In region OA particle is acccelerated, in region AB particle has uniform velocity while in region
BD particle is deceleration., Therefore, work done is positive in region OA , zero in region AB and
negative in region BC.
23. (a, c)
v2 2gR
at B acceleration of block    2g
R R
24. (a, d)
EXERCISE – III
1. (A – q)
Work energy theorem – w.d. by all forces is equal to change in K.e.
(B – s)
Negative of work done by conservative force is equal to change in potential energy
(C – r)
W.dext. + Wnon cons. = K.E. – Wcons.
= K.E. + U
= T.M.E
2. (A – r)
14
1 1
mu 2  mgR  mv 2B
2 2
 v B  7gR
(B - q)
1 1
mu 2  mg  2R  mvC2
2 2
v C  5gR
(C – p)
mv 2B
TB   7 mg
R
(D – t)
mv 2C
TC  mg 
R
m5gR
TC   mg  4mg
R
3. (A – q)
4 4
 kx 2  1
w.d   kx dx     k[42  2 2 ]   ive
2  2 2 2
(B – p)
2 2
 kx 2  1
w.d   kx dx     k[22  42 ]   ive
4  2  4 2
(C – r)
2 2
 kx 2 
w.d   kx dx    0
2  2  2
4. (A – t), (B – p), (C – s), (D – q)
1
S   2  (4) 2  16m
2
w.dgravity = – mg  16 = –1  10  16 = –160 J
w.dnormal reaction = N cos   S = m(g+a) cos2  S = 144J
w.dfriction = f  S  sin
= m(g+a) sin2  S = 48 J
w.d.forces = K.E.
1 1
 m(at)2   1 (2  4) 2  32J
2 2
5. c
work done by both against gravity = mgh
6. b
mgh 50  10 15
Average Power =   250W
t 30
7. b
15
Chemical energy expended by the physicist ends up increasing the potential energy.
8. b
As the physicist falls, gravitational potential gets converted into kinetic energy, increasing his
speed. After he hits the cushion, this kinetic energy gets converted into heat.
9. d
K.E. = K.Ef – K.Ei = K.Ef – 0
= w.d. by mg
= mgh
15
= 50  10   2500J
3
10. a
11. c
12. a
1 1
From conservation of energy at A and B, we have mv2A  mv 2B  mgR(1  sin ) … (1)
2 2
At B the string becomes slack. Therefore
mv 2
mg sin   B … (2)
R
After passing through B, the ball goes in a projectile
 v B sin t  R cos  … (3)
1
and  v B cos t  gt 2  R  R sin  … (4)
2
On solving 1, 2, 3 & 4
 = 30°
7gR gR
v and v B 
2 2
13. c
 (  - x)v +  hg dt – v2dt
= (  - (x + dx)] (v + dv), where  is mass per unit length
 hg dt = (  - x) dv
x v
dx
 hg 
  x v 0
 v dv
x 0
v2 
  hg ln
2 x

 v at B  2gh ln
h
14. a

KE x   (  x) gh ln
x
16

It is maximum, when e
x
 mgh
 KE max   hg 
e e
15. b
1 
Heat generated = (  - h) gh – . h . 2gh ln
2 h
mgh  

    h  h ln h 
16. b
17. a
18. b
19. c
U(x) = 20 + (x – 3)2
At x = 0,
T.M.E = U + K.E
= 20 + 9 + 20 = 49 J
At extreme positions, K.E = 0
 U = 49J
 20 + (x – 3)2 = 49
 x – 3 =  29
 x  3  29 i.e.,  3.4 and 7.4 m
K.E.max = T.M.E – Min. potential energy
= 49 – 20 (at x = 3)
= 29 J
Body is in equilibrium at min. potential energy
i.e., at x = 3
20. Wmg is path and rate independent
21. Total energy is conserved when there is no external and no internal non conservative force.
22. Work done for conservative forces are path independent
23. Wmg = mgh

   
24. F.  r  F . (r2  r1 )
2vsin 
25. t is time of flight and vertical displacement is zero.
g
17
EXERCISE – 4
1. WF = Fh = 80J; Wweight = –(mg)h = –40 J.
2m1m 2 m  m1
2. Tension, T  g; acceleration a = 2 g
m1  m 2 m1  m 2
1 m m (m  m )
 W = T . at 2  1 2 2 2 1 g 2 t 2
2 (m1  m 2 )
200
 J
9
x 2
3. W  (2  x) dx  3.5J
x 1
F F
4.  N  mg;  N
2 2
F  F  F 
    mg     mg  3.6
2  2 2  1
F
(a) WF  S  7.2J
2
(b) Wfriction = –WF = –7.2J
(c) Wgravity = 0
1
5. W = Area under the curve = 10  2 +  2  10
2
= 30J
6. WF = increase in potential energy = mg (1  cos )
7. dW = mg( cos + sin ) ds = mg (d  dh)
 W = mg( h)
1
8. W = KE =  (2) 202   400J
2
1
9. W  KE  m 2 v
2
1 4mg
10. a) (2m)g xm = k x 2m  x m 
2 k
2
1 1 x  x 
b) (3m)v 2  K  m   (2m) g  m 
2 2  2   2 
m
 v  2g
3k
xm g
c) 2mg  k  (3m) a  a 
4 3
 2m g
A  m
11. K   mg  m A 
 K  2
12. Let x be the extension of the spring and  the angle that the spring makes with the vertical at
break off.
18
0.4
K x cos  = mg  40x  3.2
0.4  x
 x = 0.1m; The … of B = the slide of A  (0.5)2  (0.4)2  0.3 metres = h (say)
1 1
(2m)v 2  Kx 2  mg h
2 2
 v = 1.54 m/s
v
mv 2 dv dv  t
13.  m   2  dt
R dt v  v0
v R t 0
t
1 1 t R R  v 
    v  v0 ; S   v dt  ln  1  0 t
v v0 R R  v0 t 0
  R 
1 Fb(1  sin )
14. Fb(1 sin  )  2  mv 2  v  , Fmax  2mg
2 m
15.
Conserving mechanical energy:
2
1 1  2v 
2  10  1 = 0.5  10  ( 5  1)   2  v 2   0.5  
2 2  5
 v = 3.39 m/s
2
  (4,6)
16. W   F . ds   (3x 2î  2yj)
ˆ . (dxiˆ  dy ˆj  dzk)
ˆ
1 (2,3)
6
4
(4,6)
  (3x 2 dx  2y dy)  x 3  y 2
(2,3) 2
3
= 83J
c
17. (i) W  (xy î  xy ˆj) . (dx î  dy ˆj)
0 (alongOC)
4 1
2
  x (iˆ  ˆj) . 2dx î   2x 2 dx  J
2
0 0
3
A C
(ii) W  xy (iˆ  ˆj) . dx î   xy (iˆ  ˆj) . dy ˆj
0 (along OA) A (along AC)
y 1
1
0  y dy  J
y 0
2
k 1
1
(iii) W 0  x dx  J
k0
2
18. (a)  dmg h  mgh
x
m  1
(b)   dx  gx  mg
x 0 
  2
19
  /R
m  mgR 2 
(c) 0   
Rd  g (R cos )  sin
 R
19. (a) u cos  = v
2
1 1  v 
(b) m  10  5  mv 2  m  
2 2  0.8 
40
 v m/s
41
2
3  1 D 1
20. mg  d   k    mv 2
4  2  4  2
3g k
 vd 
2d 16m
1
21. mg h min  mg 2r  m ( gr ) 2
2
5r 1
 h min  ; mg(5r) – mg 2r = mv 2
2 2
2
mv
Now, Fresul tan t   6mg
r
mv 2
22. mg (1 – cos ) = … (1)

1 3
mg  mg(1  cos )  mv 2 … (2)
2 2
From (1) and (2)
g 2
v ;   cos 1
3 3
1
23. mv02  mg(1  cos 60o )
2
 v 0  g  9.8  5  7 m / s
1 1
24. m( 5gR ) 2  mgR(1  cos )  mv 2 … (1)
2 2
2R sin  2vsin 
Also, t flight   … (2)
v cos  g
From (1) and (2)
 = 0 or  = 60°
1 1
25. mv2  mg.2R  mv 2 , where v is velocity at the highest point.
2 2
 v  u 2  4gR
Now, v tflight = 3R
4R
 u 2  4gR  3R
g
20
5
 u gR ;
2
4R
x min  v min t flight  gh  2R
g
x R
 x  1
26.  (dx) g  r sin  x   ( r ) v 2
x 0  r  2
 2 
 v  2gr   
 2
1  1
27. mgR   1  cos    mv 2 … (1)
4  2
v2
mg cos  = m … (2)
R
5
   cos 1
6
 2 
 n 1   1  n 1 0
28. mg    1  Mg n  1     1
 n 3   2  
 
m n 1
 2 1
M n 3
1
29. (F R 2  mgR)  mv 2
2
F 2 
 v  2R  g
 m 
 
x  2L
1 1
30. mv 20   (Gx)(mg) dx  k L2
2 x 0
2
k
 v  4ag 
m
 2 
31. Stretch x = 0.4 (sec 30° - 1) = 0.4   1 ; kx sin 30° =  (mg – kx cos 30°)
 3 
mg
 kx 
sin 30   cos 30o
o
1
Now, W = U = kx 2  0.09J
2
1
32. a) mg (1 – cos ) = mv 2 … (1)
2
mv 2
F + mg cos  = … (2)
R
From (1) and (2)
F = mg (2 – 3 cos )
21
N = Mg – 2F cos , = Mg – 2 mg (2 – 3 cos ) cos 

1
which is minimum when  = cos–1
3
m 3
b) N=0 
M 2
EXERCISE - V
1. (b)
The centripetal acceleration
v2
a c  k 2 rt 2 or  k 2 rt 2
r
 v = krt
dv
So, tangential acceleration, a t   kr
dt
Work is done by tangential force.
Power = Ft . v. cos 0º
= (m at) (krt)
= (mkr) (krt)
= mk2 r2t
2. (b)
The force constant of a spring is inversely proportional to the length of the spring.
Let the original length of spring be L and spring constant is K (given)
Therefore,
2L 3
KL  K'  K'  K
3 2
3. (d)
dU(x) = – Fdx
x
 U x  –  Fdx
0
kx 2 ax 4
 –
2 4
2k
U = 0 at x = 0 and at x  ;  we have potential energy zero twice (out of which one is at
a
origin).
Also, when we put x = 0 in the function,
dU
We get F= 0. But F  –
dx
dU
 At x  0;  0 i.e. the slope of the graph should be zero. These characteristics are
dx
represents by (d).
22
4. (b)
Let the maximum extension of the spring be x as shown in the figure. Work is
done by the gravitational and the spring force. There is no change in the kinetic
energy between the initial and final position of the mass.
From Work-energy theorem;
Wg + Ws = 0
Where Wg = work done by gravity

And Wsg = work done by spring
1
  Mgx – kx 2  0
2
2Mg
 x
k
5. (b)
In a conservative field work done does not depend on the path. The gravitational field is a
conservative field.
 W1 = W2 = W3
6. (b)
We know that
x x
U  –  Fdx or U  –  k xdx
0 0
2
kx
 U  x  – U 0   –
2
Given U(0) = 0
kx 2
Ux   –
2
7. (d)
v  5gL … (1)
2
v 2
   v  2gh … (2)
2
h = L(1 – cos ) … (3)
Solving Eqs. (1), (2) and (3), we get
7  7
cos    or   cos 1     151o
8  8
8. (c)
When the block B is displaced towards wall 1, only spring S1 is compressed and S2 is in its
natural state. This happens because the other end of S2 is not attached to the wall but is free.
1
Therefore the energy stored in the system  k1x 2 . When the block is released, it will come
2
back to the equilibrium position, gain momentum, overshoot to equilibrium position and move
towards wall 2. As this happens, the spring S1 comes to its natural length and S2 gets
compressed. As there are no frictional forces involved, the P.E. stored in the spring S1 gets stored
as the P.E. of spring S2 when the block B reaches its extreme position after compressing S2 by y.
23
1 1
 k 1x 2  k 2 y 2
2 2
1 1
 kx 2  4ky 2
2 2
x2 = 4y2
y 1
 
x 2
9. (b)
The forces acting on the bead as seen by the observer in the accelerated frame are: (a) N; (b)
mg; (c) ma (Pseudo force).
Let  is the angle which the tangent at P makes with the X-
axis. As the bead is in equilibrium with respect to the wire,
therefore
N sin  = ma and N cos  = mg
a
 tan   ...  i 
g
But y = kx2. Therefore,
dy
 2kx  tan  ..  ii 
dx
From (i) & (ii)
a a
2kx   x 
g 2kg
10. Let M strikes with speed v. Then, velocity of m

2
at this instant will be v cos  or v . Further
5
M will fall a distance of 1 m while m will rise
up by ( 5  1)m . From energy conservation:
decrease in potential energy of M = increase in
potential energy of m + increase in kinetic
energy of both the blocks.
2
1 1  2v 
or (2) (9.8) (1) = (0.5) (9.8) ( 5  1)   2  v 2 +  0.5   
2 2  5
Solving this equation, we get v = 3.29 m/s
24
11. Let the string slacks at point Q as shown in figure. From P to Q

path is circular and beyond Q path is parabolic. At point C,
velocity of particle becomes horizontal, therefore, QD = half the
range of the projectile.
Now, we have following equations
mv 2
(1) TQ = 0. Therefore, mg sin  = … (i)
L
(2) v2 = u2 – 2gh = u2 – 2gL (1 + sin ) … (ii)
1
(3) QD = (Range)
2
 L  v 2 sin 2(90o  ) v 2 sin 2
  L cos      … (iii)
 8 2g 2g
Eq. (iii) canbe written as
 1   v2 
 cos     sin cos 
 8   gL 
 v2 
Substituting value of    sin  from eq. (i), we get
 gL 
 1 2 2
 cos     sin     (1  cos ) cos 
 8
1
or cos    cos   cos3 
8
1 1
 cos3   or cos   or   60o
8 2
From Eq. (i) v2 = gL sin  = gL sin 60°
3
or v2  gL
2
 Substituting this value of v2 in Eq. (ii)
u2 = v2 + 2gL (1 + sin )
3  3
 gL  2gL  1  
2  2 
3 3
 gL  2gL
2
 3 3
 gL  2  
 2 
 3 3
u  gL  2  
 2 

12. The ball is moving in a circular motion. The necessary centripetal force is provided by
(mg cos – N). Therefore,
25
mv 2
mg sin  – N A  ...  i 
 d
R  2 
 
According to energy conservation
1  d
mv2  mg  R   1 – cos   ...  ii 
2  2
From (i) and (ii)
NA = mg (3 cos  – 2) …(iii)
The above equation shows that as  increases NA decreases. At a particular value of , NA will
become zero and the ball will lose contact with sphere A. This condition can be found by putting
NA = 0 in eq. (iii)
0 = mg(3cos  – 2)
2
   cos –1  
3
The graph between NA and cos
From equation (iii) when  = 0, NA = mg.
2
When    cos –1  
3
The graph is a straight line as shown

2
when   cos –1  
3
26
mv 2
N B –  mg cos   
d
R
2
mv2
 N B  mg cos   ...  iv 
 d
R  2 
 
Using energy conservation
1  d  d 
mv 2  mg  R   –  R   cos 
2  2  2 
2
mv
 2mg 1 – cos  ...  v 
 d
R  2 
 
From (iv) and (v), we get
NB + mg cos = 2mg – 2mg cos
NB = mg(2 – 3 cos)
2
When cos   , N B  0
3
When cos = –1, NB = 5 mg
13. Given m = 0.36 kg, M = 0.72 kg.

The figure shows the forces on m and M. When the system is released, let the acceleration be a.
Then
T – mg = ma
Mg – T = Ma
M – mg
 a g/3
Mm
and T = 4mg/3
For block m:
u = 0, a = g/3, t = 1, s=? T
1 1 g T
s  ut  at 2  0    11  g / 6
2 2 3 a M
 Work done by the string on m is M a
  mg 9 4  0.36  10  10
T. s  Ts  4    8J Mg
3 6 3 6
Mg
27

Circular Motion & WPE Solution

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IIT-ian’s PACE Edu.Pvt.Ltd.

Circular Motion & WPE Rg-C&WPE-11

1. Here frictional force will provide the required tangential and  v2 

3. Applying Newton’s law towards the centre of circle we get 

4. Applying Newton’s law along vertical we get Q

WORK POWER & ENERGY

W  (kx ˆj) . (ai)

20. Wmg is path and rate independent

22. Work done for conservative forces are path independent

23. Wmg = mgh

N = Mg – 2F cos , = Mg – 2 mg (2 – 3 cos ) cos 

Where Wg = work done by gravity

10. Let M strikes with speed v. Then, velocity of m

11. Let the string slacks at point Q as shown in figure. From P to Q

The graph is a straight line as shown

13. Given m = 0.36 kg, M = 0.72 kg.

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