height = y so U = potential energy = M g y
so we need y as a function of x
(R-y)^2 = R^2-x^2
R-y = +/- (R^2-x^2)^.5
y = R +/- (R^2-x^2)^.5
for x < R and bottom half of the circle
y = R - (R^2-x^2)^.5
U = M g [ R - (R^2-x^2)^.5 ]
F = -m g dU/dx
F = -Mg d/dx [ R - (R^2-x^2)^.5 ]
= -Mg[- (R^2-x^2)^.5 ]
= Mg .5(R^2-x^2)^-.5 (-2 x)
= - Mg x /sqrt(R^2-x^2)
for x<<R
F = -Mg x/R which is a lot like a mass on a spring (F = -k x)
m a = -Mg x/R
if x = A sin wt
a = -w^2 x
M (-w^2 x ) = -Mg x/R
w^2 = (2 pi f)^2 = g/R
2 pi f = sqrt (g/R)
f = (1/2pi) sqrt (g/R)
T = 1/f = 2 pi sqrt (R/g)
y = R
An ice cube of mass M is placed at rest in on a circular track of radius R, as shown at right. The position of the ice cube along the track follows that of a circle, given by: x^2 + (R – y)^2 = R^2 .
A)With respect to the bottom of the track, calculate the potential energy of the ice cube in terms of as a function of x.
B)Ignoring friction, the force governing the motion of the ice cube is conservative (it is part of gravity). Find the force acting on the ice cube at any position x. (Hint: recall relation between a conservative force and its potential function).
C)Show that for special case of x << R, the ice cube will execute simple harmonic motion. What is the restoring force?
D)What is the period of small oscillations of the ice cube as it rocks back and forth near the bottom of the bowl?
1 answer