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system of particles and rotational motion

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physics part i

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Class 10 Class 12
A body at rest explodes in three fragments. What is the velocity of the centre of mass of exploded part?

Zero.

In the absence of external force, the centre of mass moves with constant velocity. Since before explosion the body was at rest, therefore, the velocity of the centre of mass is zero before an explosion. Hence it will also be zero after the explosion.
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(a) Prove the theorem of perpendicular axes. 

(Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin perpendicular to the plane is (x+ y2). 

The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body. 

A physical body with centre O and a point mass m,in the xyplane at (xy) is shown in the following figure below. 


                         
Moment of inertia about x-axis, Ix = mx

Moment of inertia about y-axis, Iy = my

Moment of inertia about z-axis, Iz = m(x2 + y2)1/2 


Therefore, 

Ix + Iy = mx2 + my

         = m(x2 + y2

         = straight m space open square brackets square root of straight x squared plus straight y squared end root close square brackets to the power of begin inline style bevelled 1 half end style end exponent 

I
x + Iy = I

Hence, the theorem is proved.

 
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26. (b) Prove the theorem of parallel axes. 

(Hint: If the centre of mass is chosen to be the origin ∑ miri = 0).

The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes. 


                          
Suppose a rigid body is made up of n particles, having masses m1m2m3, … , mn, at perpendicular distances r1,r2r3, … , rn respectively from the centre of mass O of the rigid body. 

The moment of inertia about axis RS passing through the point O, 

straight I space subscript RS space equals space sum from straight i space equals space 1 to straight n of space straight m subscript straight i space straight r subscript straight i squared space

The space perpendicular space distance space of space mass space straight m comma
space from space the space axis space QP space equals space straight a space plus space straight r subscript straight i space

Hence comma space the space moment space of space Inertia space about space axis space QP thin space is comma space

straight I subscript QP space equals space sum from straight i space equals 1 to straight n of space straight m subscript straight i space left parenthesis thin space straight a space plus space straight r subscript straight i right parenthesis squared space

space space space space space space equals space sum from straight i space equals 1 to straight n of space straight m subscript straight i space left parenthesis thin space straight a squared space plus space straight r subscript straight i squared space plus space 2 ar subscript straight i space right parenthesis space squared space

space space space space space space equals space sum from straight i space equals 1 to straight n of space straight m subscript straight i space straight a squared space space plus space sum from straight i space equals 1 to straight n of space space straight m subscript straight i space straight r subscript straight i squared space plus space sum from straight i space equals 1 to straight n of space space straight m subscript straight i space space 2 ar subscript straight i space

space space space space space space equals space space sum from straight i space equals 1 to straight n of space straight m subscript straight i space straight a squared space plus space 2 space sum from straight i space equals 1 to straight n of space space straight m subscript straight i space straight r subscript straight i squared


The moment of Inertia of all the particles about the axis passing through the centre of mass is zero, at the centre of mass. 

That is, 

2 space sum from straight i space equals space 1 to straight n of space straight m subscript straight i space straight a space straight r subscript straight i space equals space 0 space left square bracket space because space straight a space not equal to space 0 space right square bracket

therefore space sum space straight m subscript straight i space straight r subscript straight i space equals space 0 space

Also comma space
sum from straight i space equals space 1 to straight n of space straight m subscript straight i space equals space straight M thin space

where comma space

straight M space is space the space total space mass space of space the space rigid space body. space

Therefore comma space

straight I subscript QP space equals space straight I subscript RS space plus space Ma squared

Thus the theorem is proved. 




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Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω1 and ω2 are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω1 ≠ ω2.

(a) Moment of inertia of disc I= I

Angular speed of disc I = ω

Moment of inertia of disc II = I

Angular speed of disc II = ω

Angular momentum of disc I, L1 = I1ω

Angular momentum of disc II, L2 = I2ω

Total initial angular momentum Li = I1ω1 + I2ω

When the two discs are joined together, their moments of inertia get added up. 

Moment of inertia of the system of two discs, I = I1 + I

Let ω be the angular speed of the system. 

Total final angular momentum, LT = (I1 + I2ω 

Using the law of conservation of angular momentum, we have

               Li = L

I1ω1 + I2ω2 = (I1 + I2)ω 

∴              ω =fraction numerator straight I subscript 1 straight omega subscript 1 plus space straight I subscript 2 straight omega subscript 2 over denominator straight I subscript 1 space plus space straight I subscript 2 end fraction

(b) Kinetic energy of disc I, E1 = open parentheses 1 half close parentheses I1ω1
Kinetic energy of disc II, E2= open parentheses 1 half close parentheses I2ω2
Total initial kinetic energy, Ei =open parentheses 1 half close parenthesesI1ω12 + I2ω22

When the discs are joined, their moments of inertia get added up. 

Moment of inertia of the system, I = I1 + I

Angular speed of the system = ω 

Final kinetic energy is given by,

      
E= open parentheses 1 half close parenthesesI1 + I2) ω2

          = open parentheses 1 half close parentheses ( I1 + I2open square brackets fraction numerator open parentheses straight I subscript 1 space straight omega subscript 1 space plus space straight I subscript 2 straight omega subscript 2 close parentheses over denominator open parentheses straight I subscript 1 space plus space straight I subscript 2 close parentheses end fraction close square brackets squared 

          =open parentheses 1 half close parentheses  fraction numerator open parentheses straight I subscript 1 space straight omega subscript 1 space plus space straight I subscript 2 straight omega subscript 2 close parentheses squared over denominator open parentheses straight I subscript 1 space plus space straight I subscript 2 close parentheses end fraction 

∴  Ei - Ef 1 half open square brackets fraction numerator straight I subscript 1 straight omega subscript 1 squared space plus straight I subscript 2 straight omega subscript 2 squared space space minus space open parentheses begin display style bevelled 1 half end style close parentheses space open parentheses straight I subscript 1 straight omega subscript 1 space plus space straight I subscript 2 straight omega subscript 2 close parentheses space squared over denominator straight I subscript 1 space plus space straight I subscript 2 end fraction close square brackets
Solving the equation, we get 

          = fraction numerator straight I subscript 1 space straight I subscript 2 space open parentheses straight omega subscript 1 space minus space straight omega subscript 2 close parentheses squared over denominator 2 space left parenthesis straight I subscript 1 space plus space straight I subscript 2 right parenthesis end fraction 

All the quantities on RHS are positive 

Therefore,

 
Ei - Ef > 0 

i.e.,       Ei > E

When the two discs come in contact with each other, there is a frictional force between the two. Hence there would be a loss of Kinetic Energy.
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Prove the result that the velocity of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height is given by v2 = 2gh/ [1 + (k2/R2) ].

Using dynamical consideration (i.e. by consideration of forces and torques). Note is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane. 

A body rolling on an inclined plane of height h,is shown in figure below:




Let,

m
 = Mass of the body 

= Radius of the body 

K = Radius of gyration of the body 

= Translational velocity of the body 

=Height of the inclined plane 

g = Acceleration due to gravity 

Total energy at the top of the plane, E­1= mg

Total energy at the bottom of the plane, Eb = KErot + KEtrans 

                                                        = open parentheses 1 half close parentheses I ω2 +open parentheses 1 half close parentheses mv2
But I = mk2 and ω = v / 

∴ Eb = open parentheses 1 half close parentheses (mk2)open parentheses straight v squared over straight R squared close parentheses + open parentheses 1 half close parenthesesmv2
      = open parentheses 1 half close parenthesesmv2 (1 +straight k squared over straight R squared)
From the law of conservation of energy, we have

                 ET = E

              mgh = open parentheses 1 half close parenthesesmv2 (1 +straight k squared over straight R squared)
∴                v = open parentheses fraction numerator 2 gh over denominator 1 space plus space begin display style straight k squared over straight R squared end style end fraction close parentheses 

Hence, the result. 
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A disc rotating about its axis with angular speed ωois placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. 7.41? Will the disc roll in the direction indicated?


Angular speed of the disc = ω

Radius of the disc = 

Using the relation for linear velocity, v = ωo

For point A: 

vA = Rωo, in the direction tangential to the right.

For point B: 

vB = Rωo, in the direction tangential to the left. 

For point C:

vc = open parentheses straight R over 2 close parenthesesωo, in the direction same as that of vA.

vA = Rωo

vB = Rω0

vc = open parentheses straight R over 2 close parenthesesω0
The disc will not roll.

The directions of motion of points A, B, and C on the disc are shown in the following figure 


Since the disc is placed on a frictionless table, it will not roll. This is because the presence of friction is essential for the rolling of a body.
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