Question

PART A

Q14 (10 points): If the rotational inertia about an axis through the midpoint of one end of the slab, axis 1 in the below figure, is 4.7x104 kg.m2, what is its rotational inertia about the axis through its center of mass, axis 2? Mass of the slab is 0.16 kg. (Hint: Use parallel-axis theorem.) Answer: I-3.26x I0 kgm2 a 2 cm -бст

PART B

PART C

Answer 1

I_1 = 4.7 times 10^-4 kg m^2 I_2 = ? I_2 + [m (b/2)^2] = I_1 I_2 + [0.16 times 9 times 10^-4] = 4.7 times 10^-4 I_2 = [4.7 - 1.44] times 10^-4 I_2 = 3.25 times 10^-4 kg m^2 c vector = r vector times F vector tau_net = (r_1 vector times F_1 vector) + (r_2 vector times F_2 vector) r_2 vector = -0.1 i^^- 0.2 j^^m F_2 vector = -12 j^^r_1 vector = 0.2 i^^+ 0.1 j^^F_1 = -8 i^^T_net = [0.2i^^+ 0.1 j^^] times [-8i^^] + [-0.1 i^^- 0.2 j^^] times [12 j^^] = 0.8 k^^+ [-1.2 k^^] tau = -0.4 k^^ \r\n tau_net = [1.5 times 7 - k^^] + [1.5 times 9 k^^] = (13.5 - 10.5) k^^tau_net = 3 k^^