Class 11-science H C VERMA Solutions Physics Chapter 7 - Circular Motion
Circular Motion Exercise 114
Solution 1
Speed of the moon=distance/time
=1025.4m/s
Acceleration of moon=
Ar = 2.73 × 10-3m/s2
Solution 2
Speed of particle at equator=distance/time
=465.1m/s
Acceleration of particle=
Ar = 0.038m/s2
Solution 3
(a)
Velocity of particle at t=1 sec
V=2t
V=2(1)
V=2cm/s
Radial acceleration
Ar=4cm/s2
(b)
Tangential acceleration
At=
At=2cm/s2
(c)
An= 
An= 
An =
cm/s
Solution 4
Horizontal force required=Centrifugal force
F=500N
Solution 5
Let banking angle be θ
tan
θ=
tan
θ=
= 
Solution 6
Let banking angle be θ
tan
θ =
tan
θ
=
Solution 7
Centrifugal force=Frictional force
= µN
=µmg
µ= =
µ=0.25
Solution 8
tan θ=
tan30° =
v ≅ 17m/s
Solution 9
Coulomb force=Centrifugal force
=
=(9.1 ×
10-31
)v2
V=2.2 × 106m/s
Solution 10
At highest point
T
+ mg =
For minimum speed, T=0
Solution 11
Radius
of the circle = 
= 60cm
Angular speed=1500rpm
ω
=1500 × 
ω =157 rad/sec
Force=mRω2
(0.6 )(157)2
F=14.8N
This force is exerted by friction. On particle it acts towards the centre of the circle and on fan blade outside the circle due to action-reaction pair.
So, force along the surface=14.8N
Solution 12
ω = 33
rpm = 
rpm
For mosquito to remain at rest on L.P record.
Frictional force ≥ Centrifugal force
µN ≥ mRω2
µmg ≥ mRω2
Solution 13
Tsin θ= ------------------ (1)
Tcosθ =mg -------------------(2)
Divide,
tan
θ
=
tan
θ
=
θ = 45°
Circular Motion Exercise 115
Solution 14
At lowest point
=
(0.1)(10) + 
T = 1.2N
Solution 15
At angle θ
T
= mg(1-
)+
T
= (0.1)(10)(1- 
)+
T ≃ 1.16
Solution 16
At extreme position,
T= mgcos θ0
(No centrifugal force as v=0 at extreme position.)
Solution 18
For banking angle,
tan
θ=
=
tan θ=0.5
=15m/s
=54kmph
Vmin=
=
=4m/s
Vmin=14.7kmph
Solution 19
(a)
At highest point,
For maximum speed, N=0
Vma×=
(b)
V=
=
Let at angle 𝚽 it loses its contact with road
N'=0
N'+
=mgcos𝚽
0+
= mgcos𝚽
Cos𝚽=
𝚽=60° or 
rad.
So it loses contact after a distance πR/3 along the bridge from the highest point.
(c)
Normal contact at any angle 𝚽
N'=mgcos 𝚽- ---->(1)
So, as motorcycle moves along the curve 𝚽 increases, so cos𝚽 decreases and hence N' decreases.
Normal contact will be minimum at the extreme point.
At extreme point
𝚽= 
= and N'=0
From equation (1),
0=mgcos 
Solution 20
From free body diagram
ff=
µN=
(µmg)
=
Squaring and solving,
v=[(µ2g2-a2)R2]1/4
Solution 21
(a)
For block not to slip
ff ≥ mLω2
µN ≥ mLω2
µmg ≥ mLω2
ω2 
ωma×=
(b)
Now, ruler makes uniformly accelerated circular motion at angular acceleration of α,so tangential acceleration
At=Lα
Ft = mat
Ft = mLα
Radial force Fc = mLω'2
Where ω' is maximum angular speed at which it slips
µN=
µmg=
squaring and solving,
ω'=[
- α2]1/4
Solution 22
(a)
Velocity
of cyclist
=5m/s
At point B
NB
+ 
= mg
NB
= (100)(10)- 
NB = 975N
At point D
ND
= mg +
ND
= (100)(10) +
ND = 1025N
(b)
No force acts along surface at point B and D
So,ffB=ffD=0
At point C,
∵acceleration=0
So,ffc = mgsinθ
ffc = (100)(10)sin450
ffc = 707N
(C)
Before C,
Nc+ =mgcos45°
Nc=(100)(10)
NC = 682N
After C,
N'C= + mgcos45°
N'C
= 
N'C = 732N
(d) ff=µN
For minimum friction force normal contact should be minimum.
Normal contact is minimum at point just before C.
ff=mgsinѲ = µN
(100)(10)sin45°= µ(682)
µ = 1.037
Solution 23
ω =20
rpm=20×
ω =
(rad/sec)
R=1.5m
Frictional force=mR ω2
=(15)(1.5)
=10 π2
Solution 24
Particle will have tendency to move up. So, frictional force is in downward direction.
r=Rsinθ----->(ί)
N=
mgcosθ+
sinθ------->(ii)
ff + mgsinθ=
cosθ------->(iii)
ff= µN-------->(iv)
on solving,
ωma×=
For ωmin, particle will have tendency to move down. So, frictional force acting upward direction.
r=Rsinθ----->(1)
N=
mgcosθ +
sinθ------->(ii)
Mgsinθ =
ff+ cosθ------->(iii)
ff= µN-------->(iv)
on solving,
ωmin=
Solution 25
At highest point
Aradial=
Solution 26
Since acceleration in horizontal direction is zero
ucosθ
=
vcos 
aradial=
Solution 27
(a)
(b)
ff=µN= 
(c)
Let acceleration along velocity be at for the block
0-ff=m- at
=mat
at=-
(d)
V
=
=
lnv|
=
(x)
V=v0e-2πµ
Circular Motion Exercise 116
Solution 28
Component of force mRω2 along the line AB will be responsible for motion of particle
Force= mRω2cosθ
ma= mRω2cosθ
a=Rω2cosθ
u=0
s=L
using
s=ut+at2
l=(Rω2cosθ) t2
Solution 29
(a)
N=0.2
(b)
At the time of just sliding
µ=tanθ
θ=tan-1(0.58)
θ=300
Solution 30
2mRω2-T=2ma
T- mRω2=ma
Adding both,
mRω2=3ma
T-
mRω2= 
T=
mRω2
Circular Motion Exercise 155
Solution 17
(a)
At equator, normal contact force
N+mRω2=mg
Reading N=mg-mRω2
So,
fraction less than true weight=
=
=
=3.5×10-3
(b)
Balance
reading=(True weight)
mg-mRω2=( (mg)
mRω2=
ω=
= 
T'=2×3.14×
(in sec)
T'=
(in hrs.)
T'=2hrs
