DC Pandey Solutions: Circular Motion- 1

# DC Pandey Solutions: Circular Motion- 1 | DC Pandey Solutions for NEET Physics PDF Download

``` Page 1

Introductory Exercise 7.1
1. In uniform circular motion the magnitude
of acceleration =
æ
è
ç
ö
ø
÷
v
r
2
does not change
while its direction (being always towards
the centre of the circular path) changes.
2. If w
0
and w are in rad s
-1
the value of a
-2
. But, if w
0
and w are in
degree s
-1
the value of a must also be in
degree s
-2
. Thus, it is not necessary to
express all angles in radian. One way
3. During motion of an object along a curved
path the speed and magnitude of its radial
acceleration may remain constant. Due to
change in direction of motion the velocity
of the object will change even if its speed
is constant. Further, the acceleration will
also change even if the speed is constant.
4. (i) Radial acceleration ( ) a
c
= =
-
v
r
2 1 2
2
1
( ) cms
cm
= 4  cms
-2
(ii) Tangential acceleration (a
T
)
= =
dv
dt
d
dt
t ( ) 2
=2 cms
-2
(iii) Magnitude of net acceleration
= + ( ) ( ) a a
c T
2 2
= +
- -
( ) ( ) 4 2
2 2 2 2
cms cms
=2 5 cms
-2
5. | | | | v v
1 2
® ®
= =v (say)
T
r
v
=
2p
| |
| |
v
PQ ®
=
®
av
PQ
t
= =
r
T
r
r v
2
4
2
2 / / p
=
2 2
p
v
\
| | v
®
=
av
v
2 2
p
6. w
t
t = + 0 4
Centripetal acceleration
= tangential acceleration
r r
t
w a
2
=
Þ w a
t
2
=
( ) 4 4
2
t =
t =
1
2
s
Circular Motion
7
v
1
P
r
r
O
Q
v
2
®
®
Page 2

Introductory Exercise 7.1
1. In uniform circular motion the magnitude
of acceleration =
æ
è
ç
ö
ø
÷
v
r
2
does not change
while its direction (being always towards
the centre of the circular path) changes.
2. If w
0
and w are in rad s
-1
the value of a
-2
. But, if w
0
and w are in
degree s
-1
the value of a must also be in
degree s
-2
. Thus, it is not necessary to
express all angles in radian. One way
3. During motion of an object along a curved
path the speed and magnitude of its radial
acceleration may remain constant. Due to
change in direction of motion the velocity
of the object will change even if its speed
is constant. Further, the acceleration will
also change even if the speed is constant.
4. (i) Radial acceleration ( ) a
c
= =
-
v
r
2 1 2
2
1
( ) cms
cm
= 4  cms
-2
(ii) Tangential acceleration (a
T
)
= =
dv
dt
d
dt
t ( ) 2
=2 cms
-2
(iii) Magnitude of net acceleration
= + ( ) ( ) a a
c T
2 2
= +
- -
( ) ( ) 4 2
2 2 2 2
cms cms
=2 5 cms
-2
5. | | | | v v
1 2
® ®
= =v (say)
T
r
v
=
2p
| |
| |
v
PQ ®
=
®
av
PQ
t
= =
r
T
r
r v
2
4
2
2 / / p
=
2 2
p
v
\
| | v
®
=
av
v
2 2
p
6. w
t
t = + 0 4
Centripetal acceleration
= tangential acceleration
r r
t
w a
2
=
Þ w a
t
2
=
( ) 4 4
2
t =
t =
1
2
s
Circular Motion
7
v
1
P
r
r
O
Q
v
2
®
®
Introductory Exercise 7.2
1. In uniform circular motion of a body the
body is never in equilibrium as only one
force (centripetal) acts on the body which
forces the perform circular motion.
2. v
gra
h
max
( )
= =
´ ´ 9.8 1.5/
1.5
250 2
=35 ms
-1
3. (a) T mg sinq =
T mr cosq w =
2
\    tanq
w
=
g
r
2
Þ w
q q q
= =
g
r
g
l tan ( cos ) tan
or
2p
q T
g
l
=
sin
Þ T
l
g
= 2p
q sin
=
æ
è
ç
ö
ø
÷
2
2
1
2
p
9.8
\     f = =
1
1 2
9.8
p
rve s
-1
= ´
9.8
2
60
p
rev min
-1
=29.9 rev min
-1
(b) T
mg
mg = =
sin q
2
= ´ ´ 2 5 9.8 =69.3N
4.
(a) At rest :
Required CPF = - = w N
w
2
Þ
mv
r
w
2
2
= …(i)
At dip :
Required CPF = ¢ - N w

mv
r
N w
2
= ¢ - …(ii)
Comparing Eqs. (i) and (ii),
N w
w
¢ - =
2
\ N
w
¢ = = ´
3
2
3
2
16 kN
=24 kN
(b) At crest on increasing the speed (v), the
value of N will decrease and for
maximum value of v the of N will be just
zero.
Thus,
mv
r
w
max
2
0 = -
Þ         v
wr
m
max
=
= gr (as w mg = )
= ´ 10 250
=50 ms
-1
(c) At dip :
N w
mv
r
¢ = +
2
= + w mg
= = 2 32 w kN
5. Case I. If the driver turns the vehicle
Circular Motion | 141
w'
N =
w
2
w
w
r
m mg
l
r
q
mg
T sin q T
l
T cos q
Page 3

Introductory Exercise 7.1
1. In uniform circular motion the magnitude
of acceleration =
æ
è
ç
ö
ø
÷
v
r
2
does not change
while its direction (being always towards
the centre of the circular path) changes.
2. If w
0
and w are in rad s
-1
the value of a
-2
. But, if w
0
and w are in
degree s
-1
the value of a must also be in
degree s
-2
. Thus, it is not necessary to
express all angles in radian. One way
3. During motion of an object along a curved
path the speed and magnitude of its radial
acceleration may remain constant. Due to
change in direction of motion the velocity
of the object will change even if its speed
is constant. Further, the acceleration will
also change even if the speed is constant.
4. (i) Radial acceleration ( ) a
c
= =
-
v
r
2 1 2
2
1
( ) cms
cm
= 4  cms
-2
(ii) Tangential acceleration (a
T
)
= =
dv
dt
d
dt
t ( ) 2
=2 cms
-2
(iii) Magnitude of net acceleration
= + ( ) ( ) a a
c T
2 2
= +
- -
( ) ( ) 4 2
2 2 2 2
cms cms
=2 5 cms
-2
5. | | | | v v
1 2
® ®
= =v (say)
T
r
v
=
2p
| |
| |
v
PQ ®
=
®
av
PQ
t
= =
r
T
r
r v
2
4
2
2 / / p
=
2 2
p
v
\
| | v
®
=
av
v
2 2
p
6. w
t
t = + 0 4
Centripetal acceleration
= tangential acceleration
r r
t
w a
2
=
Þ w a
t
2
=
( ) 4 4
2
t =
t =
1
2
s
Circular Motion
7
v
1
P
r
r
O
Q
v
2
®
®
Introductory Exercise 7.2
1. In uniform circular motion of a body the
body is never in equilibrium as only one
force (centripetal) acts on the body which
forces the perform circular motion.
2. v
gra
h
max
( )
= =
´ ´ 9.8 1.5/
1.5
250 2
=35 ms
-1
3. (a) T mg sinq =
T mr cosq w =
2
\    tanq
w
=
g
r
2
Þ w
q q q
= =
g
r
g
l tan ( cos ) tan
or
2p
q T
g
l
=
sin
Þ T
l
g
= 2p
q sin
=
æ
è
ç
ö
ø
÷
2
2
1
2
p
9.8
\     f = =
1
1 2
9.8
p
rve s
-1
= ´
9.8
2
60
p
rev min
-1
=29.9 rev min
-1
(b) T
mg
mg = =
sin q
2
= ´ ´ 2 5 9.8 =69.3N
4.
(a) At rest :
Required CPF = - = w N
w
2
Þ
mv
r
w
2
2
= …(i)
At dip :
Required CPF = ¢ - N w

mv
r
N w
2
= ¢ - …(ii)
Comparing Eqs. (i) and (ii),
N w
w
¢ - =
2
\ N
w
¢ = = ´
3
2
3
2
16 kN
=24 kN
(b) At crest on increasing the speed (v), the
value of N will decrease and for
maximum value of v the of N will be just
zero.
Thus,
mv
r
w
max
2
0 = -
Þ         v
wr
m
max
=
= gr (as w mg = )
= ´ 10 250
=50 ms
-1
(c) At dip :
N w
mv
r
¢ = +
2
= + w mg
= = 2 32 w kN
5. Case I. If the driver turns the vehicle
Circular Motion | 141
w'
N =
w
2
w
w
r
m mg
l
r
q
mg
T sin q T
l
T cos q
mv
r
mg
1
2
£ m
[where v
1
= maximum speed of vehicle]
Þ v gr
1
£ m
Case II. If the driver tries to stop the
vehicle by applying breaks.
Maximum retardation = m g
\ v g r
2
2 2
0 2 = + ( ) m
Þ v gr
2
2 = m
= 2
1
v
As v v
2 1
> , driver should apply breaks to
stop the vehicle rather than taking turn.
6. In the answer to question 3(a) if we
replace q by f
w =
f
g
l sin
If q is the angle made by the string with
the vertical
q + f = ° 90
i.e., f = ° - 90 q
Þ          sin cos f = q
\ w
q
=
g
lcos
Þ          cosq
w
=
g
l
2
Introductory Exercise 7.3
1. | | v
1
®
= u and | | v
2
®
= v (say)
Dv v v
® ® ®
= + -
2 1
( )
= -
® ®
v v
2 1
D D v u u v v u
® ® ® ® ® ®
× = - × - ( ) ( )
2 1 2 1
= × + ×
® ® ® ®
u v u v
2 2 1 1
- -
® ®
2
2 1
v v
| | | | v v
2
2
1
2 2 2
® ®
+ = + v u
= - + u gL u
2 2
2
| | ( ) Dv
®
= -
2 2
2 u gL
| | ( ) Dv
®
= - 2
2
u gL
2. Ball motion from A to B :
0 2 2
2 2
= + - u g R
min
( )( )
Þ       u gR
min
2
4 =
Ball motion from C to A :
v u g h
2 2
2 = + -
min
( )
= - 4 2 gR gh
Þ v g R h = - 2 2 ( )
3. Decrease in KE of bob
= Increase in PE of bob
142 | Mechanics-1
L
v
2
– v
1
Di
g
v
1
®
®
®
®
h
v
C
A
2R
B
PE of bob
= K (say)
KE of bob
2
= 1/2 mv
0
v
0
Page 4

Introductory Exercise 7.1
1. In uniform circular motion the magnitude
of acceleration =
æ
è
ç
ö
ø
÷
v
r
2
does not change
while its direction (being always towards
the centre of the circular path) changes.
2. If w
0
and w are in rad s
-1
the value of a
-2
. But, if w
0
and w are in
degree s
-1
the value of a must also be in
degree s
-2
. Thus, it is not necessary to
express all angles in radian. One way
3. During motion of an object along a curved
path the speed and magnitude of its radial
acceleration may remain constant. Due to
change in direction of motion the velocity
of the object will change even if its speed
is constant. Further, the acceleration will
also change even if the speed is constant.
4. (i) Radial acceleration ( ) a
c
= =
-
v
r
2 1 2
2
1
( ) cms
cm
= 4  cms
-2
(ii) Tangential acceleration (a
T
)
= =
dv
dt
d
dt
t ( ) 2
=2 cms
-2
(iii) Magnitude of net acceleration
= + ( ) ( ) a a
c T
2 2
= +
- -
( ) ( ) 4 2
2 2 2 2
cms cms
=2 5 cms
-2
5. | | | | v v
1 2
® ®
= =v (say)
T
r
v
=
2p
| |
| |
v
PQ ®
=
®
av
PQ
t
= =
r
T
r
r v
2
4
2
2 / / p
=
2 2
p
v
\
| | v
®
=
av
v
2 2
p
6. w
t
t = + 0 4
Centripetal acceleration
= tangential acceleration
r r
t
w a
2
=
Þ w a
t
2
=
( ) 4 4
2
t =
t =
1
2
s
Circular Motion
7
v
1
P
r
r
O
Q
v
2
®
®
Introductory Exercise 7.2
1. In uniform circular motion of a body the
body is never in equilibrium as only one
force (centripetal) acts on the body which
forces the perform circular motion.
2. v
gra
h
max
( )
= =
´ ´ 9.8 1.5/
1.5
250 2
=35 ms
-1
3. (a) T mg sinq =
T mr cosq w =
2
\    tanq
w
=
g
r
2
Þ w
q q q
= =
g
r
g
l tan ( cos ) tan
or
2p
q T
g
l
=
sin
Þ T
l
g
= 2p
q sin
=
æ
è
ç
ö
ø
÷
2
2
1
2
p
9.8
\     f = =
1
1 2
9.8
p
rve s
-1
= ´
9.8
2
60
p
rev min
-1
=29.9 rev min
-1
(b) T
mg
mg = =
sin q
2
= ´ ´ 2 5 9.8 =69.3N
4.
(a) At rest :
Required CPF = - = w N
w
2
Þ
mv
r
w
2
2
= …(i)
At dip :
Required CPF = ¢ - N w

mv
r
N w
2
= ¢ - …(ii)
Comparing Eqs. (i) and (ii),
N w
w
¢ - =
2
\ N
w
¢ = = ´
3
2
3
2
16 kN
=24 kN
(b) At crest on increasing the speed (v), the
value of N will decrease and for
maximum value of v the of N will be just
zero.
Thus,
mv
r
w
max
2
0 = -
Þ         v
wr
m
max
=
= gr (as w mg = )
= ´ 10 250
=50 ms
-1
(c) At dip :
N w
mv
r
¢ = +
2
= + w mg
= = 2 32 w kN
5. Case I. If the driver turns the vehicle
Circular Motion | 141
w'
N =
w
2
w
w
r
m mg
l
r
q
mg
T sin q T
l
T cos q
mv
r
mg
1
2
£ m
[where v
1
= maximum speed of vehicle]
Þ v gr
1
£ m
Case II. If the driver tries to stop the
vehicle by applying breaks.
Maximum retardation = m g
\ v g r
2
2 2
0 2 = + ( ) m
Þ v gr
2
2 = m
= 2
1
v
As v v
2 1
> , driver should apply breaks to
stop the vehicle rather than taking turn.
6. In the answer to question 3(a) if we
replace q by f
w =
f
g
l sin
If q is the angle made by the string with
the vertical
q + f = ° 90
i.e., f = ° - 90 q
Þ          sin cos f = q
\ w
q
=
g
lcos
Þ          cosq
w
=
g
l
2
Introductory Exercise 7.3
1. | | v
1
®
= u and | | v
2
®
= v (say)
Dv v v
® ® ®
= + -
2 1
( )
= -
® ®
v v
2 1
D D v u u v v u
® ® ® ® ® ®
× = - × - ( ) ( )
2 1 2 1
= × + ×
® ® ® ®
u v u v
2 2 1 1
- -
® ®
2
2 1
v v
| | | | v v
2
2
1
2 2 2
® ®
+ = + v u
= - + u gL u
2 2
2
| | ( ) Dv
®
= -
2 2
2 u gL
| | ( ) Dv
®
= - 2
2
u gL
2. Ball motion from A to B :
0 2 2
2 2
= + - u g R
min
( )( )
Þ       u gR
min
2
4 =
Ball motion from C to A :
v u g h
2 2
2 = + -
min
( )
= - 4 2 gR gh
Þ v g R h = - 2 2 ( )
3. Decrease in KE of bob
= Increase in PE of bob
142 | Mechanics-1
L
v
2
– v
1
Di
g
v
1
®
®
®
®
h
v
C
A
2R
B
PE of bob
= K (say)
KE of bob
2
= 1/2 mv
0
v
0

1
2
0
2
mv mgh =
v gh
0
2
2 =
v gl
0
2 1 = - ( cos ) q
= ´ ´ ´ - ° 2 5 1 60 9.8 ( cos )
=7 ms
-1
AIEEE Corner
Subjective Questions (Level 1)
1. v t = 4
2
\
dv
dt
t =8
i.e.,      a
T
= ´ = 8 3 24 ms
-2
v = ´ = 4 3
2
36 ms
-1
(at t = 3 s)
\     a
v
c
= =
2 2
4
36
54
( )
= 24 ms
-2
Angle between a
®
net
and a
t
®
q = = = °
-
®
®
-
tan
| |
| |
tan
1 1
1 45
a
a
c
T
2. v = 16 ms
-1
and r = 50 m
\ a
v
r
c
= = =
2 2
16
50
( )
5.12 ms
-2
a a a
c T net
= +
2 2
= + ( ) 5.12
2 2
8
(given a
T
=
-
8
2
ms )
=9.5 ms
-2
3. Speed ( ) v at the highest point ( ) P
v u = cosq
Now,      a g
c
=
\
v
R
g
2
=
i.e.,        R
v
g
u
g
= =
2 2 2
cos q
4. (a) a a
c
= ° = ´ cos30 25
3
2
= 21.65 ms
-2
\
v
R
2
=21.65
i.e., v = ´ 21.65 2.5 (Q R =2.5 m)
= 7.36 ms
-1
(b) a a
T
= ° = ´ sin 30 25
1
2
= 12.5 ms
-2
5. (a) v t = = ´ 2.0 2.0 1
=2.0 cms
-1
\   a
v
R
c
= = =
2 2
1
4
( ) 2.0
cms
-2

(b) v t =2.0
\
dv
dt
=2.0
i.e., a
T
=
-
2.0cms
2
(c) a a a
C T net
= +
2 2
= 4.47 cms
-2
6. R uT u
h
g
= =
2
\ u R
g
h
=
2
i.e., u
R g
h
2
2
2
=
Circular Motion | 143
q
u cos q
u
R
a = g
c
v = u cos q
Centre of curvature
r
P
Range (R) = 10 m
h
h = 2.9 m
u
h
PE of bob = K + mgh
Page 5

Introductory Exercise 7.1
1. In uniform circular motion the magnitude
of acceleration =
æ
è
ç
ö
ø
÷
v
r
2
does not change
while its direction (being always towards
the centre of the circular path) changes.
2. If w
0
and w are in rad s
-1
the value of a
-2
. But, if w
0
and w are in
degree s
-1
the value of a must also be in
degree s
-2
. Thus, it is not necessary to
express all angles in radian. One way
3. During motion of an object along a curved
path the speed and magnitude of its radial
acceleration may remain constant. Due to
change in direction of motion the velocity
of the object will change even if its speed
is constant. Further, the acceleration will
also change even if the speed is constant.
4. (i) Radial acceleration ( ) a
c
= =
-
v
r
2 1 2
2
1
( ) cms
cm
= 4  cms
-2
(ii) Tangential acceleration (a
T
)
= =
dv
dt
d
dt
t ( ) 2
=2 cms
-2
(iii) Magnitude of net acceleration
= + ( ) ( ) a a
c T
2 2
= +
- -
( ) ( ) 4 2
2 2 2 2
cms cms
=2 5 cms
-2
5. | | | | v v
1 2
® ®
= =v (say)
T
r
v
=
2p
| |
| |
v
PQ ®
=
®
av
PQ
t
= =
r
T
r
r v
2
4
2
2 / / p
=
2 2
p
v
\
| | v
®
=
av
v
2 2
p
6. w
t
t = + 0 4
Centripetal acceleration
= tangential acceleration
r r
t
w a
2
=
Þ w a
t
2
=
( ) 4 4
2
t =
t =
1
2
s
Circular Motion
7
v
1
P
r
r
O
Q
v
2
®
®
Introductory Exercise 7.2
1. In uniform circular motion of a body the
body is never in equilibrium as only one
force (centripetal) acts on the body which
forces the perform circular motion.
2. v
gra
h
max
( )
= =
´ ´ 9.8 1.5/
1.5
250 2
=35 ms
-1
3. (a) T mg sinq =
T mr cosq w =
2
\    tanq
w
=
g
r
2
Þ w
q q q
= =
g
r
g
l tan ( cos ) tan
or
2p
q T
g
l
=
sin
Þ T
l
g
= 2p
q sin
=
æ
è
ç
ö
ø
÷
2
2
1
2
p
9.8
\     f = =
1
1 2
9.8
p
rve s
-1
= ´
9.8
2
60
p
rev min
-1
=29.9 rev min
-1
(b) T
mg
mg = =
sin q
2
= ´ ´ 2 5 9.8 =69.3N
4.
(a) At rest :
Required CPF = - = w N
w
2
Þ
mv
r
w
2
2
= …(i)
At dip :
Required CPF = ¢ - N w

mv
r
N w
2
= ¢ - …(ii)
Comparing Eqs. (i) and (ii),
N w
w
¢ - =
2
\ N
w
¢ = = ´
3
2
3
2
16 kN
=24 kN
(b) At crest on increasing the speed (v), the
value of N will decrease and for
maximum value of v the of N will be just
zero.
Thus,
mv
r
w
max
2
0 = -
Þ         v
wr
m
max
=
= gr (as w mg = )
= ´ 10 250
=50 ms
-1
(c) At dip :
N w
mv
r
¢ = +
2
= + w mg
= = 2 32 w kN
5. Case I. If the driver turns the vehicle
Circular Motion | 141
w'
N =
w
2
w
w
r
m mg
l
r
q
mg
T sin q T
l
T cos q
mv
r
mg
1
2
£ m
[where v
1
= maximum speed of vehicle]
Þ v gr
1
£ m
Case II. If the driver tries to stop the
vehicle by applying breaks.
Maximum retardation = m g
\ v g r
2
2 2
0 2 = + ( ) m
Þ v gr
2
2 = m
= 2
1
v
As v v
2 1
> , driver should apply breaks to
stop the vehicle rather than taking turn.
6. In the answer to question 3(a) if we
replace q by f
w =
f
g
l sin
If q is the angle made by the string with
the vertical
q + f = ° 90
i.e., f = ° - 90 q
Þ          sin cos f = q
\ w
q
=
g
lcos
Þ          cosq
w
=
g
l
2
Introductory Exercise 7.3
1. | | v
1
®
= u and | | v
2
®
= v (say)
Dv v v
® ® ®
= + -
2 1
( )
= -
® ®
v v
2 1
D D v u u v v u
® ® ® ® ® ®
× = - × - ( ) ( )
2 1 2 1
= × + ×
® ® ® ®
u v u v
2 2 1 1
- -
® ®
2
2 1
v v
| | | | v v
2
2
1
2 2 2
® ®
+ = + v u
= - + u gL u
2 2
2
| | ( ) Dv
®
= -
2 2
2 u gL
| | ( ) Dv
®
= - 2
2
u gL
2. Ball motion from A to B :
0 2 2
2 2
= + - u g R
min
( )( )
Þ       u gR
min
2
4 =
Ball motion from C to A :
v u g h
2 2
2 = + -
min
( )
= - 4 2 gR gh
Þ v g R h = - 2 2 ( )
3. Decrease in KE of bob
= Increase in PE of bob
142 | Mechanics-1
L
v
2
– v
1
Di
g
v
1
®
®
®
®
h
v
C
A
2R
B
PE of bob
= K (say)
KE of bob
2
= 1/2 mv
0
v
0

1
2
0
2
mv mgh =
v gh
0
2
2 =
v gl
0
2 1 = - ( cos ) q
= ´ ´ ´ - ° 2 5 1 60 9.8 ( cos )
=7 ms
-1
AIEEE Corner
Subjective Questions (Level 1)
1. v t = 4
2
\
dv
dt
t =8
i.e.,      a
T
= ´ = 8 3 24 ms
-2
v = ´ = 4 3
2
36 ms
-1
(at t = 3 s)
\     a
v
c
= =
2 2
4
36
54
( )
= 24 ms
-2
Angle between a
®
net
and a
t
®
q = = = °
-
®
®
-
tan
| |
| |
tan
1 1
1 45
a
a
c
T
2. v = 16 ms
-1
and r = 50 m
\ a
v
r
c
= = =
2 2
16
50
( )
5.12 ms
-2
a a a
c T net
= +
2 2
= + ( ) 5.12
2 2
8
(given a
T
=
-
8
2
ms )
=9.5 ms
-2
3. Speed ( ) v at the highest point ( ) P
v u = cosq
Now,      a g
c
=
\
v
R
g
2
=
i.e.,        R
v
g
u
g
= =
2 2 2
cos q
4. (a) a a
c
= ° = ´ cos30 25
3
2
= 21.65 ms
-2
\
v
R
2
=21.65
i.e., v = ´ 21.65 2.5 (Q R =2.5 m)
= 7.36 ms
-1
(b) a a
T
= ° = ´ sin 30 25
1
2
= 12.5 ms
-2
5. (a) v t = = ´ 2.0 2.0 1
=2.0 cms
-1
\   a
v
R
c
= = =
2 2
1
4
( ) 2.0
cms
-2

(b) v t =2.0
\
dv
dt
=2.0
i.e., a
T
=
-
2.0cms
2
(c) a a a
C T net
= +
2 2
= 4.47 cms
-2
6. R uT u
h
g
= =
2
\ u R
g
h
=
2
i.e., u
R g
h
2
2
2
=
Circular Motion | 143
q
u cos q
u
R
a = g
c
v = u cos q
Centre of curvature
r
P
Range (R) = 10 m
h
h = 2.9 m
u
h
PE of bob = K + mgh
Thus, centripetal acceleration of the stone
(at point P) while in circular motion
= = ×
u
r
R g
h r
2 2
2
1
=
´
´ ´
10
2
2
9.8
2.9 1.5
=112.6 ms
-2
7. v =18 km/h
18.5
ms =
-
18
1
=
-
5
1
ms
Angle of banking ( ) q =
-
tan
1
2
v
rg
=
´
-
tan
( )
1
2
5
10 10
=
-
tan
1
1
40

8.
mv
r
mg
2
= m
i.e., m = =
´
= =
v
gr
2 2
5
10 10
1
4
( )
0.25
9.     v rg = tanq
= ´ ´ ° 50 10 30 tan
=
-
17
1
ms
10.      mr N w
min
2
=  …(i)
and  mg N = m        …(ii)
Solving Eqs. (i) and (ii),
w
m
min
=
g
r
=
´
10
3 0.15
-1
11. (a) ( ) cos T T mr
1 2
2
+ = q w           …(i)
and ( )sin T T mg
1 2
- = q …(ii)
i.e., T mr T
2
2
1
cos cos q w q = -
and T T mg
2 1
sin sin q q = -
Þ   cos
cos
sin
q
w q
q
=
-
-
mr T
T mg
2
1
1
or T mg mr T
1
2
1
cos cos cos q q w q - = -
or w
q q
=
- 2
1
T mg
mr
cos cos
=
´ ´
æ
è
ç
ö
ø
÷ - ´ ´
æ
è
ç
ö
ø
÷
´
2 200
3
5
4 10
3
4
4 3
=
-
1
=
´ 8.37 rev
min
1
2
1
60
p
=39.94 rev min
-1
.
(b) From Eq. (ii),
T T
mg
2 1
= -
sin q
= -
´
æ
è
ç
ö
ø
÷
200
4 100
4
5
=150 N
12. (a) For the block not to slip
mL N mg w m m
max
2
= =
\ w
m
max
=
g
L

(b)If the angular speed is gradually
increased, the block will also have
translational acceleration ( ) =aL
besides centripetal acceleration
(=Lw
2
).
144 | Mechanics-1
mN
mg
N
T sin q
1
T sin q
2
mg
4m
4m
T cos q
1
T cos q
2
T
2
T
1
aL
At the time ( t) of
just slipping
At t = 0
2
Lw
a
net
N
mg
L
mN
```

122 docs

## FAQs on DC Pandey Solutions: Circular Motion- 1 - DC Pandey Solutions for NEET Physics

 1. What is circular motion?
Ans. Circular motion refers to the movement of an object along a circular path. In this type of motion, the object continuously changes its direction but maintains a constant distance from a fixed point called the center of the circle.
 2. What are the key characteristics of circular motion?
Ans. The key characteristics of circular motion include the presence of a centripetal force that acts towards the center of the circle, the constant speed of the object, and the continuous change in the direction of motion.
 3. How is centripetal force related to circular motion?
Ans. Centripetal force is the force that acts towards the center of the circle and keeps an object moving in circular motion. It is responsible for continuously changing the direction of motion and preventing the object from moving in a straight line.
 4. Can an object in circular motion have a constant speed but changing velocity?
Ans. Yes, an object in circular motion can have a constant speed but changing velocity. Velocity is a vector quantity that includes both speed and direction. In circular motion, even if the speed remains constant, the direction of the object's velocity continuously changes, resulting in a changing velocity.
 5. How can we calculate the acceleration of an object in circular motion?
Ans. The acceleration of an object in circular motion can be calculated using the formula a = v^2/r, where a is the acceleration, v is the speed of the object, and r is the radius of the circular path. This formula shows that the acceleration is directly proportional to the square of the speed and inversely proportional to the radius of the circle.

## DC Pandey Solutions for NEET Physics

122 docs

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