Unit 6 - SHM

A disk with a mass M, a radius R, and a rotational inertia of I = 1⁄2 MR^2 is attached to a horizontal spring which has a spring constant of k as shown
in the diagram. When the spring is stretched by a distance x and then released from rest, the disk rolls without slipping while the spring is attached to the frictionless axle within the center of the disk.


(a) Calculate the maximum translational velocity of the disk in terms of M,R, x, k.

(b) What would happen to the period of this motion if the spring constant of the spring increased? Justify your

© What would happen to the period of this motion if the surface was now frictionless and the disk was not allowed to roll? Justify your answer.

(a) Conservation of energy
For A i don’t know if kinetic energy should be added to Ke rot
(b) If you increase the spring constant then the period will decrease because in T=2pisq(m/K) the spring constant in inversely related to the period so if you incease K, then T will decrease.
© If the surface was now frictionless and the disk was not allowed to roll then all the PEE would turn into heat so the period would be smaller because there is a greater velcoity now since there is no rotational velocity so the period will be smaller

A. PEs=Krot+Ktrans. 1/2kx^2=1/2mv^2+1/2Iw^2. V=squareroot kx^2-Iw^2/m.

B. Period = 2pi *square root m/k. If k increase then the period would decrease, k is inversely proportional to period of mass spring system.

C. If the disk couldn’t roll, then there would be no Krot, thus all of the elastic potential energy would be translated to Ktrans. Thus the k would increase, as by conservation of mechanical energy, PEs=Ktrans, 1/2kx^2=1/2mv^2, and k is equal to mv^2/x^2. the mass and distance from equilibrium would still be the same, but now the velocity would increase, thus making k increase for v is in the numerator and is directly proportional. The period of a mass spring system is 2pi times square root m/k, so as k increases, the period will decrease, inversely proportional.

a) Conservation of Energy at equilibrium point
PEs = Ktrans + Krot <-- 2 types of KE
1/2kx^2 = 1/2Mv^2 + 1/2Iw^2
kx^2 = Mv^2 + 1/2MR^2(v/R)^2
kx^2 = Mv^2 + 1/2Mv^2
kx^2 = 3/2Mv^2
v = sqrt(2kx^2/3M)

b) The period of oscillation on a spring is described by the equation: Tsp = 2pi sqrt(m/k). A larger k value will result in a smaller period since k is in the denominator of the fraction. Also the derivation in part a shows that the velocity of the sphere is proportional to k, a larger k results in a greater v which gives a shorter time to travel the distance of 4*amplitude

c) The period would also decrease when the sphere is on a frictionless surface. No friction means no rolling (since friction provides the torque needed to rotate) which means that all the energy from the spring goes into translational kinetic energy. This results in the sphere moving faster which reduces the period.

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