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?. (a) What is the moment of inertia of the sign for rotation about the side of length ?h?? (b) If ?M? = 5.40 kg, ?b? = 1.60 m, and ?h? = 1.20 m, what is the kinetic energy of the sign when it is rotating about an axis along the 1.20-m side at 2.00 rev/s?

Solution 79P Let us have a look at the following figure. Let ABC be the right triangle. (a) Moment of inertia of AB = Mb /12 2 2 Moment of inertia of BC = Mb /12 Moment of inertia of AC= 0 Therefore, total moment of inertia I = Mb /12 + Mb /12 = Mb /6 2 (b) Now, angular speed = 2.00 rev/s = 2.00 × 2 rad/s = 4.0 rad/s Give, mass M = 5.40 kg , b = 1.60 m 1 2 Kinetic energy about the axis of rotation K = I 2 2 K = 1 × Mb × 2 2 6 Substituting M , b and values in this equation, we get 5.40 kg×(1.60) m K = 1 × × (4.0) J 2 6 K = 182 J So, the kinetic energy of the sign when it is rotating about the given axis is 182 J.