Get Answers to all your Questions

header-bg qa

A person of 80 \; kg  mass is standing on a circular platform of mass 200 \; kg rotating about its axis at 5 rpm (revolutions per minute). The person now starts moving towards the center of the platform. What will be the rotational speed in rpm of the platform when the person reaches its center.
Option: 1 3
Option: 2 6
Option: 3 9
Option: 4 12

Answers (1)

best_answer

Applying conservation of angular momentum - 

 

I_1 N_1 = I_2 N_2

Now

I_1 = M_1R^2 + \frac{M_2R^2}{2} = {80 \times R^2} + \frac{200 \times R^2}{2}

N_{1}=5 \; rpm

When the person reaches center - 

I_2 = \frac{M_2R^2}{2} = \frac{200 \times R^2}{2}

So,

I_1 N_1 = I_2 N_2

\frac{360 \times R^2}{2} \times 5 = \frac{200 \times R^2}{2} \times N_2

N_2 = 9 \ \ rpm

 

 

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE