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A woman of mass 55 kg stands on the rim of a frictionless merry-go-round of radius 2.0m and rotational inertia 1250 kg-m2 that is not moving. She throws a rock of mass 350g 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 2.0m/s. Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the woman

Sagot :

Answer:

a)   [tex]\omega=9.10*10^{-4}rad/sec[/tex]

b)   [tex]V=1.8217*10^{-3}[/tex]

Explanation:

From the question we are told that:

Mass of woman [tex]M_w=55kg[/tex]

Radius of merry go round [tex]r=2.0m[/tex]

Rotational inertia [tex]i= 1250 kg-m2[/tex]

Mass of rock [tex]M_r=350g \approx 0.350kg[/tex]

Speed of rock [tex]V_r=2.0m/s[/tex]

Tangent angle to the outer edge [tex]\theta=1[/tex]

a)

Generally the equation for conservation of momentum is mathematically given by

[tex]M_r(ucos\theta)r=(I+M_wr^2)\omega[/tex]

[tex]0.350(2.0cos1)(2.0)=(1250+(55)(2.0)^2)\omega[/tex]

[tex]1.3998=1470\omega\\\omega=\frac{1.339}{1470}[/tex]

[tex]\omega=9.10*10^{-4}rad/sec[/tex]

b)

Generally the equation for linear speed V is mathematically given by

[tex]V=r\omega\\V=2.0*9.10*10^{-4}[/tex]

[tex]V=1.8217*10^{-3}[/tex]