A solid cylinder of mass M = 7.3 kg is attached to a horizontal massless spring

Question

A solid cylinder of mass M = 7.3 kg is attached to a horizontal massless spring so that it can roll without slipping along a horizontal surface, as shown in the Figure. The force constant of the spring is k = 132 N/m. The system is released from rest at a position in which the spring is stretched by a distance d = 12.2 cm. What is the translational kinetic energy of the cylinder when it passes through the equilibrium position?

What is the rotational kinetic energy of the cylinder when it passes through the equilibrium position?

Under these conditions the center of mass of the cylinder executes a simple harmonic motion. What is the period of this motion?

Explanation / Answer

Given that m = 7.3 kg d = 12.2 cm = 0.122 m k = 132 N/m ------------------------------------------------------------------------------------------------- The total energy E = KEtran + KErot + PE = [ 0.5 m v^2 + 0.5 I w^2 ] + 0.5 k d^2 We know that I = 0.5 m r^2 and v = w r Therefore E = [ 0.5 m v^2 + 0.5 *(0.5 m r^2)* (v / r)^2 ] + 0.5 k d^2 = 3mv^2 / 4 + 0.5 kd^2 Now dE/dt = 0 ==> dv/dt = -2kd / 3m So that w = [2k / 3m]^1/2 Therefore T = 2 * pi / w = 2 * pi * [ 3m/2k]^1/2 Substitute values.

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