A solid cylinder of mass M is placed on a table, on which it can roll without sl
ID: 1837449 • Letter: A
Question
A solid cylinder of mass M is placed on a table, on which it can roll without slipping. A spring is attached to its center. The other end of the spring is tethered to a wall. The force constant of the spring is k. The spring is stretched and released, and the system undergoes oscillatory motion.
a) What is the rotational kinetic energy when the translational velocity is v?
b) Write the total energy E as a sum of kx2/2, the translational and the rotational kinetic energy (which are proportional to each other). Using the fact that dE/dt = 0, obtain the equation of motion for the cylinder and the frequency of oscillation.
Explanation / Answer
a) Rotational KE when translational velocity is v = 0.5*I*w^2
but in pure rolling, v = w*r
and I = 0.5mr^2 ( for cylinder)
so, RKE = 0.25mv^2
b) Net energy = Spring PE + KE + PE = 0.5kx^2 + 0.5mv^2 + 0.25mv^2 = 0.5kx^2 + 0.75mv^2
c) nOW, dE/dt = 0
0.5k*2x*v + 0.75*m*2v*a = 0
kxv + 1.5mva = 0
kx = -1.5ma .. (1)
from torque balance on the cylinder
restoring torque = f*r
where kx - f = ma
f = -ma + kx
so, (-ma + kx)r = 0.5mr^2*alpha
for pure rolling, alpha = a/r
so, (-ma + kx)r = 0.5mr*a
(-ma + kx) = 0.5ma
kx = 1.5 ma
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