Page 1
7.1
SOLUTIONS TO CONCEPTS circular motion;;
CHAPTER 7
1. Distance between Earth & Moon
r = 3.85 × 10
5
km = 3.85 × 10
8
m
T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10
6
sec
v =
T
r 2 ?
=
6
8
10 36 . 2
10 85 . 3 14 . 3 2
?
? ? ?
= 1025.42m/sec
a =
r
v
2
=
8
2
10 85 . 3
) 42 . 1025 (
?
= 0.00273m/sec
2
= 2.73 × 10
–3
m/sec
2
2. Diameter of earth = 12800km
Radius R = 6400km = 64 × 10
5
m
V =
T
R 2 ?
=
3600 24
10 64 14 . 3 2
5
?
? ? ?
m/sec = 465.185
a =
R
V
2
=
5
2
10 64
) 5185 . 46 (
?
= 0.0338m/sec
2
3. V = 2t, r = 1cm
a) Radial acceleration at t = 1 sec.
a =
r
v
2
=
1
2
2
= 4cm/sec
2
b) Tangential acceleration at t = 1sec.
a =
dt
dv
= ) t 2 (
dt
d
= 2cm/sec
2
c) Magnitude of acceleration at t = 1sec
a =
2 2
2 4 ? = 20 cm/sec
2
4. Given that m = 150kg,
v= 36km/hr = 10m/sec, r = 30m
Horizontal force needed is
r
mv
2
=
30
) 10 ( 150
2
?
=
30
100 150 ?
= 500N
5. in the diagram
R cos ? = mg ..(i)
R sin ? =
r
mv
2
..(ii)
Dividing equation (i) with equation (ii)
Tan ? =
rmg
mv
2
=
rg
v
2
v = 36km/hr = 10m/sec, r = 30m
Tan ? =
rg
v
2
=
10 30
100
?
= (1/3)
? ? = tan
–1
(1/3) ?
6. Radius of Park = r = 10m
speed of vehicle = 18km/hr = 5 m/sec
Angle of banking tan ? =
rg
v
2
? ? = tan
–1
rg
v
2
= tan
–1
100
25
= tan
–1
(1/4)
R
mv
2
/R
mg
Page 2
7.1
SOLUTIONS TO CONCEPTS circular motion;;
CHAPTER 7
1. Distance between Earth & Moon
r = 3.85 × 10
5
km = 3.85 × 10
8
m
T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10
6
sec
v =
T
r 2 ?
=
6
8
10 36 . 2
10 85 . 3 14 . 3 2
?
? ? ?
= 1025.42m/sec
a =
r
v
2
=
8
2
10 85 . 3
) 42 . 1025 (
?
= 0.00273m/sec
2
= 2.73 × 10
–3
m/sec
2
2. Diameter of earth = 12800km
Radius R = 6400km = 64 × 10
5
m
V =
T
R 2 ?
=
3600 24
10 64 14 . 3 2
5
?
? ? ?
m/sec = 465.185
a =
R
V
2
=
5
2
10 64
) 5185 . 46 (
?
= 0.0338m/sec
2
3. V = 2t, r = 1cm
a) Radial acceleration at t = 1 sec.
a =
r
v
2
=
1
2
2
= 4cm/sec
2
b) Tangential acceleration at t = 1sec.
a =
dt
dv
= ) t 2 (
dt
d
= 2cm/sec
2
c) Magnitude of acceleration at t = 1sec
a =
2 2
2 4 ? = 20 cm/sec
2
4. Given that m = 150kg,
v= 36km/hr = 10m/sec, r = 30m
Horizontal force needed is
r
mv
2
=
30
) 10 ( 150
2
?
=
30
100 150 ?
= 500N
5. in the diagram
R cos ? = mg ..(i)
R sin ? =
r
mv
2
..(ii)
Dividing equation (i) with equation (ii)
Tan ? =
rmg
mv
2
=
rg
v
2
v = 36km/hr = 10m/sec, r = 30m
Tan ? =
rg
v
2
=
10 30
100
?
= (1/3)
? ? = tan
–1
(1/3) ?
6. Radius of Park = r = 10m
speed of vehicle = 18km/hr = 5 m/sec
Angle of banking tan ? =
rg
v
2
? ? = tan
–1
rg
v
2
= tan
–1
100
25
= tan
–1
(1/4)
R
mv
2
/R
mg
Chapter 7
7.2
7. The road is horizontal (no banking)
R
mv
2
= ?N
and N = mg
So
R
mv
2
= ? mg v = 5m/sec, R = 10m
?
10
25
= ?g ? ? =
100
25
= 0.25 ?
8. Angle of banking = ? = 30°
Radius = r = 50m
tan ? =
rg
v
2
? tan 30° =
rg
v
2
?
3
1
=
rg
v
2
? v
2
=
3
rg
=
3
10 50 ?
? v =
3
500
= 17m/sec.
9. Electron revolves around the proton in a circle having proton at the centre.
Centripetal force is provided by coulomb attraction.
r = 5.3 ?t 10
–11
m m = mass of electron = 9.1 × 10
–3
kg.
charge of electron = 1.6 × 10
–19
c.
r
mv
2
=
2
2
r
q
k ? v
2
=
rm
kq
2
=
31 11
38 9
10 1 . 9 10 3 . 5
10 6 . 1 6 . 1 10 9
? ?
?
? ? ?
? ? ? ?
=
13
10
23 . 48
04 . 23
?
? v
2
= 0.477 × 10
13
= 4.7 × 10
12
? v =
12
10 7 . 4 ? = 2.2 × 10
6
m/sec
10. At the highest point of a vertical circle
R
mv
2
= mg
? v
2
= Rg ? v = Rg
11. A celling fan has a diameter = 120cm.
?Radius = r = 60cm = 0/6m
Mass of particle on the outer end of a blade is 1g.
n = 1500 rev/min = 25 rev/sec
? = 2 ??n = 2 ? ×25 = 157.14
Force of the particle on the blade = Mr ?
2
= (0.001) × 0.6 × (157.14) = 14.8N
The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle
also exerts a force of 14.8N on the blade along its surface.
12. A mosquito is sitting on an L.P. record disc & rotating on a turn table at
3
1
33 rpm.
n =
3
1
33 rpm =
60 3
100
?
rps
?? = 2 ? n = 2 ? ×
180
100
=
9
10 ?
rad/sec
r = 10cm =0.1m, g = 10m/sec
2
?mg ? mr ?
2
? ? =
g
r
2
?
?
10
9
10
1 . 0
2
?
?
?
?
?
? ?
?
? ? ?
81
2
?
?
R
mg
??g mv
2
/R
Page 3
7.1
SOLUTIONS TO CONCEPTS circular motion;;
CHAPTER 7
1. Distance between Earth & Moon
r = 3.85 × 10
5
km = 3.85 × 10
8
m
T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10
6
sec
v =
T
r 2 ?
=
6
8
10 36 . 2
10 85 . 3 14 . 3 2
?
? ? ?
= 1025.42m/sec
a =
r
v
2
=
8
2
10 85 . 3
) 42 . 1025 (
?
= 0.00273m/sec
2
= 2.73 × 10
–3
m/sec
2
2. Diameter of earth = 12800km
Radius R = 6400km = 64 × 10
5
m
V =
T
R 2 ?
=
3600 24
10 64 14 . 3 2
5
?
? ? ?
m/sec = 465.185
a =
R
V
2
=
5
2
10 64
) 5185 . 46 (
?
= 0.0338m/sec
2
3. V = 2t, r = 1cm
a) Radial acceleration at t = 1 sec.
a =
r
v
2
=
1
2
2
= 4cm/sec
2
b) Tangential acceleration at t = 1sec.
a =
dt
dv
= ) t 2 (
dt
d
= 2cm/sec
2
c) Magnitude of acceleration at t = 1sec
a =
2 2
2 4 ? = 20 cm/sec
2
4. Given that m = 150kg,
v= 36km/hr = 10m/sec, r = 30m
Horizontal force needed is
r
mv
2
=
30
) 10 ( 150
2
?
=
30
100 150 ?
= 500N
5. in the diagram
R cos ? = mg ..(i)
R sin ? =
r
mv
2
..(ii)
Dividing equation (i) with equation (ii)
Tan ? =
rmg
mv
2
=
rg
v
2
v = 36km/hr = 10m/sec, r = 30m
Tan ? =
rg
v
2
=
10 30
100
?
= (1/3)
? ? = tan
–1
(1/3) ?
6. Radius of Park = r = 10m
speed of vehicle = 18km/hr = 5 m/sec
Angle of banking tan ? =
rg
v
2
? ? = tan
–1
rg
v
2
= tan
–1
100
25
= tan
–1
(1/4)
R
mv
2
/R
mg
Chapter 7
7.2
7. The road is horizontal (no banking)
R
mv
2
= ?N
and N = mg
So
R
mv
2
= ? mg v = 5m/sec, R = 10m
?
10
25
= ?g ? ? =
100
25
= 0.25 ?
8. Angle of banking = ? = 30°
Radius = r = 50m
tan ? =
rg
v
2
? tan 30° =
rg
v
2
?
3
1
=
rg
v
2
? v
2
=
3
rg
=
3
10 50 ?
? v =
3
500
= 17m/sec.
9. Electron revolves around the proton in a circle having proton at the centre.
Centripetal force is provided by coulomb attraction.
r = 5.3 ?t 10
–11
m m = mass of electron = 9.1 × 10
–3
kg.
charge of electron = 1.6 × 10
–19
c.
r
mv
2
=
2
2
r
q
k ? v
2
=
rm
kq
2
=
31 11
38 9
10 1 . 9 10 3 . 5
10 6 . 1 6 . 1 10 9
? ?
?
? ? ?
? ? ? ?
=
13
10
23 . 48
04 . 23
?
? v
2
= 0.477 × 10
13
= 4.7 × 10
12
? v =
12
10 7 . 4 ? = 2.2 × 10
6
m/sec
10. At the highest point of a vertical circle
R
mv
2
= mg
? v
2
= Rg ? v = Rg
11. A celling fan has a diameter = 120cm.
?Radius = r = 60cm = 0/6m
Mass of particle on the outer end of a blade is 1g.
n = 1500 rev/min = 25 rev/sec
? = 2 ??n = 2 ? ×25 = 157.14
Force of the particle on the blade = Mr ?
2
= (0.001) × 0.6 × (157.14) = 14.8N
The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle
also exerts a force of 14.8N on the blade along its surface.
12. A mosquito is sitting on an L.P. record disc & rotating on a turn table at
3
1
33 rpm.
n =
3
1
33 rpm =
60 3
100
?
rps
?? = 2 ? n = 2 ? ×
180
100
=
9
10 ?
rad/sec
r = 10cm =0.1m, g = 10m/sec
2
?mg ? mr ?
2
? ? =
g
r
2
?
?
10
9
10
1 . 0
2
?
?
?
?
?
? ?
?
? ? ?
81
2
?
?
R
mg
??g mv
2
/R
Chapter 7
7.3
13. A pendulum is suspended from the ceiling of a car taking a turn
r = 10m, v = 36km/hr = 10 m/sec, g = 10m/sec
2
From the figure T sin ? =
r
mv
2
..(i)
T cos ? = mg ..(ii)
?
?
?
cos
sin
=
rmg
mv
2
? tan ? =
rg
v
2
? ? = tan
–1
?
?
?
?
?
?
?
?
rg
v
2
= tan
–1
10 10
100
?
= tan
–1
(1) ? ? = 45° ?
14. At the lowest pt.
T = mg +
r
mv
2
Here m = 100g = 1/10 kg, r = 1m, v = 1.4 m/sec
T = mg +
r
mv
2
=
10
) 4 . 1 (
8 . 9
10
1
2
? ? = 0.98 + 0.196 = 1.176 = 1.2 N
15. Bob has a velocity 1.4m/sec, when the string makes an angle of 0.2 radian.
m = 100g = 0.1kg, r = 1m, v = 1.4m/sec.
From the diagram,
T – mg cos ? =
R
mv
2
? T =
R
mv
2
+ mg cos ?
? T =
?
?
?
?
?
?
?
?
?
? ? ? ?
?
2
1 8 . 9 ) 1 . 0 (
1
) 4 . 1 ( 1 . 0
2 2
? T = 0.196 + 9.8 ×
?
?
?
?
?
?
?
?
?
2
) 2 (.
1
2
( ? cos ? =
2
1
2
?
? for small ?)
? T = 0.196 + (0.98) × (0.98) = 0.196 + 0.964 = 1.156N ? 1.16 N ?
16. At the extreme position, velocity of the pendulum is zero.
So there is no centrifugal force.
So T = mg cos ?
o
?
17. a) Net force on the spring balance.
R = mg – m ?
2
r
So, fraction less than the true weight (3mg) is
=
mg
) r m mg ( mg
2
? ? ?
=
g
2
?
=
10
10 6400
3600 24
2
3
2
?
? ?
?
?
?
?
?
?
?
= 3.5 × 10
–3
b) When the balance reading is half the true weight, ?
mg
) r m mg ( mg
2
? ? ?
= 1/2
?
2
r = g/2 ? ????
r 2
g
???
3
10 6400 2
10
? ?
rad/sec
? Duration of the day is
T =
?
? 2
=
8 . 9
10 6400 2
2
3
? ?
? ? sec =
49
10 64
2
6
?
? ? sec =
3600 7
8000 2
?
? ?
hr = 2hr
mv
2
/R
mg
? ?
mg ?
mv
2
/r ?
T
mg sin ? ?
mg cos ? ?
T
mg sin ? ?
mg cos
? ?
T
m ?
2
/R
mg
R
Page 4
7.1
SOLUTIONS TO CONCEPTS circular motion;;
CHAPTER 7
1. Distance between Earth & Moon
r = 3.85 × 10
5
km = 3.85 × 10
8
m
T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10
6
sec
v =
T
r 2 ?
=
6
8
10 36 . 2
10 85 . 3 14 . 3 2
?
? ? ?
= 1025.42m/sec
a =
r
v
2
=
8
2
10 85 . 3
) 42 . 1025 (
?
= 0.00273m/sec
2
= 2.73 × 10
–3
m/sec
2
2. Diameter of earth = 12800km
Radius R = 6400km = 64 × 10
5
m
V =
T
R 2 ?
=
3600 24
10 64 14 . 3 2
5
?
? ? ?
m/sec = 465.185
a =
R
V
2
=
5
2
10 64
) 5185 . 46 (
?
= 0.0338m/sec
2
3. V = 2t, r = 1cm
a) Radial acceleration at t = 1 sec.
a =
r
v
2
=
1
2
2
= 4cm/sec
2
b) Tangential acceleration at t = 1sec.
a =
dt
dv
= ) t 2 (
dt
d
= 2cm/sec
2
c) Magnitude of acceleration at t = 1sec
a =
2 2
2 4 ? = 20 cm/sec
2
4. Given that m = 150kg,
v= 36km/hr = 10m/sec, r = 30m
Horizontal force needed is
r
mv
2
=
30
) 10 ( 150
2
?
=
30
100 150 ?
= 500N
5. in the diagram
R cos ? = mg ..(i)
R sin ? =
r
mv
2
..(ii)
Dividing equation (i) with equation (ii)
Tan ? =
rmg
mv
2
=
rg
v
2
v = 36km/hr = 10m/sec, r = 30m
Tan ? =
rg
v
2
=
10 30
100
?
= (1/3)
? ? = tan
–1
(1/3) ?
6. Radius of Park = r = 10m
speed of vehicle = 18km/hr = 5 m/sec
Angle of banking tan ? =
rg
v
2
? ? = tan
–1
rg
v
2
= tan
–1
100
25
= tan
–1
(1/4)
R
mv
2
/R
mg
Chapter 7
7.2
7. The road is horizontal (no banking)
R
mv
2
= ?N
and N = mg
So
R
mv
2
= ? mg v = 5m/sec, R = 10m
?
10
25
= ?g ? ? =
100
25
= 0.25 ?
8. Angle of banking = ? = 30°
Radius = r = 50m
tan ? =
rg
v
2
? tan 30° =
rg
v
2
?
3
1
=
rg
v
2
? v
2
=
3
rg
=
3
10 50 ?
? v =
3
500
= 17m/sec.
9. Electron revolves around the proton in a circle having proton at the centre.
Centripetal force is provided by coulomb attraction.
r = 5.3 ?t 10
–11
m m = mass of electron = 9.1 × 10
–3
kg.
charge of electron = 1.6 × 10
–19
c.
r
mv
2
=
2
2
r
q
k ? v
2
=
rm
kq
2
=
31 11
38 9
10 1 . 9 10 3 . 5
10 6 . 1 6 . 1 10 9
? ?
?
? ? ?
? ? ? ?
=
13
10
23 . 48
04 . 23
?
? v
2
= 0.477 × 10
13
= 4.7 × 10
12
? v =
12
10 7 . 4 ? = 2.2 × 10
6
m/sec
10. At the highest point of a vertical circle
R
mv
2
= mg
? v
2
= Rg ? v = Rg
11. A celling fan has a diameter = 120cm.
?Radius = r = 60cm = 0/6m
Mass of particle on the outer end of a blade is 1g.
n = 1500 rev/min = 25 rev/sec
? = 2 ??n = 2 ? ×25 = 157.14
Force of the particle on the blade = Mr ?
2
= (0.001) × 0.6 × (157.14) = 14.8N
The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle
also exerts a force of 14.8N on the blade along its surface.
12. A mosquito is sitting on an L.P. record disc & rotating on a turn table at
3
1
33 rpm.
n =
3
1
33 rpm =
60 3
100
?
rps
?? = 2 ? n = 2 ? ×
180
100
=
9
10 ?
rad/sec
r = 10cm =0.1m, g = 10m/sec
2
?mg ? mr ?
2
? ? =
g
r
2
?
?
10
9
10
1 . 0
2
?
?
?
?
?
? ?
?
? ? ?
81
2
?
?
R
mg
??g mv
2
/R
Chapter 7
7.3
13. A pendulum is suspended from the ceiling of a car taking a turn
r = 10m, v = 36km/hr = 10 m/sec, g = 10m/sec
2
From the figure T sin ? =
r
mv
2
..(i)
T cos ? = mg ..(ii)
?
?
?
cos
sin
=
rmg
mv
2
? tan ? =
rg
v
2
? ? = tan
–1
?
?
?
?
?
?
?
?
rg
v
2
= tan
–1
10 10
100
?
= tan
–1
(1) ? ? = 45° ?
14. At the lowest pt.
T = mg +
r
mv
2
Here m = 100g = 1/10 kg, r = 1m, v = 1.4 m/sec
T = mg +
r
mv
2
=
10
) 4 . 1 (
8 . 9
10
1
2
? ? = 0.98 + 0.196 = 1.176 = 1.2 N
15. Bob has a velocity 1.4m/sec, when the string makes an angle of 0.2 radian.
m = 100g = 0.1kg, r = 1m, v = 1.4m/sec.
From the diagram,
T – mg cos ? =
R
mv
2
? T =
R
mv
2
+ mg cos ?
? T =
?
?
?
?
?
?
?
?
?
? ? ? ?
?
2
1 8 . 9 ) 1 . 0 (
1
) 4 . 1 ( 1 . 0
2 2
? T = 0.196 + 9.8 ×
?
?
?
?
?
?
?
?
?
2
) 2 (.
1
2
( ? cos ? =
2
1
2
?
? for small ?)
? T = 0.196 + (0.98) × (0.98) = 0.196 + 0.964 = 1.156N ? 1.16 N ?
16. At the extreme position, velocity of the pendulum is zero.
So there is no centrifugal force.
So T = mg cos ?
o
?
17. a) Net force on the spring balance.
R = mg – m ?
2
r
So, fraction less than the true weight (3mg) is
=
mg
) r m mg ( mg
2
? ? ?
=
g
2
?
=
10
10 6400
3600 24
2
3
2
?
? ?
?
?
?
?
?
?
?
= 3.5 × 10
–3
b) When the balance reading is half the true weight, ?
mg
) r m mg ( mg
2
? ? ?
= 1/2
?
2
r = g/2 ? ????
r 2
g
???
3
10 6400 2
10
? ?
rad/sec
? Duration of the day is
T =
?
? 2
=
8 . 9
10 6400 2
2
3
? ?
? ? sec =
49
10 64
2
6
?
? ? sec =
3600 7
8000 2
?
? ?
hr = 2hr
mv
2
/R
mg
? ?
mg ?
mv
2
/r ?
T
mg sin ? ?
mg cos ? ?
T
mg sin ? ?
mg cos
? ?
T
m ?
2
/R
mg
R
Chapter 7
7.4
18. Given, v = 36km/hr = 10m/s, r = 20m, ? = 0.4
The road is banked with an angle,
? = tan
–1
?
?
?
?
?
?
?
?
rg
v
2
= tan
–1
?
?
?
?
?
?
?10 20
100
= tan
–1 ?
?
?
?
?
?
2
1
or tan ? = 0.5
When the car travels at max. speed so that it slips upward, ?R
1
acts downward as shown in Fig.1
So, R
1
– mg cos ? –
r
mv
2
1
sin ? = 0 ..(i)
And ?R
1
+ mg sin ? –
r
mv
2
1
cos ? = 0 ..(ii)
Solving the equation we get,
V
1
=
? ? ?
? ? ?
tan 1
tan
rg =
2 . 1
1 . 0
10 20 ? ? = 4.082 m/s = 14.7 km/hr
So, the possible speeds are between 14.7 km/hr and 54km/hr.
19. R = radius of the bridge
L = total length of the over bridge
a) At the highest pt.
mg =
R
mv
2
? v
2
= Rg ? v = Rg
b) Given, v = Rg
2
1
suppose it loses contact at B. So, at B, mg cos ? =
R
mv
2
? v
2
= Rg cos ?
?
2
2
Rv
?
?
?
?
?
?
?
?
= Rg cos ? ?
2
Rg
= Rg cos ? ? cos ? = 1/2 ? ? = 60° = ?/3
? =
r
?
? l = r ? =
3
R ?
So, it will lose contact at distance
3
R ?
from highest point
c) Let the uniform speed on the bridge be v.
The chances of losing contact is maximum at the end of the bridge for which ? =
R 2
L
.
So,
R
mv
2
= mg cos ? ? v = ?
?
?
?
?
?
R 2
L
cos gR
?
20. Since the motion is nonuniform, the acceleration has both radial & tangential
component
a
r
=
r
v
2
a
t
=
dt
dv
=a
Resultant magnitude =
2
2
2
a
r
v
?
?
?
?
?
?
?
?
?
Now ?N = m
2
2
2
a
r
v
?
?
?
?
?
?
?
?
?
? ? mg = m
2
2
2
a
r
v
?
?
?
?
?
?
?
?
?
? ?
2
g
2
=
2
4
a
2 r
v
?
?
?
?
?
?
?
?
?
? v
4
= ( ?
2
g
2
– a
2
) r
2
? v = [( ?
2
g
2
– a
2
) r
2
]
1/4
? ?
?R 1
? ?
mv
1
2
/r
mg
R 1
? ?
mv
2
2
/r
?R 2
? ?
mg
R 2
? ?
2 ? ?
mv
2
/R
mg
2 ?= L/R ?
? ?
mv
2
/R
mg
2 ? ?
2 ?= L/R ?
? ?
mv
2
/R
2 ? ?
2 ?= L/R ?
mv
2
/R ? mg
m dv/dt
m
N
mv
2
/R
m
Page 5
7.1
SOLUTIONS TO CONCEPTS circular motion;;
CHAPTER 7
1. Distance between Earth & Moon
r = 3.85 × 10
5
km = 3.85 × 10
8
m
T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10
6
sec
v =
T
r 2 ?
=
6
8
10 36 . 2
10 85 . 3 14 . 3 2
?
? ? ?
= 1025.42m/sec
a =
r
v
2
=
8
2
10 85 . 3
) 42 . 1025 (
?
= 0.00273m/sec
2
= 2.73 × 10
–3
m/sec
2
2. Diameter of earth = 12800km
Radius R = 6400km = 64 × 10
5
m
V =
T
R 2 ?
=
3600 24
10 64 14 . 3 2
5
?
? ? ?
m/sec = 465.185
a =
R
V
2
=
5
2
10 64
) 5185 . 46 (
?
= 0.0338m/sec
2
3. V = 2t, r = 1cm
a) Radial acceleration at t = 1 sec.
a =
r
v
2
=
1
2
2
= 4cm/sec
2
b) Tangential acceleration at t = 1sec.
a =
dt
dv
= ) t 2 (
dt
d
= 2cm/sec
2
c) Magnitude of acceleration at t = 1sec
a =
2 2
2 4 ? = 20 cm/sec
2
4. Given that m = 150kg,
v= 36km/hr = 10m/sec, r = 30m
Horizontal force needed is
r
mv
2
=
30
) 10 ( 150
2
?
=
30
100 150 ?
= 500N
5. in the diagram
R cos ? = mg ..(i)
R sin ? =
r
mv
2
..(ii)
Dividing equation (i) with equation (ii)
Tan ? =
rmg
mv
2
=
rg
v
2
v = 36km/hr = 10m/sec, r = 30m
Tan ? =
rg
v
2
=
10 30
100
?
= (1/3)
? ? = tan
–1
(1/3) ?
6. Radius of Park = r = 10m
speed of vehicle = 18km/hr = 5 m/sec
Angle of banking tan ? =
rg
v
2
? ? = tan
–1
rg
v
2
= tan
–1
100
25
= tan
–1
(1/4)
R
mv
2
/R
mg
Chapter 7
7.2
7. The road is horizontal (no banking)
R
mv
2
= ?N
and N = mg
So
R
mv
2
= ? mg v = 5m/sec, R = 10m
?
10
25
= ?g ? ? =
100
25
= 0.25 ?
8. Angle of banking = ? = 30°
Radius = r = 50m
tan ? =
rg
v
2
? tan 30° =
rg
v
2
?
3
1
=
rg
v
2
? v
2
=
3
rg
=
3
10 50 ?
? v =
3
500
= 17m/sec.
9. Electron revolves around the proton in a circle having proton at the centre.
Centripetal force is provided by coulomb attraction.
r = 5.3 ?t 10
–11
m m = mass of electron = 9.1 × 10
–3
kg.
charge of electron = 1.6 × 10
–19
c.
r
mv
2
=
2
2
r
q
k ? v
2
=
rm
kq
2
=
31 11
38 9
10 1 . 9 10 3 . 5
10 6 . 1 6 . 1 10 9
? ?
?
? ? ?
? ? ? ?
=
13
10
23 . 48
04 . 23
?
? v
2
= 0.477 × 10
13
= 4.7 × 10
12
? v =
12
10 7 . 4 ? = 2.2 × 10
6
m/sec
10. At the highest point of a vertical circle
R
mv
2
= mg
? v
2
= Rg ? v = Rg
11. A celling fan has a diameter = 120cm.
?Radius = r = 60cm = 0/6m
Mass of particle on the outer end of a blade is 1g.
n = 1500 rev/min = 25 rev/sec
? = 2 ??n = 2 ? ×25 = 157.14
Force of the particle on the blade = Mr ?
2
= (0.001) × 0.6 × (157.14) = 14.8N
The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle
also exerts a force of 14.8N on the blade along its surface.
12. A mosquito is sitting on an L.P. record disc & rotating on a turn table at
3
1
33 rpm.
n =
3
1
33 rpm =
60 3
100
?
rps
?? = 2 ? n = 2 ? ×
180
100
=
9
10 ?
rad/sec
r = 10cm =0.1m, g = 10m/sec
2
?mg ? mr ?
2
? ? =
g
r
2
?
?
10
9
10
1 . 0
2
?
?
?
?
?
? ?
?
? ? ?
81
2
?
?
R
mg
??g mv
2
/R
Chapter 7
7.3
13. A pendulum is suspended from the ceiling of a car taking a turn
r = 10m, v = 36km/hr = 10 m/sec, g = 10m/sec
2
From the figure T sin ? =
r
mv
2
..(i)
T cos ? = mg ..(ii)
?
?
?
cos
sin
=
rmg
mv
2
? tan ? =
rg
v
2
? ? = tan
–1
?
?
?
?
?
?
?
?
rg
v
2
= tan
–1
10 10
100
?
= tan
–1
(1) ? ? = 45° ?
14. At the lowest pt.
T = mg +
r
mv
2
Here m = 100g = 1/10 kg, r = 1m, v = 1.4 m/sec
T = mg +
r
mv
2
=
10
) 4 . 1 (
8 . 9
10
1
2
? ? = 0.98 + 0.196 = 1.176 = 1.2 N
15. Bob has a velocity 1.4m/sec, when the string makes an angle of 0.2 radian.
m = 100g = 0.1kg, r = 1m, v = 1.4m/sec.
From the diagram,
T – mg cos ? =
R
mv
2
? T =
R
mv
2
+ mg cos ?
? T =
?
?
?
?
?
?
?
?
?
? ? ? ?
?
2
1 8 . 9 ) 1 . 0 (
1
) 4 . 1 ( 1 . 0
2 2
? T = 0.196 + 9.8 ×
?
?
?
?
?
?
?
?
?
2
) 2 (.
1
2
( ? cos ? =
2
1
2
?
? for small ?)
? T = 0.196 + (0.98) × (0.98) = 0.196 + 0.964 = 1.156N ? 1.16 N ?
16. At the extreme position, velocity of the pendulum is zero.
So there is no centrifugal force.
So T = mg cos ?
o
?
17. a) Net force on the spring balance.
R = mg – m ?
2
r
So, fraction less than the true weight (3mg) is
=
mg
) r m mg ( mg
2
? ? ?
=
g
2
?
=
10
10 6400
3600 24
2
3
2
?
? ?
?
?
?
?
?
?
?
= 3.5 × 10
–3
b) When the balance reading is half the true weight, ?
mg
) r m mg ( mg
2
? ? ?
= 1/2
?
2
r = g/2 ? ????
r 2
g
???
3
10 6400 2
10
? ?
rad/sec
? Duration of the day is
T =
?
? 2
=
8 . 9
10 6400 2
2
3
? ?
? ? sec =
49
10 64
2
6
?
? ? sec =
3600 7
8000 2
?
? ?
hr = 2hr
mv
2
/R
mg
? ?
mg ?
mv
2
/r ?
T
mg sin ? ?
mg cos ? ?
T
mg sin ? ?
mg cos
? ?
T
m ?
2
/R
mg
R
Chapter 7
7.4
18. Given, v = 36km/hr = 10m/s, r = 20m, ? = 0.4
The road is banked with an angle,
? = tan
–1
?
?
?
?
?
?
?
?
rg
v
2
= tan
–1
?
?
?
?
?
?
?10 20
100
= tan
–1 ?
?
?
?
?
?
2
1
or tan ? = 0.5
When the car travels at max. speed so that it slips upward, ?R
1
acts downward as shown in Fig.1
So, R
1
– mg cos ? –
r
mv
2
1
sin ? = 0 ..(i)
And ?R
1
+ mg sin ? –
r
mv
2
1
cos ? = 0 ..(ii)
Solving the equation we get,
V
1
=
? ? ?
? ? ?
tan 1
tan
rg =
2 . 1
1 . 0
10 20 ? ? = 4.082 m/s = 14.7 km/hr
So, the possible speeds are between 14.7 km/hr and 54km/hr.
19. R = radius of the bridge
L = total length of the over bridge
a) At the highest pt.
mg =
R
mv
2
? v
2
= Rg ? v = Rg
b) Given, v = Rg
2
1
suppose it loses contact at B. So, at B, mg cos ? =
R
mv
2
? v
2
= Rg cos ?
?
2
2
Rv
?
?
?
?
?
?
?
?
= Rg cos ? ?
2
Rg
= Rg cos ? ? cos ? = 1/2 ? ? = 60° = ?/3
? =
r
?
? l = r ? =
3
R ?
So, it will lose contact at distance
3
R ?
from highest point
c) Let the uniform speed on the bridge be v.
The chances of losing contact is maximum at the end of the bridge for which ? =
R 2
L
.
So,
R
mv
2
= mg cos ? ? v = ?
?
?
?
?
?
R 2
L
cos gR
?
20. Since the motion is nonuniform, the acceleration has both radial & tangential
component
a
r
=
r
v
2
a
t
=
dt
dv
=a
Resultant magnitude =
2
2
2
a
r
v
?
?
?
?
?
?
?
?
?
Now ?N = m
2
2
2
a
r
v
?
?
?
?
?
?
?
?
?
? ? mg = m
2
2
2
a
r
v
?
?
?
?
?
?
?
?
?
? ?
2
g
2
=
2
4
a
2 r
v
?
?
?
?
?
?
?
?
?
? v
4
= ( ?
2
g
2
– a
2
) r
2
? v = [( ?
2
g
2
– a
2
) r
2
]
1/4
? ?
?R 1
? ?
mv
1
2
/r
mg
R 1
? ?
mv
2
2
/r
?R 2
? ?
mg
R 2
? ?
2 ? ?
mv
2
/R
mg
2 ?= L/R ?
? ?
mv
2
/R
mg
2 ? ?
2 ?= L/R ?
? ?
mv
2
/R
2 ? ?
2 ?= L/R ?
mv
2
/R ? mg
m dv/dt
m
N
mv
2
/R
m
Chapter 7
7.5
21. a) When the ruler makes uniform circular motion in the horizontal
plane, (fig–a)
? mg = m ?
?
?
L
?
?
?
??
L
g ?
?
b) When the ruler makes uniformly accelerated circular motion,(fig–b)
? mg =
2 2 2
2
) mL ( ) L m ( ? ? ? ? ?
2
4
+ ?
2
=
2
2 2
L
g ?
? ?
2
=
4 / 1
2
2
L
g
?
?
?
?
?
?
?
?
? ? ?
?
?
?
?
? ?
(When viewed from top)
22. Radius of the curves = 100m
Weight = 100kg
Velocity = 18km/hr = 5m/sec
a) at B mg –
R
mv
2
= N ? N = (100 × 10) –
100
25 100 ?
= 1000 – 25 = 975N
At d, N = mg +
R
mv
2
= 1000 + 25 = 1025 N
b) At B & D the cycle has no tendency to slide. So at B & D, frictional force is zero.
At ‘C’, mg sin ? = F ? F = 1000 ×
2
1
= 707N
c) (i) Before ‘C’ mg cos ? – N =
R
mv
2
? N = mg cos ? –
R
mv
2
= 707 – 25 = 683N
(ii) N – mg cos ? =
R
mv
2
? N =
R
mv
2
+ mg cos ? = 25 + 707 = 732N
d) To find out the minimum desired coeff. of friction, we have to consider a point just before C. (where
N is minimum)
Now, ? N = mg sin ? ? ? × 682 = 707
So, ? = 1.037 ?
23. d = 3m ? R = 1.5m
R = distance from the centre to one of the kids
N = 20 rev per min = 20/60 = 1/3 rev per sec
??= 2 ?r = 2 ?/3
m = 15kg
? Frictional force F = mr ?
2
= 15 × (1.5) ×
9
) 2 (
2
?
= 5 × (0.5) × 4 ?
2
= 10 ?
2
? Frictional force on one of the kids is 10 ?
2
?
24. If the bowl rotates at maximum angular speed, the block tends to slip upwards. So, the frictional force
acts downward.
Here, r = R sin ?
From FBD –1
R
1
– mg cos ? – m ?
?
2
(R sin ?) sin ? = 0 ..(i) [because r = R sin ?]
and ?R
1
mg sin ? – m ?
1
2
(R sin ?) cos ? = 0 ..(ii)
Substituting the value of R
1
from Eq (i) in Eq(ii), it can be found out that
?
1
=
2 / 1
) sin (cos sin R
) cos (sin g
?
?
?
?
?
?
? ? ? ? ?
? ? ? ?
Again, for minimum speed, the frictional force
?R
2
acts upward. From FBD–2, it can be proved
that,
??? 1
2
L ? mg
mg
? ? ? ?
R
L
(Fig–a)
(Fig–b)
m ? 2
2
L ? mg
mL ? ?
?R 1
? ?
m ? 1
2
r
R 1
(FBD – 1)
?R 2
? ?
m ? 2
2
r
R 2
(FBD – 2)
?
E
D
C
B
A
mv
2
/R
mg
N
B
15kg
mg
m ?
2
r ? F
15kg
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