Q25 of 33 Page 178

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 w1 and w2 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 your account for this loss in energy? Take .

Given:

Moment of inertia of disc 1 = I1


Angular speed of disc 1 = ω1


∴ Angular momentum of disc 1 , L1 = I1ω1


Angular speed of disc 2 = ω2


∴ Angular momentum of disc 2, L2 = I2ω2


⇒ Total initial angular momentum, Li = I1ω1 + I2ω2


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 + I2


Let ω be the angular speed of the system.


Let total final angular momentum, Lf = (I1 + I2) × ωf


Using the law of conservation of angular momentum, we have:


Li = Lf


⇒ I1ω1 + I2ω2 = (I1 + I2) × ωf


⇒ ωf = (I1ω1 + I2ω2) / (I1 + I2) …(i)


(b) Kinetic energy of disc I, E1 = 1/2 I1ω12


Kinetic energy of disc II, E2 = 1/2 I2ω22


⇒ Total initial kinetic energy, Ei = 1/2 I1ω12 + 1/2 I2ω22 …(ii)


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


Moment of inertia of the system, I = I1 + I2


Angular speed of the system = ωf


Final kinetic energy,


Ef = 1/2 (I1 + I2)ωf2


Substituting the value of ω from equation (i)


⇒ Ef = 1/2 (I1 + I2) {(I1ω1 + I2ω2) / (I1 + I2)}2 …(iii)


From (ii) and (iii)


Ei- Ef = [1/2 I1ω12 + 1/2 I2ω22] - 1/2 (I1 + I2) {(I1ω1 + I2ω2) / (I1 + I2)}2


By solving the equation we get,


Ei- Ef = I1I2(ω1-ω2)2/ 2(I1 + I2)


Since all Quantities on RHS are positive so we conclude that,


Ei-Ef> 0


i.e. Ei> Ef


The loss can be attributed to the frictional forces that arise when discs come into contact with each other.


More from this chapter

All 33 →
23

A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m2.

A. What is his new angular speed? (Neglect friction.)


B. Is kinetic energy conserved in the process? If not, from where does the change come about?

 24

A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.

 26

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 and perpendicular to the plane is x2 + y2).


B. Prove the theorem of parallel axes.


(Hint: If the centre of mass is chosen to be the origin ).

27

Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by


using dynamical consideration (i.e. by consideration of forces and torques). Note k 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.