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A solid cylinder attached to a horizontal spring (???? = 3.00 N/m) rolls without slipping along a horizontal surface. If the system is released from rest when the spring is stretched by 0.250 m, find (a) the translational kinetic energy and (b) the rotational kinetic energy of the cylinder as it passes through the equilibrium position. (c) Show that under these conditions the cylinder’s center of mass executes simple harmonic motion with period

User Kino Lucky
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Answer:

a. Ek = 62.5 x 10⁻³ J

b. Ek = 31.25 x 10 ⁻³ J

Step-by-step explanation:

E = [ Us + (M * v²) / 2 ] + [ (I * ω² ) / 2 ]

T = 2π * √ 3M / 2 k , K = 3.0 N / m , d = 0.25m

a.

Ek = ¹/₂ * m * v² = ¹/₃ * k * d²

Ek = ¹/₃ * k * d² = ¹/₃ * 3.0 N / m * 0.25²m

Ek = 62.5 x 10⁻³ J

b.

Ek = ¹/₂ * I * ω² = ¹/₄ * M * v²

Ek = ¹/₆ * k * Xm² = ¹/₆ * 3.0 N / m * 0.25²m

Ek = 31.25 x 10 ⁻³ J

c.

d Emech / dt = d / dt * [ 3 m * v² / 4 + k * x² / 2 ]

acm = - ( 2 k / 3M)

ω = √ 2k / 3m ⇒ T = 2π * √ K / m

User Kreys
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