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Transcript of ALP Sol P Circular Motion E
8/18/2019 ALP Sol P Circular Motion E
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SOLN_Circular Motion - 1
TOPIC : CIRCULAR MOTION
PART - I
1. at = 0 = constant
2. Velocity chnages as its direction changeAcceleration changes as its direction change.
3. |v| = constant
4. T = m!2RT + 60 = m (3!)2R
m!2
R = 8
60
!2R = 2.5 m/s2
9!2R = 22.5 m/s2
5. (C) as # = 0 $ at = 0
% FC = ma
C = m!2R = constant
6. mg = constant vector
7. When critical velocities are taken T = 6mgmax
A mg
TB
BTD
D
T = 0A
C
TA + T
C = 6 mg = constant
TB + T
D = 6 mg = constant
8. Let v be the speed of particle at B, just when it is about to loose contact.From application of Newton's second law to the particle normal to the spherical surface.
r
mv 2
= mg sin & .......... (1)
Applying conservation of energy as the block moves from A to B..
2
1 mv2 = mg (r cos # – r sin &) .......... (2)
Solving 1 and 2 we get3 sin & = 2 cos #
9. % T2 # d3 $
3
13
122
2
1
10
10
T
T
'
'
(
)
*
*
+
, -
'
'
(
)
*
*
+
, $
2
1
T
T =
1010
1
% T1 : T
2 = 1 : 1010
10. Initial extension will be equal to 6 m.
% Initial energy =2
1 (200) (6)2 = 3600 J.
Reaching A :2
1mv2 = 3600 J
$ mv2 = 7200 J
From F.B.D. at A :
N =5
7200
R
mv 2- = 1440 N
PHYSICS SOLUTIONS OF"ADVANCED LEVEL PROBLEMS"
Target : ISEET (IIT-JEE)
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SOLN_Circular Motion - 2
11. Since F!
. V!
, the particle will move along a circle.
% F =R
mv 2
& / =R
S$ / = 2mv
FS
12. x2 = 4ayDifferentiating w.r.t. y, we get
dx
dy
= a2
x
% At (2a, a),dx
dy = 1 $ hence / = 45°
the component of weight along tangential direction is mg sin /.
hence tangential acceleration is g sin / =2
g
13. As tangential acceleration a = dV/dt = !dr/dt
but ! = 20n= 40 rad/s anddt
dr = (2 × 0.75) × 10 –2 m/s = 1.5 × 10 –2 m/s (reel is turned uniformly at the rate
of 2 r.p.s.)
% a = 40 × 1.5 × 10 –2 m/s2 = 60 × 10 –2 m/s2
Now by the F.B.D. of the mass.
w
T
T – W = ag
W% T = W (1 + a/g) put a = 60 × 10 –2 % T = 1.019 W
14. For A :
T – m!2r – ma = 0 .............(i)Seen from object itself
T – 3
mg = ma .............(i)
For B :mg – T = ma .............(ii)(i) – (ii)
2T =3
4mg
T =3
2mg
15. |v| = constant
16. ! = constant
% # = 0 = constant
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SOLN_Circular Motion - 3
17. F = 22t
2 )mg(Ff 11 > mg#
f
F
mg
(f = m r)!2
(asdt
Vd = constant % F
t = constant)
Now when the angular speed of the rod isincreasing at const. rate the resultant force
will be more inclined towards f! .
Hence the angle between F!
and horizontal plane decreases so as with the rod due to increases in
f = m!2r only.
PART - II
1. (a) The system is in equilibrium whenm
1gsin/ = m
2g
or sin/ =1
2
m
m
(b) Let the tangential acceleration of m1 be a.% m
2g – m
1g sin/ = (m
1 + m
2) a
a =9
2540 2 =
9
15m/s2
th e normal acceleration of m1 is zero. # speed of m
1 is zero.
% The magnitude of acceleration of m1 =
9
15m/s2 .
2. (a) Applying conservation of energy between initial and final position isLoss in gravitational P.E. of the bead of mass m = gain in spring P. E.
% mg R =
2
1 K (2R – 2 R)2
or K =)223(R
mg
2(b) At t = 0a
t = g
ac = 0
at lowest pointa
t = 0
ac = 0
The centripetal acceleration of bead at the initial and final position is zero because its speed at both positionis zero.The tangential acceleration of the bead at initial position is g.
The tangential acceleration of the bead at lower most position is zero.
3. Since belt is not slipping, speed at rim of A and B is same
!Ar
A = !
Br
B
!A = 100 ×
10
25 = 250 rpm = 250 ×
60
20 rad/sec.
=3
250 rad/sec.
! = !0 + #t
t = 2 /
03
25
0
20
= 3
50
sec.
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SOLN_Circular Motion - 4
4. !0
2 = 900 (rad/sec)2 $ !0 = 30 rad/sec.
!2 = 1600 (rad/sec)2 $ ! = 40 rad/sec.
/ = t2
0 ' (
)*+
, !1!
t =3040
21002
1033
=7
400sec.
5. (a) As a rod AB moves, the point ‘P’ will always lie on the circle.
% its velocity will be along the circle as shown by ‘VP’ in the figure.
If the point P has to lie on the rod ‘ AB’ also then it should have
component in ‘x’ direction as ‘V’.
% VP sin / = V $ V
P = V cosec /
here cos/ =R
x =
R
1 .
5
R3 =
5
3
% sin/ =5
4% cosec / =
4
5
% VP = 45 V ...Ans.
(b) ! =R
VP =
R4
V5
ALTERNATIVE SOLUTION :(a) Let ‘P’ have coordinate (x, y)
x = R cos /, y = R sin /.
VX =
dt
dx = – R sin /
dt
d/ = V $
dt
d/ =
/2sinR
V
and VY = R cos /
dtd/ = R cos / ''
( )**
+ , /2 sinR
V = – V cot /
% VP = 2
y2x VV 1 = /1 222 cotVV = V cosec / ...Ans.
(b) ! =R
VP =
R4
V5
6. Taking a small element at DN sin/ = dmg
N =/sin
dmg
2T sin4 –
N cos/ = dm!2
R2T sin4 = dm(!2R + g cot/)
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SOLN_Circular Motion - 5
But 4 is very small, sin4 5 4
2T 4 =R2
md
0$
(!2R + g cot/)
' (
)*+
, R2
dT2 $
=R2
md
0$
(!2R + g cot/)
T = '' (
)
**+
,
/1
!
0 cotg
R
2
mg 2
.
7. The particle velocity has two components.(i) v
0 sin# vertical which move it in vertical direction.
(ii) v0 cos# in horizontal direction and along the cylindrical surface which cause it to move in circle.
So, N =R
)cosv(m 20 #
=R
cosmv 220 #
8. amF!!
- or )aa(mF yx
!!!
1- (# a2 = 0)
x = a sin !t
vx
=dt
dx = aw cos (!t)
ax = 2
2
dt
xd = – a!2 sin(!t)
vy =
dt
dy = – b !sin(!t)
ay = 2
2
dt
yd = – b !2 cos(!t)
So ) j tcosbitsina(mF 22 !!2!!2-!
) jˆ tcosbi
ˆtsina(mF 2 !1!!2-
!
) j yix(mF 2 1!2-!
222 yxm|F| 1!-!
direction tan# =x
y =
a
bcot(!t) (from x-axis)
or )] jyix[( 1 is position vector of the particle in corrdinate system. Because of negative sign force is
opposite to it and always acting towards the orzon.
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SOLN_Circular Motion - 6
9. In x-direction,
ax = g sin# –
g sin# cos6ax = g sin# (1 – cos6)
vx = g sin# 7 62
t
0
dt)cos1( .........(i)
In tangential direction, As a rod AB moves, the point ‘P’ will always lie on the circle.
at = g sin# cos6 – 8g cos#
at = –g sin# (1 – cos6)
v – v0 = – gsin# 7 62
t
0
dt)cos1( .........(ii)
from (i) & (ii)v – v
0 = –v
x = – v cos6
v =)cos1(
v0
61 .
10. Net tangential force acting on the element due to gravity is
d mg sin/ gS
Total external force on chain along the length is
F = 7 /sindmg
F = //7 dRsingm
R /
0
$
$ $ a = 7 //-R /
0
dsin gR
m
F $
$
a = 9 : R / 0cos
gR $
$/2 $ a = ;
<
=>?
@' (
)*+
, 2
Rcos1
gR $
$.
11. !t = a
So, v = at = as2
also, !N =
R
v2
bt4 =R
ta 22
t2 =bR
a2
and bt4 =R
as2
22
bR
a b '
' (
)**+
, =
R
as2
R =bs2
a3
$ ! = 2N
2t !1! = (# !
N=
R
v2
=R
ta 22
= 2
4
bR
a)
! =
2
2
42
bR
a
a '' (
)
**+
,
1 $ ! =
2
3
2
a
bS4
1a ''
(
)
**
+
,
1 .
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SOLN_Circular Motion - 7
12. (a) Parabola y = ax2 is shown. It is clear from diagram that at x = 0 velocity is along x-axis and constant aN
is along y-axis. So,
aN = 2
2
dt
yd
dt
dy = 2a ×
dt
dx = 2aVx
2
2
dt
yd = 2av
dt
dx = 2av2 (# 0
dt
xd2
2
- )
aN = 2av2
R = 2
2
av2
v =
a2
1.
(b) 1b
y
a
x2
2
2
-' (
)*+
, 1'
(
)*+
,
Here again at x = 0 particle is at (0, ± b) moving along positive or negative x-axis hence aN is along y-axis
only.
aN = 2
2
dt
yd
0dt
dy
b
y2
dt
dx
a
x222
-1
0dt
dy
b
y2
a
vx222 -'
(
)*+
, 1
0dt
yd
b
y2
dt
dy
b
2
dt
dx
a
v22
2
2
2
22 -'
' (
)**+
, 1'
(
)*+
, 1 [# v = const. along x-axis only
dt
dy = 0]
'' (
)**+
, 2-
2
2
22
2
dt
yd
b
)b(2
a
v2a
N = 2
2
a
bv2 R =
N
2
a
v2 =
b
a2
13. at =
dt
dv = a = 0.50 m/s2
at = #R
/ =2t
2
1#
$ s = R/ =
2
1 #Rt2 $ s =
2
1 a
tt2
(0.1) 20R =2
1 (0.5)t2 $ t =
5
R40
v = a . t = 0.55
R40
ac =
R
v2
=5
425.0 03 = 0.2 0
a = 2t
2t aa 1 = A B A B2
5
2
21 01 =
2510
41 1 =
2013 m/s2
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SOLN_Circular Motion - 8
14. V = 10 m/s
tan / =Rg
v2
$ / = tan –1 º451010
1010-'
(
)*+
, 33
15. (i) At # = 0º
A
T
mg
anet
= CR
v2
acceleration is vertically up(ii) At # = 90º is at B
v = 0
mg
Acceleration is vertically down.(iii) Horizontally
#||
|
|
atotal
mga = gsint #
a =C
v2
R
u=0
#
R cos #
tan # = #sing
R / v2
$ g sin # . tan # =R
v2
.....(1)
Using energy conservation :
#- cosmgRmv2
1 2....(2)
By (1) & (2)
tan # = 2
1
% cos # = 3
1
% # = cos –1 '' (
)**+
,
3
1
16. (a) N1 =
R
mv2
=5010
1001
33
8N
mg
N1
CN1 = 0.2 N
(b) N =R
mv2
cos / ....(1)
for just slipping
8N =R
mv2
sin / .....(2)
from eqn (1) & (2)
tan / = 8 =3
1=
58.0
1 = 1.724 $ / = 30º Ans.
17. a =12
12
mm
mm
12
!2R =3
R2!
T =12
21
mm
mm2
1 !2R =3
4 m!2R
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SOLN_Circular Motion - 9
18. / =4
0
tan 45º =c
t
a
a$ a
t = a
C
$ g sin # =
t
v2
...(1)
#
$ $ – cos #
3g$
v=?
Using energy conservation
2
1 m 3g$ –
2
1 mv2 = mg $ (1 – cos #)
$ mv2 = 3mg$ – 2mg$ + 2mg$ cos #mv2 = mg$ + 2mg$ cos # .............(2)
By eq. (1) and (2)sin # = 1 + 2 cos # $ # = 90º
19.mgcosN
RmsinN 2
-/!-/
tan / =g
R2!
$hr
R
2 =
g
R2!$ !2 = g/(r – h) ......(1)
(a) h > 0 $ r – 2
g
! > 0
$ ! > r / g =1.0
8.9 = 98 = 27 rad/second Ans.
(b) g = !2 (r – h)
h
h
g
g D2-
D = –
h
10 42
maximum value of h is 0.1 so that Dg = –h
10g 423 = – 9.8 × 10 –3 m/s2 Ans.
20. N – mg sin 30º = m!2R .....(1)
30ºN
!
mg cos30º
8N
mg sin30º
mg cos 30º = 8N .....(2)
! = 20 rad/s
$2
mg3 = 8 ;
<
=>?
@!1 Rm
2
mg 2
$ R)3(2
g 2!-82
R = 22
)3(g
!
82
R = 2)2(2
6.03(8.9
0
2 = 0.24 m Ans.
For minimum angular velocity, normal sould be zero at heighest pointm!2 R = mg
! = R
g
= 24.0
8.9
= 6.4 rad/second
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SOLN_Circular Motion - 10
Also, condition for which block will not slip on cylinder isN – mg cos/ = m!2RN = mg cos/ + m!2Rfr max
= 8N = 8(mg cos/ + m!2R)For the block does not slip over cylinder,
mg sin/ E frmax
mg sin/ E 8 mg cos/ + 8 m!2R
R
cosmg –sing
8
//F!
)cos(sinR
g/82/
8F!
block will not shift anywhere if ! is greater than maximum possible value of RHS which is
2 / 12 )1(R
g81
8F!
! F 8.9 rad/sec.
!min
= 8.9 rad/sec.
21. # = kt
a
ac
/ atat = #r = ktr
# =dt
d! = kt $ 77 -!
! t
00
ktdtd
! =2
kt2
, aC = !2r = r
4
tk 42
tan/ =t
c
a
a =
ktr
4 / rtk 42
=4
kt3
$ t =
3 / 1
k
tan4' (
)*+
, /
22. Centripetal acceleration at A = !2R
acceleration along AB = !
2
R cos /Time taken to reach the point B
L = 0 +2
1 (!2 R cos /)t2 (L << R)
t =/! cosR
L22 Ans.
23. T = kx = 147 (0.1 sec / – 0.1)
T sin / = m !2 r
$ 147(0.1 sec / – 0.1)sin / = 0.3 × (14)2 (0.1 tan /)
1 – cos / = 0.4
$ cos/ =5
3$ / = 53º
T = 147 (0.1 sec 53 – 0.1) = 9.8 N
N = T cos/ – mg = 9.8 ×5
3 – 0.3 × 9.8 = 2.94 N
N = 2.94 N Ans.
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SOLN_Circular Motion - 1
TOPIC : CIRCULAR MOTION
PART - I
1. at = 0 = constant
2. Velocity chnages as its direction changeAcceleration changes as its direction change.
3. |v| = constant
4. T = m!2RT + 60 = m (3!)2R
m!2
R = 8
60
!2R = 2.5 m/s2
9!2R = 22.5 m/s2
5. (C) as # = 0 $ at = 0
% FC = ma
C = m!2R = constant
6. mg = constant vector
7. When critical velocities are taken T = 6mgmax
A mg
TB
BTD
D
T = 0A
C
TA + T
C = 6 mg = constant
TB + T
D = 6 mg = constant
8. Let v be the speed of particle at B, just when it is about to loose contact.From application of Newton's second law to the particle normal to the spherical surface.
r
mv 2
= mg sin & .......... (1)
Applying conservation of energy as the block moves from A to B..
2
1 mv2 = mg (r cos # – r sin &) .......... (2)
Solving 1 and 2 we get3 sin & = 2 cos #
9. % T2 # d3 $
3
13
122
2
1
10
10
T
T
'
'
(
)
*
*
+
, -
'
'
(
)
*
*
+
, $
2
1
T
T =
1010
1
% T1 : T
2 = 1 : 1010
10. Initial extension will be equal to 6 m.
% Initial energy =2
1 (200) (6)2 = 3600 J.
Reaching A :2
1mv2 = 3600 J
$ mv2 = 7200 J
From F.B.D. at A :
N =5
7200
R
mv 2- = 1440 N
PHYSICS SOLUTIONS OF"ADVANCED LEVEL PROBLEMS"
Target : ISEET (IIT-JEE)
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SOLN_Circular Motion - 2
11. Since F!
. V!
, the particle will move along a circle.
% F =R
mv 2
& / =R
S$ / = 2mv
FS
12. x2 = 4ayDifferentiating w.r.t. y, we get
dx
dy
= a2
x
% At (2a, a),dx
dy = 1 $ hence / = 45°
the component of weight along tangential direction is mg sin /.
hence tangential acceleration is g sin / =2
g
13. As tangential acceleration a = dV/dt = !dr/dt
but ! = 20n= 40 rad/s anddt
dr = (2 × 0.75) × 10 –2 m/s = 1.5 × 10 –2 m/s (reel is turned uniformly at the rate
of 2 r.p.s.)
% a = 40 × 1.5 × 10 –2 m/s2 = 60 × 10 –2 m/s2
Now by the F.B.D. of the mass.
w
T
T – W = ag
W% T = W (1 + a/g) put a = 60 × 10 –2 % T = 1.019 W
14. For A :
T – m!2r – ma = 0 .............(i)Seen from object itself
T – 3
mg = ma .............(i)
For B :mg – T = ma .............(ii)(i) – (ii)
2T =3
4mg
T =3
2mg
15. |v| = constant
16. ! = constant
% # = 0 = constant
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SOLN_Circular Motion - 3
17. F = 22t
2 )mg(Ff 11 > mg#
f
F
mg
(f = m r)!2
(asdt
Vd = constant % F
t = constant)
Now when the angular speed of the rod isincreasing at const. rate the resultant force
will be more inclined towards f! .
Hence the angle between F!
and horizontal plane decreases so as with the rod due to increases in
f = m!2r only.
PART - II
1. (a) The system is in equilibrium whenm
1gsin/ = m
2g
or sin/ =1
2
m
m
(b) Let the tangential acceleration of m1 be a.% m
2g – m
1g sin/ = (m
1 + m
2) a
a =9
2540 2 =
9
15m/s2
th e normal acceleration of m1 is zero. # speed of m
1 is zero.
% The magnitude of acceleration of m1 =
9
15m/s2 .
2. (a) Applying conservation of energy between initial and final position isLoss in gravitational P.E. of the bead of mass m = gain in spring P. E.
% mg R =
2
1 K (2R – 2 R)2
or K =)223(R
mg
2(b) At t = 0a
t = g
ac = 0
at lowest pointa
t = 0
ac = 0
The centripetal acceleration of bead at the initial and final position is zero because its speed at both positionis zero.The tangential acceleration of the bead at initial position is g.
The tangential acceleration of the bead at lower most position is zero.
3. Since belt is not slipping, speed at rim of A and B is same
!Ar
A = !
Br
B
!A = 100 ×
10
25 = 250 rpm = 250 ×
60
20 rad/sec.
=3
250 rad/sec.
! = !0 + #t
t = 2 /
03
25
0
20
= 3
50
sec.
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SOLN_Circular Motion - 4
4. !0
2 = 900 (rad/sec)2 $ !0 = 30 rad/sec.
!2 = 1600 (rad/sec)2 $ ! = 40 rad/sec.
/ = t2
0 ' (
)*+
, !1!
t =3040
21002
1033
=7
400sec.
5. (a) As a rod AB moves, the point ‘P’ will always lie on the circle.
% its velocity will be along the circle as shown by ‘VP’ in the figure.
If the point P has to lie on the rod ‘ AB’ also then it should have
component in ‘x’ direction as ‘V’.
% VP sin / = V $ V
P = V cosec /
here cos/ =R
x =
R
1 .
5
R3 =
5
3
% sin/ =5
4% cosec / =
4
5
% VP = 45 V ...Ans.
(b) ! =R
VP =
R4
V5
ALTERNATIVE SOLUTION :(a) Let ‘P’ have coordinate (x, y)
x = R cos /, y = R sin /.
VX =
dt
dx = – R sin /
dt
d/ = V $
dt
d/ =
/2sinR
V
and VY = R cos /
dtd/ = R cos / ''
( )**
+ , /2 sinR
V = – V cot /
% VP = 2
y2x VV 1 = /1 222 cotVV = V cosec / ...Ans.
(b) ! =R
VP =
R4
V5
6. Taking a small element at DN sin/ = dmg
N =/sin
dmg
2T sin4 –
N cos/ = dm!2
R2T sin4 = dm(!2R + g cot/)
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SOLN_Circular Motion - 5
But 4 is very small, sin4 5 4
2T 4 =R2
md
0$
(!2R + g cot/)
' (
)*+
, R2
dT2 $
=R2
md
0$
(!2R + g cot/)
T = '' (
)
**+
,
/1
!
0 cotg
R
2
mg 2
.
7. The particle velocity has two components.(i) v
0 sin# vertical which move it in vertical direction.
(ii) v0 cos# in horizontal direction and along the cylindrical surface which cause it to move in circle.
So, N =R
)cosv(m 20 #
=R
cosmv 220 #
8. amF!!
- or )aa(mF yx
!!!
1- (# a2 = 0)
x = a sin !t
vx
=dt
dx = aw cos (!t)
ax = 2
2
dt
xd = – a!2 sin(!t)
vy =
dt
dy = – b !sin(!t)
ay = 2
2
dt
yd = – b !2 cos(!t)
So ) j tcosbitsina(mF 22 !!2!!2-!
) jˆ tcosbi
ˆtsina(mF 2 !1!!2-
!
) j yix(mF 2 1!2-!
222 yxm|F| 1!-!
direction tan# =x
y =
a
bcot(!t) (from x-axis)
or )] jyix[( 1 is position vector of the particle in corrdinate system. Because of negative sign force is
opposite to it and always acting towards the orzon.
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SOLN_Circular Motion - 6
9. In x-direction,
ax = g sin# –
g sin# cos6ax = g sin# (1 – cos6)
vx = g sin# 7 62
t
0
dt)cos1( .........(i)
In tangential direction, As a rod AB moves, the point ‘P’ will always lie on the circle.
at = g sin# cos6 – 8g cos#
at = –g sin# (1 – cos6)
v – v0 = – gsin# 7 62
t
0
dt)cos1( .........(ii)
from (i) & (ii)v – v
0 = –v
x = – v cos6
v =)cos1(
v0
61 .
10. Net tangential force acting on the element due to gravity is
d mg sin/ gS
Total external force on chain along the length is
F = 7 /sindmg
F = //7 dRsingm
R /
0
$
$ $ a = 7 //-R /
0
dsin gR
m
F $
$
a = 9 : R / 0cos
gR $
$/2 $ a = ;
<
=>?
@' (
)*+
, 2
Rcos1
gR $
$.
11. !t = a
So, v = at = as2
also, !N =
R
v2
bt4 =R
ta 22
t2 =bR
a2
and bt4 =R
as2
22
bR
a b '
' (
)**+
, =
R
as2
R =bs2
a3
$ ! = 2N
2t !1! = (# !
N=
R
v2
=R
ta 22
= 2
4
bR
a)
! =
2
2
42
bR
a
a '' (
)
**+
,
1 $ ! =
2
3
2
a
bS4
1a ''
(
)
**
+
,
1 .
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SOLN_Circular Motion - 7
12. (a) Parabola y = ax2 is shown. It is clear from diagram that at x = 0 velocity is along x-axis and constant aN
is along y-axis. So,
aN = 2
2
dt
yd
dt
dy = 2a ×
dt
dx = 2aVx
2
2
dt
yd = 2av
dt
dx = 2av2 (# 0
dt
xd2
2
- )
aN = 2av2
R = 2
2
av2
v =
a2
1.
(b) 1b
y
a
x2
2
2
-' (
)*+
, 1'
(
)*+
,
Here again at x = 0 particle is at (0, ± b) moving along positive or negative x-axis hence aN is along y-axis
only.
aN = 2
2
dt
yd
0dt
dy
b
y2
dt
dx
a
x222
-1
0dt
dy
b
y2
a
vx222 -'
(
)*+
, 1
0dt
yd
b
y2
dt
dy
b
2
dt
dx
a
v22
2
2
2
22 -'
' (
)**+
, 1'
(
)*+
, 1 [# v = const. along x-axis only
dt
dy = 0]
'' (
)**+
, 2-
2
2
22
2
dt
yd
b
)b(2
a
v2a
N = 2
2
a
bv2 R =
N
2
a
v2 =
b
a2
13. at =
dt
dv = a = 0.50 m/s2
at = #R
/ =2t
2
1#
$ s = R/ =
2
1 #Rt2 $ s =
2
1 a
tt2
(0.1) 20R =2
1 (0.5)t2 $ t =
5
R40
v = a . t = 0.55
R40
ac =
R
v2
=5
425.0 03 = 0.2 0
a = 2t
2t aa 1 = A B A B2
5
2
21 01 =
2510
41 1 =
2013 m/s2
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SOLN_Circular Motion - 8
14. V = 10 m/s
tan / =Rg
v2
$ / = tan –1 º451010
1010-'
(
)*+
, 33
15. (i) At # = 0º
A
T
mg
anet
= CR
v2
acceleration is vertically up(ii) At # = 90º is at B
v = 0
mg
Acceleration is vertically down.(iii) Horizontally
#||
|
|
atotal
mga = gsint #
a =C
v2
R
u=0
#
R cos #
tan # = #sing
R / v2
$ g sin # . tan # =R
v2
.....(1)
Using energy conservation :
#- cosmgRmv2
1 2....(2)
By (1) & (2)
tan # = 2
1
% cos # = 3
1
% # = cos –1 '' (
)**+
,
3
1
16. (a) N1 =
R
mv2
=5010
1001
33
8N
mg
N1
CN1 = 0.2 N
(b) N =R
mv2
cos / ....(1)
for just slipping
8N =R
mv2
sin / .....(2)
from eqn (1) & (2)
tan / = 8 =3
1=
58.0
1 = 1.724 $ / = 30º Ans.
17. a =12
12
mm
mm
12
!2R =3
R2!
T =12
21
mm
mm2
1 !2R =3
4 m!2R
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SOLN_Circular Motion - 9
18. / =4
0
tan 45º =c
t
a
a$ a
t = a
C
$ g sin # =
t
v2
...(1)
#
$ $ – cos #
3g$
v=?
Using energy conservation
2
1 m 3g$ –
2
1 mv2 = mg $ (1 – cos #)
$ mv2 = 3mg$ – 2mg$ + 2mg$ cos #mv2 = mg$ + 2mg$ cos # .............(2)
By eq. (1) and (2)sin # = 1 + 2 cos # $ # = 90º
19.mgcosN
RmsinN 2
-/!-/
tan / =g
R2!
$hr
R
2 =
g
R2!$ !2 = g/(r – h) ......(1)
(a) h > 0 $ r – 2
g
! > 0
$ ! > r / g =1.0
8.9 = 98 = 27 rad/second Ans.
(b) g = !2 (r – h)
h
h
g
g D2-
D = –
h
10 42
maximum value of h is 0.1 so that Dg = –h
10g 423 = – 9.8 × 10 –3 m/s2 Ans.
20. N – mg sin 30º = m!2R .....(1)
30ºN
!
mg cos30º
8N
mg sin30º
mg cos 30º = 8N .....(2)
! = 20 rad/s
$2
mg3 = 8 ;
<
=>?
@!1 Rm
2
mg 2
$ R)3(2
g 2!-82
R = 22
)3(g
!
82
R = 2)2(2
6.03(8.9
0
2 = 0.24 m Ans.
For minimum angular velocity, normal sould be zero at heighest pointm!2 R = mg
! = R
g
= 24.0
8.9
= 6.4 rad/second
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SOLN_Circular Motion - 10
Also, condition for which block will not slip on cylinder isN – mg cos/ = m!2RN = mg cos/ + m!2Rfr max
= 8N = 8(mg cos/ + m!2R)For the block does not slip over cylinder,
mg sin/ E frmax
mg sin/ E 8 mg cos/ + 8 m!2R
R
cosmg –sing
8
//F!
)cos(sinR
g/82/
8F!
block will not shift anywhere if ! is greater than maximum possible value of RHS which is
2 / 12 )1(R
g81
8F!
! F 8.9 rad/sec.
!min
= 8.9 rad/sec.
21. # = kt
a
ac
/ atat = #r = ktr
# =dt
d! = kt $ 77 -!
! t
00
ktdtd
! =2
kt2
, aC = !2r = r
4
tk 42
tan/ =t
c
a
a =
ktr
4 / rtk 42
=4
kt3
$ t =
3 / 1
k
tan4' (
)*+
, /
22. Centripetal acceleration at A = !2R
acceleration along AB = !
2
R cos /Time taken to reach the point B
L = 0 +2
1 (!2 R cos /)t2 (L << R)
t =/! cosR
L22 Ans.
23. T = kx = 147 (0.1 sec / – 0.1)
T sin / = m !2 r
$ 147(0.1 sec / – 0.1)sin / = 0.3 × (14)2 (0.1 tan /)
1 – cos / = 0.4
$ cos/ =5
3$ / = 53º
T = 147 (0.1 sec 53 – 0.1) = 9.8 N
N = T cos/ – mg = 9.8 ×5
3 – 0.3 × 9.8 = 2.94 N
N = 2.94 N Ans.