DC Pandey Solutions: Circular Motion- 1 Notes | Study DC Pandey Solutions for NEET Physics - NEET

NEET: DC Pandey Solutions: Circular Motion- 1 Notes | Study DC Pandey Solutions for NEET Physics - NEET

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 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
must be in rad s
-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
change rad into degree using p rad = ° 180 .
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
must be in rad s
-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
change rad into degree using p rad = ° 180 .
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
must be in rad s
-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
change rad into degree using p rad = ° 180 .
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
must be in rad s
-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
change rad into degree using p rad = ° 180 .
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
must be in rad s
-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
change rad into degree using p rad = ° 180 .
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
          = 4.7 rads
-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
      =
-
8.37 rads
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
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