Q23 of 33 Page 178

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?

Given:

(a) Moment of Inertia of the man-platform system = 7.6 Kgm2


Moment of inertia when the man stretches his hands to a distance of 90 cm.


We know, I = 2 × mr2


Where,


m = mass


r = distance of mass from axis of rotation


Using the above formula, we get,


⇒ I = 2 × 5Kg × (0.9m)2


⇒ I = 8.1 Kgm2


Now, the Initial moment of inertia of the system,


Ii = 7.6 + 8.1 = 15.7 Kgm2


Angular speed, ωi = 30 rev/min


Angular momentum, Li = Iiωi


Li = 15.7Kgm2 × 30rev/min


⇒ Li = 471 Kgm2/sec …(i)


Moment of inertia when the man folds his hands to a distance of 20 cm,


I = 2 × mr2


Where,


m = mass


r = distance of mass from axis of rotation


⇒ I = 2 × 5Kg × (0.2m)2


⇒ I = 0.4 Kgm2


Final moment of inertia, If = 7.6 + 0.4 = 8 Kgm2


Final angular speed = ωf


Final angular momentum = Lf = Ifωf


Lf = 0.79 ωf …(ii)


From the conservation of angular momentum, We have


Iiωi = Ifωf


From equation (i) and (ii),


ωf = 15.7 Kg m2 × ((30rev/sec)/8 Kgm2)


ωf = 58.88 rev/min


(b) Kinetic energy is not conserved in the given process. In fact, with the decrease in moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands towards himself.


More from this chapter

All 33 →
21

A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.

A. How far will the cylinder go up the plane?


B. How long will it take to return to the bottom?

 22

As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m/s2)

(Hint: Consider the equilibrium of each side of the ladder separately.)


 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.

 25

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 .