Allen career institute
PHYSICS
SHEET SOLUTION
CHAPTER - CIRCULAR MOTION
Index
Page no
1. Solution of exercise S1
1 - 17
2. Solution of exercise S2
18 - 22
3. Solution of exercise O1
23 - 53
4. Solution of exercise O2
55 - 71
5. Solution of exercise JM
72 - 74
6. Solution of exercise JA
75 - 82
Exercise S1
@
olution
Seo
Revolution
14
:
,
i.e
Time
.
w
Lince
,
=
is
Stone
centripetal
Acceleration
LIT
=
period
=
=
T
2g 44)
moving
acceleration
of
2% See
will take
=
=
2,5g
stone
=
=
Ac
Sec
22dg Says
a-
with constant speed
Acceleration
Ac
takes 25 See
Revolution
I
1
Therefore
it
only
Centripetal Acceleration
.
=
802
=
0.8
( 2za 2
9 89
.
acE9-9m
Towards Centre
(
Radially
in wards
)
direction
ee
has
2
{oluliogfnikaffngulalvefocity-wo-12radf.ec
Radius
Afters
we
8=
w=4¥RPM
see
w=2g
know
Im
where
n=
No
-
of Revolutions
minute
per
w=2a(4¥)
.
see
-
60
Hec
16rad
w=
Now
,
j
w
Wo -12T
-
16=12+2
dis
(2)
L=2rad/secQ
v
-
-
rw
=r(wotdt)
✓
=
I
(12+27)
v=l2t2t#B
at
t-o.si see
centripetal Acceleration
ac=r(wotdt)
'
=rw
'
Ac
l
=
(
Rt 2 Co
435=169 m/s
=
Acceleration
Tangential
3
.
at
=
2
84
=
Az
Total Acceleration
at
=
at
=
1×2
=
m/s
2
=
M/s
-
2
Nafta
=
Etc
ATI169.am/sQ
Since
After
Angular
its
Angulate velocity
Acceleration
a Sec
we wot
i
w
ooo
Ay
for
first
After
In next
Total
=
at
12+2 (2)
=
now
Acp
2
=
2 Sec
row ?
Seconds
,
( 16)
Sec
=
remains
O
,
=
=
122
16×1
Once
-
m/s
t 2
28rad
constant
wt =
Angulate displacement
256
2
Wo 't -240,
w?
16 ?
w
16radIsec
=
constant
remains
O,
one
and
centripetal Acceleration
only
2 Sec
Hence
a Sec
constant
is
,
it will have
After
only for
is
=
,
=
(2) O ,
.
Therefore
16rad
.
.
O, -14=28+16
Qet=44e€
4
Given
foliation
Now
V
:
-
at
Variable
→
so
.
it
Velocity
Uniform Circular
Motion
ft Gt) 2%2
Non
is
.
Tangential Acceleration ( at)= dat
Centripetal Acceleration
(9)
(b)
Radial
=
Cad
=
Acceleration Cat )
(C) Total Acceleration (at)
=
I It
=
Acceleration C-G)
Tangential
=
=
Cac )
=
=
2 m/s
4
m/s
=
4th
4mF
2
'
4¥42
=
ME
q=fom/s
A sin 370
a÷t÷
→
Rowton
at
:
?
paan:aaiswsat
.
A X
4g
=
90×204
a=
2
Rate
of change of speed
at
=
96in 370
=
=
125 X
ifagnitude of
acceleration
Acceleration
of
Tangent
At=75m
'
}=
5
foliation
:
Given
at
let this instant
Wo
w
=
28rad
w2=
282
=
)
'
12rad
-
o
)Sec
it 220
282-122
=
1602
x
again
t
km/s
122+24 ( 80 )
l God
,
=
is
=
0=80 had
see
Wo
angular velocity
instant
one
4%14=4 racket
-
wot at
wa
t
(28-112) (28-12)
40×16
=
-
28
=
12+4 t
=
-
I
4
AHCMehod-o-fw.LI)t
do
-_@)t
t=
/t=4
6
y
.
a
Angular
displacement
for
OA
A
-
=
i.e
.
out
wat
2W
XIW
completed
A
OB
Wpt
=
ie
Therefore
Relative
JB
→
VBA
=
be
=
VIA
at
Rotation
a
opposite
diagonally
Velocity of
B
w r t
-
-
end
-
'
B
A
I
-
VA
Lewy Lewy
-
=
-
-
VBA
25
one
w
=
B will
.
=
=
-
4h04
VBA
=
4kW
v↳a=24m①
7
Goblin
wn.vefocitypoepcndiadartolineofjoir.in#
Length of
line
of joining
j.yq.ws
'"
✓ Sin 450
w-usioh.us
w
-
YI
DIE
BBBBBBB.gg
B
•
M
8
q
SEM : After
Ye
-
t
I see
Vn=U
Vy Uytayt
Vy gt
'
km/s
"=u
-
-
's
woman
Tangential
E.gr
Acceleration
.
i'
.÷÷÷÷ #
"
'
Centripetal
at
-_
=
Component
off along Velocity
gain
:÷÷÷
Acceleration
-_
Component of velocity
perpendicular
aegcoso
a
-
g #)
s
ac=2Fzm
to V
9
son
'
÷::*: .
at
. .in
are
E-
b
f
-
-
J
Let
ai
=
t
b
(
aint
Angle
J
a
.
arsj
Hw
then
.
velocity
tan O
=
O
Now
,
Tangential
vector and
Bag
=
Acceleration
At
=
Go
=
=
x-axis
Vs
°
any Cos 200
.
b cos 35
At=% Ans
Centripetal Acceleration
=
an a- Sins :
"b① Ans
is
O
10
Solution
a
de
-
-
B
dt
⑧
a
=d£xdIo
x
=
w
off dude
x
de
do
wdw
Ado
=
t
w
MBB
=
Jwdw
Jado
=
O
Wo
!
@
WI wog
-
=
WI OI
-
WZ
=
=
↳
a
Area under
Area
-
O
grafts
of Triangles
in a o
-
noon
graph
18
18
36
w=srad
Arda of
=
-
-
o
gray
tzX9X 4=18
11
obeli on
:
Nw
Nw
Normal
Normal
=
provides
→
reach on
Reaction
from
centripetal Acceleration
fII÷=N
Sfolution Usafe Frg formula of
=
:
✓
safe
BBQ
tf
ence
,
Pina
Cyclist
Wall
=
-
fExio
Vgiven
=
speed on
'
ugbugnrkeed
Fs m/s
> V'
safe
will Not be able to Turn
Without@
upping
12
Sgolutioni
provides
Tension
centripetal
w
-
-
Lan
=
sf
v
Acceleration
2X a-
XIE
Go
w=zsor
1- =
1- =
IOI
mrwz
0.25×1-5
X
T=6
Tma ,
200N
=
Tmax
200
=
=
Wimax
mrwniax
0.25×1.5 X
16000.25×1.5
2002
=
Wmax
y
Wimax
=
-
3
JE
Z
rw
-
=
=
i.
spot
2
y=
1.5×413
=
34.6
yI35m⑦fIo
m/s
13
Sedition for
m
-
T=
Mr W
T
e
for
2M
F-
2mg
f
'
2mg
=
mrw
w
-
ft
w=f⑤ friso
ang
.
14
Sgolution
oA=r
.
Gino
F-
-
I
TT
Lsino
Tsin D=
Tgino
mrw
-
mlsinow
Vertical
Tcoso
-
TEE
Afg
mg
TSIHO
direction
mg
-_
-
-
mg
m/LwYI =Xg
1/2
'
F- MLW
'
FEZ
#O
-
F- MLWZ
In
L
ysino
÷n÷
15
¥7
B.
.
→
cos
of
Centre
300
=
Circle
w
O
rt
ABL
0A
Iz
AB
-
-
tf
B
,
=
0.1
TH
mg
z=r
Now
T
.
In
1×0.1×13×4032
=
=
ION
Vertical
.
direction
NtT2ine=
Nt
10×12
=
A
-
1- Cosa -_mrw2
Tfsz
§%sinx
my
10
µ=5NW Friso
16
Solution ;
y
Km/hr
720
=
M/S
720
y=
-
v
Now
=
tano
200
M/s
=
Rg
tan
150=(2-00)-2
R( lo)
13=200×20/0
101 X
tourist
: :÷÷÷÷÷÷÷D
tank
tank
-
tan
-
.
(45-30)
4-212=2
R
=
.
-
B
40002
B
-
RIl5Ando
17
Gobelin
and
:
In This situation
Centripetal
friction
Acceleration
f- mafia
yrlng
-
Mgp
Craft Cruz 5
?
(O 5×1032=(1×3)
'
4
=
25-9
Wh
w
=
=
w=
2
=
Zt
=
(
tangent al
mfErwT5
ml craters
-
W
Hence
=
=
provides
2)
Z
x w
16
IG
2 had Sec
/
Not at
0 t 3t
t=4 Into
Exercise S2
O
a=2g
18
19
LT
O
20
①
y
21
o
22
:
Exercise O1
23
24
25
26
0
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
EXERCISE O2
o
54
55
LT
O
56
57
0
58
59
•
60
61
62
63
64
65
66
67
68
69
70
71
72
EXERCISE JM
&
I
olution
-
=
centripetal
Vector
form
:
Acceleration
ai
-
-
acaesoti)
Ii
,
"
qq.is
accost
II
i
casino ti )
a,
,
-
Daimon
I
Poletti
s
on :
v
a.
=
=
Ets
dots
=
÷÷÷÷
"
:/ anet-i.EE#z9net--F4Il4m/sT8
3+2
:
¥
Ac
=
3¥
=
72
.
-
73
Gobelin
:
Circle completes
w
;
o
in Same
=
2ft
Angular Velocity of
Ac
=
rw
both
Seo
Time
will be
same
2
E÷=:÷
aa÷.
Lobation
"
:
"
Ouest on
of
wook
Vertical
circular
power Energy
in
Motion
last
-
will be
in
teach
Topic
÷÷÷÷:÷÷÷
::
w
=
26.738 RPM
w=a7RPM(app8
"
74
Lolli
:
2¥
f-
F-
Constantly
=
MRW
Const
?
Rn
E-
@nmst)(
WE
hurt
Gn=k
KRNTI
w=l
Rnttz
2¥ Inez
-
-
TaRn
Let
K' ' EC
75
Rotation
tforizontal
:
↳
Mac
F-
•
w
ya
circle
-
mRw2
Fma
,
w
=
-
FEI
-
.
w=36re
¥
.
76
Rowton
:
when
this
heeled
fo
According
Both
will
was
question
to
asked
answers
two
Our
in
Given
are
question Answer
÷÷÷¥
Horizontal velocity of
fromSince
Centre
he
&
is
Comparison
Horizontal
less So
to
④ milady
A
the
.
P
is
more
point p
Go
only
will
land
in
logic of
9
option
C
"
.
Travel
A
more
is less
move
On shaded
.
is
.
it will
velocity of point
portion
"
:
fo
no
.
is
region
unshaded
land in
JEE
more
distance
fer )
w
in
Region
.
77
Holiday
:
Att
-
-
o
Ip
.
Relative
Relative
Velocity
after Half
-
I
Rwtj )
Velocity
ve
"
=
Ye
Rwti)
Rwtj)
o
-
Rw
Ci)
circle
To Rwlj )
-
Rwy
-
Vq=o
after
-
-
Vr=2Rw
Vpa
3
-
-
circle
After qualm
Ip
-
quarter
Ve
circle
=
up
LRN
-_
Root
te=RwtD
78
is
-
After fuel
And
circle
Beame
Method - 2
i.
At
fort
genoeaf
as
for
f- o
Time t
.
Q
.
Can check ophon
Seo Now
,
.
¥
-
rwcosoj
-
Rwsinot
%
Rw
.
÷..
i→
Hw
.
-
was
rwwsojirwsinoi
an
*
in
.
Ivie -11
grw
Kind
.
79
Rowton
Initially
Go
less
fo
from
Ta
force
to
Speed
Normal
action
bead
Ac increases
due
Circular
Question of Vertical
Wok Power and
:
wire
Meads
on
moves
Normal
from
will be
lesser
bead
on
Energy
by
force
down
will
Ed
radially
will Covered
Motion
.
of
fo requirement
aide
on
will
Wire
how
on
outwards
.
.
.
(9--12)
radially
Newton
of
do
be
outward
inwards
be
will
bead
acuiul
radially
be
a peed increases
act
in
on
inwards
wire
80
Lobito
of Integration12h you
Higher integration
Term
of force along
question
is
St
q
slot
3rd
will
class
in
Luch
from
,
learn
81
.
.
.
F-
2
@
m
=
Mudge
=
Udo
new
=
w
2
Ede
e
v
Judo Wylde
=
RI z
O
CEI
Es
-
se
(E)ne
-
w
-
o
-
u!
u
-
w
(E
-
-
w
Cee
-
-
RI)
wife
1¥ WITH
%g¥%
in
in
-
{§!7 =
Edt
ftp.T.LI#g/--wt
/ e+F
Et
JE
=
=
wt
Rzsewt
)
internet
Fry
Squaring
rzewt
-
we
Both sides
f- Rza
ER
-
82
e
get
.
R÷e2wtteyder ewt
=
ewt
=
Rf (
I
+
ezwt)
e=rq(e-wttew
⑥
from fanfare
from
From Lmfao Heng
two
Reaction is
Ii ,
Trot
=
-
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rw
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Na
ng
=
-
and
-
Stott
.
Na
wtf wya)
amfwda
(eat
e-
mRg
(ewt
e-
-
T
of
mg
HI
-
wall
wtf
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