Question

A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 3.4 m and moment of inertia 940 kg⋅m2 . The platform rotates without friction with angular velocity 1.6 rad/s . The person walks radially to the edge of the platform.

Calculate the angular velocity when the person reaches the edge. In rad/sec

Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk. In J.

Answer 1

a) The moment of inertia of the person is, person = mR2 Here m is the mass of the person, and R is the radius of the platfom. From law of conservation of angular momentum we have, plapormperson Solve the equation for co, lplatfromd0i =(lplatform+/person)of platfrom person platfrom playom +mR2 940kg m (1.60rad/s) 940kg m+(70kg) (340m) -0.859 rad/s

b) The rotational kinetic energy of the system of platform plus person before person's walk ー(940kg·㎡)(1.60 rad/s)2 -1203.2J The rotational kinetic energy of the system of platform plus person after person's walk is, plaformpercon 940kg m (70kg) (3.40m) (0.859rad/s) 2 -751.2814.J

Answer 2

SOLUTION  :

a.

M. I. of  platform = 940 kg-m^2

M. I. of the person when he stands at the centre initially = m r^2 = 70 * 0^2 = 0  kg0-m^2

M. I. of the person when he stands at the edge finally = m r^2 = 70 * 3.4^2 = 809.20 kg-m^2 

Angular velocity of platform initially = 1.6 rad/s 

Angular momentum remains constant whether the person is at centre or at the edge.

So,

Angular momentum initially = Angular momentum finally

M. I. Initially * angular velocity initially = M. I. finally * angular velocity finally

=> (940 + 0) * 1.6 = (940 + 809.20) * angular velocity finally

=> angular velocity finally = (940 * 1.6) / (1749.20) 

=> angular velocity finally = 0.8598 = 0.86  rad/s (ANSWER) 

b.

Rotational K.E. before person walks to the edge

= 1/2 M.I. initially * (angular velocity initially)^2

= 1/2 * 940 * 1.6^2 

= 1203.2 J (ANSWER).

Rotational K.E. after person walks and reaches the edge

= 1/2 M.I. finally * (angular velocity finally)^2

= 1/2 * 1749.20 * 0.8598 

= 751.98 J  (ANSWER).