Answer to Question #177273 in Mechanics | Relativity for Abbas

Question

A 25.0-kg boy stands 2.00 m from the center of a frictionless playground merry-goround, which has a moment of inertia of 200. . The boy begins to run in a circular path with a speed of 0.600 ⁄ relative to the ground.

Answer 1

Solution.

$m_b=25.0kg;$

$R=2.00m;$

$I=200kgm^2;$

$v_b=0.6m/s;$

$\omega-?;$

$v_{1b}-?;$

$I=\dfrac{1}{2}mR^2\implies m=\dfrac{2I}{R^2}$ ;

$m=\dfrac{2\sdot 200kgm^2}{4.00m^2}=100kg;$

$m_bv_b=mv\implies v=\dfrac{m_bv_b}{m};$

$v=\dfrac{25.0kg\sdot 0.6m/s}{100kg}=0.15m/s;$

$\omega =\dfrac{v}{R};$

$\omega=\dfrac{0.15 m/s}{2m}=0.075s^{-1};$

$v_{1b}=v_b-v;$

$v_{1b}=0.6m/s-0.15m/s=0.45m/s;$

Answer: $\omega=0.075s^{-1};$ $v_{1b}=0.45m/s.$