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A metal sign for a car dealership is a thin, uniform right triangle with base length b and height h. the sign has mass m.

required:
a. what is the moment of inertia of the sign for rotation about the side of length h?
b. if m = 4.40 kg ,b = 1.30 m and h = 1.30 m , what is the kinetic energy of the sign when it is rotating about an axis along the 1.30-m side at 2.40 rev/s ?

Respuesta :

onyebuchinnaji

(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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