Class 11-science H C VERMA Solutions Physics Chapter 10 - Rotational Mechanics
Rotational Mechanics Exercise 195
Solution 1
t=4 sec;
ω=100 =100×2π
Using, ω=
=50π or
=25π
Using, =400π rad
Solution 2
t=4 sec;
Using,
=8π or =4π
Using, ω=
ω=40π or ω=20
Solution 3
Angular displacement=Area under curve
=
θ=800 rad.
Rotational Mechanics Exercise 196
Solution 4
ω=15 rad/sec; =1
Substituting values in,
θ =100 rad
Solution 5
; =2 ;
θ=5 rev=5×2π rad=10π rad
Substituting values in,
ω=4π rad/sec
ω=2 rev/sec
Solution 6
(a)v= rω= (0.1) (20)
v= 2 m/s
(b) v= rω=
v=1 m/s
Solution 7
; = 4 rad/; t=1 sec
ω=+t
ω=4 rad/sec
Radial acceleration=R
= (0.01)
= 0.16
= 16
Tangential acceleration
= (0.01)(4)
= 0.04
= 4
Solution 8
Block speed=Rim speed of the disc
v=Rω = (0.2)(10)
v= 2 m/s
Solution 9
(a) = m
= 0.2
= 1.5 Kg-
(b) =
=2× 0.2
=4 Kg-
Solution 10
The perpendicular bisector passes from 50cm mark. So, there will be 49 particles on LHS and 50 particles on RHS.
I=
= (100) []
= (100) [] gm-
I≈ 0.43 kg/
[∵ Sum of square of n natural numbers, ]
Solution 11
By parallel axis theorem,
Solution 12
By parallel axis theorem,
0.1=
X = 0.34 m
Solution 13
m
∴k=
Solution 14
⇒
⇒ d=
Solution 15
If we squeeze rod along and the square becomes rod of side 'a'
So,
By perpendicular axis theorem,
=
[∵ As square is symmetric]
Solution 16
Surface density of disc is varying radially so considering an element of thickness 'dr' at a distance 'r' from center of disc.
σ=A+Br=
dm=(A+Br)(2πrdr)
=
=2[A]
Solution 17
= mg ()
= mg ()
Torque acts perpendicular to plane of the motion.
Solution 18
Torque will be zero at the lowest point of suspension as θ will be zero.
Solution 19
Torque in both the cases will be equal
()(8)=(F)(16)
F=1.6 N
Solution 20
Torque=
Consider anticlockwise torque as negative and clockwise torque as positive.
= (10) (4)-(15) (6)+5(0)-20(4)
τ=0.54 N-m (anticlockwise direction)
Solution 21
Since,
acceleration=0
ff=mg
The normal contact will shift as block is not toppling.
So,
=mg
Solution 22
Torque=
τ=F
I∝
∝=
Now,
θ=
θ=
Solution 23
Squeeze rod along axis A so it will behave as rod
I=
=
N-m
Solution 24
M.I. of square plate about its diagonal
I=
=
=0.5×N-m
Solution 25
ω=
α=
(a)
=5()
=-1 N-m
(b) =∆KE
=I
=
=9 KJ
(c)
Angular momentum
L=I
=5(12)
=60
Solution 26
=
=
=
N-m
Rotational Mechanics Exercise 197
Solution 27
For t=5sec,
=10+(-1)(5)
Solution 28
θ=10 rev= 20π rad
ω=0
Using,
F= ()()
=
F=0.87 N
Solution 29
Let after time 't' their angular velocity be 'ω'
cylinder :- ; α=-1;
ω=
ω=50-t -(i)
cylinder :- α=1
ω=
ω=t -(ii)
From (i) and (ii)
ω=25 rev/sec
t=25 sec
Solution 30
ω=
Now, wheel accelerates and when angular velocity becomes 20 rad/sec it will have same KE.
ω=
20=0+
T=
T=20 sec
Solution 31
N-m
15= []α
α=8.57
Solution 32
(a)
(50×0.5-20×0.5) = [(2)+5+]α
α= 8.1
(b)
N
N
Solution 33
Mg-=Ma -(i)
-(ii)
-(iii)
a=rα -(iv)
Solving , (i),(ii),(iii) and (iv)
a=
Solution 34
Solution 35
Solution 36
Solution 37
Translatory Motion Equation
-(i)
mg-()=m -(ii)
Rotational Motion Equation
Iα
-(iii)
For mass m
I=
(0.2) =
m=10 kg -(iv)
Using, (i),(ii),(iii) and (iv)
a=10 m/
Solution 38
Here,
I=0.5 kg-
R=0.1m
Translatory Motion Equation
4g -(i)
-(ii)
Rotational Motion Equation
-(iii)
Using (i),(ii) and (iii)
a=0.25
Solution 39
Here, I=0.5 Kg- ; R=0.1 m ; ff=
Translatory Motion Equation
4g -(i)
-(ii)
Rotational Motion Equation
(iii)
Using (i),(ii) and (iii)
a=0.125
Solution 40
Translatory Equilibrium Equation
-(i)
Rotational Equilibrium Equation at point A
clockwise Torque= anticlockwise Torque
(0.2g)+(0.5)+(0.02g)(0.7)=(1)
N
From (i) N
Solution 41
Translatory Equilibrium Equation
(Horizontal direction)
(Vertical direction)
Rotational Equilibrium at bottom of ladder
Clockwise Torque =Anticlockwise Torque
Solving,
412=
Rotational Mechanics Exercise 198
Solution 42
Here ,
Translatory Equilibrium
-(i)
- (ii)
Rotational Equilibrium about point 'O'
-(iii)
Solving (i),(ii) and (iii)
m=44 kg
Solution 43
By Pythagoras theorem,
B=2.5 m
(a)
τ=60g ()
=60g()
N-m
(b) (Vertical Translatory Equation)
Rotational Equilibrium at O
Solution 44
Magnitude of forces by hinges are equal.
-(i)
Rotational Equilibrium at point O
By translational Equilibrium
-(ii)
-(iii)
Solving (i),(ii),(iii)
Magnitude of force=
=
=43 N
Solution 45
Translatory equilibrium
-(i)
-(ii)
Rotational Equilibrium about bottom
-(iii)
-(iv)
Solving (i),(ii),(iii) and (iv)
Solution 46
(a) L=Iω
=
=
=0.05 kg-
(b) v=Rω
=(0.25)(2)
v=0.5 m/s
(c) KE=
=
=
KE=0.05 J
Solution 47
τ=Iα
0.1=
0.1=
ω= ⇒ ω=30 rad/sec
Angular momentum , L= Iα
=[]
=0.5 kg-
KE=
=
KE= 75 J
Solution 48
Iω
=()()
= -(i)
Iω
=
= -(ii)
Divide (i) by (ii)
Solution 49
Distance of COM from will be
Distance of COM from will be
L= Iω
L=()ω
L=[]
L= where
Solution 50
τ= Iω
F×r=
5(0.25) = 2()
τ=0.1
τ
Solution 51
Since ,
(0.5)(20)=[0.5+(0.2)]
Solution 52
Since,
6(2) = 5
rad/sec
Solution 53
Since,
6(120) = (2) (
rev/sec
Solution 54
Since,
0+0= -(i)
So, Angular speed of umbrella with respect to boy will be in the opposite direction.
Let angular speed of boy be ω
Now,
-2=
In equation (i),
0=
rev/sec
Solution 55
Taking wheels as a system
(0.1)(160)+ (300)=(0.1+)(200)
kg-
Solution 56
Considering ball, boy and platform as system
mvR=[I+(M+m)]ω
ω=
Solution 57
Considering ball, boy and platform as system
0+0+0=mvR+I(-ω)+(M)(- ω)
ω=
Solution 58
Let angular velocity of platform after kid start running ω'
So, angular velocity of kid with respect to earth=( ω'+)
∵
∴
ω=I ω'+
ω'= ω-
Rotational Mechanics Exercise 199
Solution 59
(a) F=ma
acceleration, a= ; u=0; t
v=u+at
v=
(b) τ=Iα
F()=. α
α= ; ; t=t
αt
α=
(c) KE=
=
=
(d) L=I
=()
L=
Solution 60
Momentum conservation
mu+0=m(0)+M(v)
v=
Angular Momentum conservation
0+mu()=()ω+0
ω=
Time to rotate by
ω=
t=
t=
distance travelled by rod
distance=speed×time
=
s=
Now, if M=4m then
=
Hence proved.
Solution 61
(a)
mv=(M+m)v'
v'=
(b) velocity of the particle with respect to COM
=v-v'
=v-
=
(c) velocity of the rod wrt the centre of mass=
(d) Distance of COM from the particle
Angular momentum of body about COM=mvr
=m
=
Angular momentum of rod about COM
=M
=
(e) M.I. about COM=
I=m
I=
(f) About COM
∴ Iω=mvr ⇒ ω=
Solution 62
∵
[M]ω=[3m] ω'
ω'=
Solution 63
(a) The light rod will exert a force on the ball B only along its length. So collision will not effect its velocity.
Momentum conservation for ball A
m=2m
(b)
=
(c) velocity of A with respect to centre of mass=
velocity of B with respect to centre of mass=
Applying, m
(2m)+ (m)=[2m]
Solution 64
(a) If we consider two bodies P and B as system
then and
m()=2mv'
v'=
m()()=Iω
mL=() ω
ω=
(b) When the mass 2m and m are at the top most position and at the lowest point respectively, they will automatically rotate.
Change in Potential Energy=Change in kinetic Energy
2mg()-mg()= I
mg()=
h=
Solution 65
Translatory Motion Equation
0.4g-=0.4a -(i)
-0.2g=0.2a -(ii)
Rotational Motion Equation,
-(iii)
Solving (i),(ii) and (iii)
a=g/5
Speed of block=
=
=1.4 m/s
Total kinetic energy of system
=
=
=0.98 J
Solution 66
By energy conservation
mgh=
1
v=0.5 m/s
Solution 67
By energy conservation
mg
mg
ω=5.42 rad/sec
Solution 68
Velocity of ball just before the collision:-
u=
u=
u= m/sec
Now, Torque about hinge is zero
mul+0=Iω
(0.1)=()ω
(0.1)=[(0.1)] ω
ω=
Now, By energy conservation
Solution 69
By energy conservation
So, centrifugal force
=
Let angular acceleration of the rod be α when it makes an angle of with the vertical.
τ=Iα
mg()=
α=0.9()
So, tangential force
=
So,
Solution 70
For pure rolling
Speed at highest point=v+
=2v
=2(25)
=50 m/s
Rotational Mechanics Exercise 200
Solution 71
v=Rω (Pure rolling)
K.E=
=
K.E=
Solution 72
mg-T=ma -(i)
τ=Iα
-(ii)
From (i) and (ii)
a=
Solution 73
By energy conservation,
mgh=
mgh=
Solution 74
By energy conservation,
Solution 75
By energy conservation,
Solution 76
(a) -(i)
-(ii)
-(iii)
Solving (i),(ii) and (iii)
(b)
Translatory Motion Equation
Time to travel ,
Rotational Motion Equation
K.E=
=
K.E=
Solution 77
By conservation of energy
Normal contact at bottom
Solution 78
For minimum u, the ball will just be in contact with surface at top. So, N=0
Now, by energy conservation
Solution 79
(a)
Energy conservation
(b)
=
-(i)
Radial acceleration
Differentiate equation (i) for tangential acceleration
(c)
Horizontal direction
=
N=5
Vertical direction
=
Solution 80
(Pure Rolling)
By angular momentum conservation
Solution 81
By angular momentum conservation at bottom point since
Solution 82
By angular momentum conservation at bottom most point
Solution 83
Time to complete one rotation
By,
For translatory motion
Solution 84
-(i)
-(ii)
-(iii)
Solving (i),(ii) and (iii)
F=3.3 N
Solution 85
(a)
Conserving angular momentum about bottom point
(b)
Conserving angular momentum
Solution 86