Question

A girl of mass 33.6 kg stands on the rim of a frictionless merry-go-round of radius 1.75 m and rotational inertia 453 kg·m2 that is not moving. She throws a rock of mass 824 g horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is 3.92 m/s. Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the girl?

Answer 1

a) 0.15rad/s

b) 0.26m/s

Step-by-step explanation:

a) To find the angular speed it is necessary to consider the conservation of energy. after the rock is dropped by the girl, the kinetic energy of the rock must equal the rotational energy of the merry-go-round joint the girl. In other words you have

`E_(ir)=E_(fk)\\`

rotational energy and kinetic energy will given by:

`E_(ir)=(1)/(2)I\omega^2\\\\E_(fk)=(1)/(2)mv^2`

`I=I_(merry-go-round)+I_(girl)=453kgm^2+m_gR^2`

I: rotational inertia

w = angular speed

m: mass of the block

m_g: mass of the girl

R: radius of the merry-go-round

By replacing you have:

`I\omega^2=mv^2\\\\(453kgm^2+(33.6kg)(1.75m)^2)\omega^2=(0.824kg)(3.92m/s)^2\\\\\omega=√(0.022)rad/s=0.15rad/s`

the angular speed is 0.15rad/s

b) the linear speed is given by:

`\omega=(v)/(r)\\\\v=\omega r=(0.15rad/s)(1.75m)=0.26m/s`

hence, the linear speed of the girl is 0.26m/s