A merry-go-round is at rest, pivoted on a frictionless axis. A child of mass m runs along the path tangential to the rim with speed v and jumps on to merry-go-round. If R is the radius of the merry-go-round and I is the moment of inertia, then the angular velocity of the merry-go-round and the child is

Question

A merry-go-round is at rest, pivoted on a frictionless axis. A child of mass m runs along the path tangential to the rim with speed v and jumps on to merry-go-round. If R is the radius of the merry-go-round and I is the moment of inertia, then the angular velocity of the merry-go-round and the child is (A) (m v R/m R2 + I) (B) (m v R/I) (C) (m R2 + I/m v R) (D) (I/m v R)

Tardigrade Admin · Accepted Answer

By law of conservation of angular momentum textitmvR=(( textitI) textitsystem) textitÏ ⇒ textitmvR=( textitI + ( textitmR)2) textitÏ ⇒ textitÏ=( textitmvR/ textitI + textitmR2)

$\frac{m v R}{m R ^{2} + I}$

$\frac{m v R}{I}$

$\frac{m R ^{2} + I}{m v R}$

$\frac{I}{m v R}$

By law of conservation of angular momentum
$mvR = ((I)_{system}) ω$
$\Rightarrow mvR = (I + (mR)^{2}) ω$
$\Rightarrow ω = \frac{mvR}{I + mR ^{2}}$

mR2+ImvR​

ImvR​

mvRmR2+I​

mvRI​

By law of conservation of angular momentum mvR=((I)system​)ω ⇒mvR=(I+(mR)2)ω ⇒ω=I+mR2mvR​

$\frac{m v R}{m R ^{2} + I}$

$\frac{m v R}{I}$

$\frac{m R ^{2} + I}{m v R}$

$\frac{I}{m v R}$

By law of conservation of angular momentum
$mvR = ((I)_{system}) ω$
$\Rightarrow mvR = (I + (mR)^{2}) ω$
$\Rightarrow ω = \frac{mvR}{I + mR ^{2}}$