As shown in the figure below, the axle of a solid cylindrical wheel is attached
ID: 1772668 • Letter: A
Question
As shown in the figure below, the axle of a solid cylindrical wheel is attached to two identical springs with spring constant k and negligible mass. The wheel with radius R and total mass M rolls without slipping. The center of mass of the wheel oscillates in simple harmonic motion in the horizontal direction about its equilibrium point x 0 (a) Find an expression for the total energy in terms of k, M, R, and x(t). You may assume that the moment of inertia for rotation of the wheel about its axle is MR2. b) What is the angular frequency of small oscillations about equilibrium?Explanation / Answer
given solid cylindrical wheel, of mass M, radius R
it rolls without slipping
spring constant of the springs = k
a. so at time t, let the mass be displaced by x in horizontal direction
PE stored in springs, PE = 0.5kx^2*2 = kx^2
translational speed = v
so KE of mass = 0.5Mv^2
and rotational angular velocity about COM = w
then Rotational KE = 0.5Iw^2
here I is moment of inertia of the object = 0.5MR^2
but v = w*R
so, Rotational KE = 0.5*0.5MR^2*v^2/R^2 = 0.25Mv^2
hence total energy
E = kx^2 + 0.75Mv^2 [ where v = dx/dt]
b. angular frequency of small osscilations be w
then
E = kx^2 + 0.75Mv^2
but for SHM
dE/dt = 0
so,
2kx*dx/dt + 1.5Mv*dv/dt = 0
dx/dt = v
so, 2kxv + 1.5Mv*dv/dt = 0
2kx = -1.5Ma [ dv/dt = a]
so w = sqroot(2k/1.5M) = sqroot(4k/3M)
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