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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=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

 

  

  

  

   

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