Question

A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.79 rad/s . Its total moment of inertia is 1790 kg⋅m2 . Four people standing on the ground, each of mass 70 kg , suddenly step onto the edge of the merry-go-round.

Answer 1

Answer:

$w_2=0.467rad/s$

Explanation:

Four people standing on the ground each of mass and usually this questions have to find the final angular velocity

$m_t=4*70kg=280kg$

The radius $r=4.2/2=2.1m$

Angular velocity  $w_1=0.79rad/s$

The moment of inertia total is $I_t=1790 kg/m^2$

Momento if inertia

$I_1=m_t*r^2$

$I_1=280kg*(2.1m)^2=1234.8kg*m^2$

Angular momentum

$I_1*w_1=I_t*w_2$

Solve to w2

$w_2=\frac{I_1*w_1}{I_t}$

$w_2=\frac{1790kg*m^2*0.79rad/s}{3024.8kg*m^2}$

$w_2=0.467rad/s$