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As part of a carnival game, a mb=0.628 kg ball is thrown at a stack of 15.3 cm tall, mo=0.423 kg objects and hits with a perfectly horizontal velocity of ????b,i=12.1 m/s. Suppose the ball strikes the topmost object. Immediately after the collision, the ball has a horizontal velocity of ????b,f=5.10 m/s in the same direction, the topmost object has an angular velocity of ????o=2.03 rad/s about its center of mass, and all the remaining objects are undisturbed. If the object's center of mass is located ????=10.7 cm below the point where the ball hits, what is the moment of inertia ????o of the object about its center of mass?

Respuesta :

Answer:

[tex]I = 0.23 kg m^2[/tex]

Explanation:

As we know that there is no external torque on the system

So here we can say that the angular momentum of the system will remain conserved

So we can say

[tex]m_b v_i L = I\omega + m_b v_f L[/tex]

here we know that

[tex]m_b = 0.628 kg[/tex]

[tex]v_i = 12.1 m/s[/tex]

[tex]v_f = 5.10 m/s[/tex]

[tex]\omega = 2.03 rad/s[/tex]

L = 10.7 cm = 0.107 m

now we will have

[tex]0.628(12.1)(0.107) = I(2.03) + (0.628)(5.10)(0.107)[/tex]

[tex]I = 0.23 kg m^2[/tex]