(10) [7 points] You construct a desktop toy out of two uniform rods, each of mass m and length L. You join them together at a right angle to form an L-shaped object. You balance the object on a sharp point. When you displace the object, it wobbles back and forth. (a) Find a symbolic expression for the wobbling angular frequency, a). (b) You would like to design your toy to have a period of 0.75 seconds. You make a prototype out of two drinking straws cut to the appropriate lengthto what length do you cut? (a) To find the wobbling angular frequency, we can use the formula for the moment of inertia of a uniform rod rotating about its center: 1= (1/12) mL^2 Since we have two rods, one horizontal and one vertical, the total moment of inertia of the object can be found by summing the individual moments of inertia: I_total = |_horizontal + |_vertical = (1/12) mL 2+ (1/12) mL*2 = (1/6) mL^2 Next, we can use the fact that the object is balanced on a sharp point to model it as a simple harmonic oscillator. The restoring torque acting on the object is proportional to the angular displacement from equilibrium, and can be found using the gravitational potential energy of the object when it is displaced: U = mgh = mg(L/2)sin(theta) where h is the height of the center of mass above the point of support, and theta is the angle of displacement from equilibrium. Taking the derivative of U with respect to theta gives the torque: T=-dU/dtheta = -mg(L/2)cos(theta) Using the equation for the torque of a simple harmonic oscillator: (=-I_total * d*2(theta)/dt*2We can set these two expressions equal to each other and solve for the wobbling angular frequency, c: -I_total * d*2(theta)/dt*2=-mg(L/2)cos(theta) d*2(theta)/dt*2 + (mgL/21_total)cos(theta) = 0 c = sqrt(gL/21_total) Substituting in the expression for |_total, we get: c = sqrt(3g/2L)(b) We want the period of oscillation to be 0.75 seconds. The period of a simple harmonic oscillator is given by: T = 2*pi/c Substituting in the expression for c that we found in part (a), we get: T =2*pi/sqrt(3g/2L) = 0.75 seconds Solving for L, we get: L = (2pisqrt(3g)*T^2)/3 Substituting in g =9.81 m/s^2 and T =0.75 seconds, we get: L =0.492 meters So the length of each rod should be approximately 49.2 cm