Controlling a Rolling Ball On a Tilting Plane - STEM2

The kinetic energy of the ball is
T = 1 2 Mv2 + 1 2 Iω2
The potential energy is
where
= 1 2 Mv2 + 1 2 I(v/r)2
= 1 2 (M + I/r2 )v 2
V = hgM,
h = s sin(θ)
is the height of the ball. The Lagrangian is
L = T − V = 1 2 (M + I/r2 )v 2 − s sin(θ)gM.
∂L
∂v =(M + I/r2 )v
d ∂L
dt ∂v =(M + I/r2 ) dv
dt
∂L
∂s = − sin(θ)gM.
The equation of motion is
So
or
d ∂L
dt ∂v − ∂L
∂s =0
(M + I/r 2 ) dv
dt
dv
dt
+sin(θ)gM =0
M
= −g sin(θ)
M + I/r 2
So the effective acceleration has been reduced by the factor
M
M + I/r 2
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