Question

A playground merry-go-round of radius $R=2m$ has a moment of inertia $I=250kg{m}^2$ and is rotating at $10rev/min$ about a frictionless, vertical axle. Facing the axle, a $25kg$ child hops onto the merry-go-round and manages to sit down on the edge. The new angular speed of the merry-go-round is

• A. $10rev/min$
• B. $7.14rev/min$
• C. $3.14rev/min$
• D. $8.45rev/min$

Answer 1

The correct option is B $7.14rev/min$

Given: $R=2m,I=250kg{m}^{{}^2},\omega =10rev/min,m=25kg$

Initially merry-go-round was rotating with constant angular speed then as boy jumped on the round moment of inertia of system changed and then to keep the angular momentum constant round decreased its speed

Moment of inertia of boy about centre will be $=m{R}^2$

From conservation of angular momentum,

$I_i{\omega }_i=I_f{\omega }_f$

$250\times 10=[250+(25\left)\right(2{)}^2]{\omega }_f$

${\omega }_f=7.14rev/min$