(a) The moment of inertia of the sign for rotation about the side of length h is 1.24 kgm².
(b) The kinetic energy of the sign when it is rotating about an axis is 141.37 J.
<h3>Moment of inertia of the triangular sign</h3>
The moment of inertia of the sign for rotation about the side of length h is calculated as follows;
I = 2 x ¹/₃M(b/2)²
I = 2 x ¹/₃Mb² x ¹/₄
I = ¹/₆Mb²
I = ¹/₆ x 4.4 x 1.3²
I = 1.24 kgm²
<h3>Rotational kinetic energy of the ball</h3>
K.E(rot) = ¹/₂Iω²
where;
- ω is angular speed = 2.4 rev/s = 2.4 x (2π rad)/s = 15.1 rad/s
K.E(rot) = ¹/₂(1.24)(15.1)²
K.E(rot) = 141.37 J
Learn more about rotational kinetic energy here: brainly.com/question/25803184
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OPTION D is the correct answer.
Refer to the attachment for complete calculation...
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lol hope it helps^^ I'm no expert but I'm just giving info
Answer:
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