The 3.00 kg cube has edge lengths d = 6.00 cm and is mounted on an axle through
ID: 2216458 • Letter: T
Question
The 3.00 kg cube has edge lengths d = 6.00 cm and is mounted on an axle through its center. A spring ( k = 1200 n/m) connects the cube's upper corner to a rigid wall. Initially the spring is at its rest length. If the cube is rotated 3 degrees and released, what is the period of the resulting SMH?
The 3.00 kg cube has edge lengths d = 6.00 cm and is mounted on an axle through its center. A spring ( k = 1200 n/m) connects the cube's upper corner to a rigid wall. Initially the spring is at its rest length. If the cube is rotated 3 degrees and released, what is the period of the resulting SMH?Explanation / Answer
The mass moment of inertia is given by J = 1/6*d^2 The torque from the spring will be T = -d/sqrt(2)*sin(theta) * k because d/sqrt(2)*sin(theta) will be the amount the spring lengthens as the square rotates by an angle theta. J*theta_dd = -d/sqrt(2)*sin(theta)*k for small rotations, sin(theta) ~theta (with theta in radians!) J*theta_dd + 1/sqrt(2)*d*k*theta = 0 so the natural frequency for the linear model (assuming small rotations) is w_nat = sqrt[ (d*k/sqrt(2)) / J) d*k/sqrt(2) = 0.06*1200/ sqrt(2) =50.91 J = 1/6*m*d^2 = 1/6 * 3 * 0.06 * 0.06 = 0.0018 w_nat = sqrt(50.91/.0018) = 168.17 rad/sec f_nat = w_nat/2/pi = 26.78 Hz T = 1/f_nat = 0.037
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