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Sagot :
(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.
Moment of inertia of the triangular sign
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
Rotational kinetic energy of the ball
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: https://brainly.com/question/25803184
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