Four identical particles (mass of each = 0.24 kg) are placed at the vertices of

Question

Four identical particles (mass of each = 0.24 kg) are placed at the vertices of a rectangle (2.0 m ´ 3.0 m) and held in those positions by four light rods which form the sides of the rectangle. What is the moment of inertia of this rigid body about an axis that passes through the center of mass of the body and is parallel to the shorter sides of the rectangle?
Answer
2.4 kg × m2
2.2 kg × m2
1.9 kg × m2
2.7 kg × m2
8.6 kg × m2

A 1.2-kg mass is projected up a rough circular track (radius = 0.80 m) as shown. The speed of the mass at point A is 8.4 m/s, and at point B, it is 5.6 m/s. How much work is done on the mass between A and B by the force of friction?

-2.7 J -8.8 J -4.7 J -6.7 J -19 J

Explanation / Answer

Mass of each particle is m = 0.24 kg Area of the rectangle is A = 2.0 m *3.0m Each particle distance is d = w / 2                                                                                    = 3.0 m / 2                                             = 1.5 m The moment of inertia of the each particle is I = m r^2                                                                          = 0.24 kg ( 1.5m)^2                                                                          = 0.54 kg m^2 For Four identical particles is I = 4 * 0.54 kgm^2                                                   ˜ 2.2 kg m^ 2 b) Given that mass m = 1.2 kg radius r = 0.80 m    Work done is W = P.E + K.E                                = mg h + 1/2 m ( v_f^2 - v_i^2 )               = 1.2 kg * 9.8 m/s^2 * 0.80 m + 1/2 * 1.2 kg ( 8.4^2 ) - 1/2 * 1.2 *5.6 ^2 m/s )               = - 14.1 J

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