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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

         =  

I
x + Iy = I

Hence, the theorem is proved.
 
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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 = ωo, in the direction same as that of vA.

vA = Rωo

vB = Rω0

vc = ω0
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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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 

                                                        =  I ω2 + mv2
But I = mk2 and ω = v / 

∴ Eb =  (mk2) + mv2
      = mv2 (1 +)
From the law of conservation of energy, we have

                 ET = E

              mgh = mv2 (1 +)
∴                v =  

Hence, the result. 

238 Views

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, 




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, 



Thus the theorem is proved. 

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Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.

(b) What is the force of friction after perfect rolling begins?


A torque is required to roll the given disc. According to the definition of torque, the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is required for making the disc roll.

(a) Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward. The sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.

(b) Frictional force acts opposite to the direction of velocity at point B. Therefore, perfect rolling will begin when the velocity at that point becomes equal to zero. This will make the frictional force acting on the disc zero.
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