Question

Solve the previous problem if the pulley has a moment of inertia I about its axis and the string does not slip over it.

Answer 1

Refer  attachment Pulley 1.1 for answer

Explanation:

Using energy method, let us solve the problem.

The displacement is shown as δ, which is from the mean position. Then, the spring’s initial extension from the mean position is shown as δ =  mgk.

During oscillation, x be any position under the equilibrium.

Let us consider that v is the velocity of angular velocity ω, mass m, and r be the radius of the pulley. Then, v = rω.

For simple harmonic motion, the total energy stays constant. This can be written as

+12Iω2 + 12Mv2 + 12k(x+δ)2-δ2-Mgx = Constant

⇒12Mv2 + 12kx2 +12Iv2r2 = Constant      

Therefore, δ = Mgk

v = dxdt and a = dvdtaM + Ir2 = -kx        

⇒T = 2πM + Ir2k

Answer 2

Answer:

Explanation:

Using energy method, let us solve the problem.

The displacement is shown as δ, which is from the mean position. Then, the spring’s initial extension from the mean position is shown as δ =  mgk.

During oscillation, x be any position under the equilibrium.

Let us consider that v is the velocity of angular velocity ω, mass m, and r be the radius of the pulley. Then, v = rω.

For simple harmonic motion, the total energy stays constant. This can be written as

+12Iω2 + 12Mv2 + 12k(x+δ)2-δ2-Mgx = Constant

⇒12Mv2 + 12kx2 +12Iv2r2 = Constant      

Therefore, δ = Mgk

v = dxdt and a = dvdtaM + Ir2 = -kx        

⇒T = 2πM + Ir2k

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