A solid cylinder attached to a horizontal spring (k = 1.20 N/m) rolls without sl

Question

A solid cylinder attached to a horizontal spring (k = 1.20 N/m) rolls without slipping along a horizontal surface. If the system is released from rest when the spring is stretched by 0.130 m, find (a) the translational kinetic energy and (b) the rotational kinetic energy of the cylinder as it passes through the equilibrium position. (c) If the cylinder has mass M = 6.50 kg, find the period of the simple harmonic motion that the cylinder's center of mass executes. (Hint: Find the time derivative of the total mechanical energy.)

Explanation / Answer

Tough: PE.initial = KE.t + KE.r 1/2 k (0.145m)^2 = 1/2 m v^2 + 1/2 I w^2 w= v/r I = 1/2 m r^2 KE.r = 1/4 mv^2 = 1/2 KE.t This means 1/3 energy rotational, 2/3 translational PE. i = 1/2 k (0.145)^2 PE.i = 4.478 J a. KE.t = 2.986 J b. KE.r = 1.493 J c. Because the aceleration will be slower, this system will act as if the mass is 50% greater. P = 2 pi sqrt (1.5m/k) P = 0.84 s

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