SHM motion - Rolling Cylinder A solid cylinder of mass M = 6.6 kg is attached to

Question

SHM motion - Rolling Cylinder

A solid cylinder of mass M = 6.6 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 = 486 N/m. The system is released from rest at a position in which the spring is stretched by a distance d = 10.7 cm.

A) What is the translational kinetic energy of the cylinder when it passes through the equilibrium position?

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

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

Explanation / Answer

k =486 N/m , M =6.6 kg , d =10.7 cm =0.107 m

Initially the cylinder is at rest and therefore the total mechanical energy of the system is equal to the potential energy stored in the spring,

Etot = (1/2)kd^2 = (1/2)(486)(0.107*0.107) = 2.782 J

When the cylinder passes through its equilibrium position, the potential energy stored in the spring vanishes, and therefore this total mechanical energy is all in the form of kinetic energy, both translational and rotational. In terms of the speed of the cylinder v, its mass, and its moment of inertia, I, and angular speed w, we have that

Etot = (1/2)Mv^2+(1/2)Iw^2

For solid cylinder moment of inertia I = (1/2)MR^2

For without slipping condition v=Rw

Etot = (1/2)Mv^2 +(1/2)(1/2)MR2w2 =(3/4)Mv^2

Etot = (3/4)Mv^2 = 2.782

(1/2)Mv^2 = (2/3)(2.782) =1.855 J

Translational kinetic energy = 1.855 J

(b) Rotational kinetic energy = 2.782 -1.855 = 0.927 J

(c) T =2pi[3M/2k]^1/2

T = (2*3.14) [ (3*6.6)/(2*486)]1/2

T =0.896 s

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