Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
283 views
in Physics by (91.8k points)
closed by
A man stands on a rotating platform, with his arms stretched horizontal 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 `90 cm` to `20 cm`. moment of inertia of the man together with the platform may be taken to be constant and equal t `7.6 kg m^(2)`. (a) What is the his new angular speed ? (Neglect friction.)
(b) Is kinetic energy conserved in the process ? If not, from where does the change come about ?

1 Answer

0 votes
by (91.2k points)
selected by
 
Best answer
Here, `l_(1)=7.6+2xx5(0.9)^(2)=15.7 kg m^(2)`
`w_(1)=30 rpm`
`l_(2)=7.6+2xx5(0.2)2=8.0 kg m^(2)`
`w_(2)=?`
According to the principle of conservation of angular momentum, `l_(2)w_(2)=l_(1)w_(1)`
No, kinetic energy is not conserved in the process. In fact, as moment of inertia decreases, K.E. of rotation increases. This change comes about as work is done by the man in bringing his arms closer to his body.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...