Not sure where to go with this problem, any help is appreciated!
A bike is turned upside down for repairs. The front wheel can be described as a hoop with
radius R=0.6 m and mass M=1.0 kg and a small point mass m=0.005 kg attached to the
perimeter to represent the valve for inflating the tire (see Fig. 1). We consider rotation of
the wheel about its frictionless axle.
(a) Describe the two equilibrium positions for the wheel and identify the one of these that
is a stable equilibrium.
(b) Calculate the period for small angle oscillations of the wheel about the stable
equilibrium.
Thanks in advance!
LOL
valve straight down is stable (pendulum)
valve straight up is unstable (balancing pencil on nose)
I = mr^2 = 1.005(.36) (about .36)
for angle x in radians, restoring moment is about .005*9.81 x r
= .003*9.81 x
oh well call g = 10 instead of 9.81
= .03 x
.03x = -I alpha = -I omega^2 x
omega^2 = .03/.36 = 3/36
omega = (1/6) sqrt 3 = 2 pi/T
T = 12 pi/sqrt 3
check my arithmetic, use calculator (I did in head)
To solve this problem, we can use the principles of rotational equilibrium and simple harmonic motion (SHM). Let's break it down step by step:
(a) To describe the two equilibrium positions for the wheel, we need to consider the forces acting on it. In this case, there are two forces involved: the gravitational force and the tension force from the valve.
The gravitational force (mg) acts at the center of mass of the system (the hoop and the point mass). Since the wheel is upside down, the gravitational force points upwards.
The tension force acts at the point mass and points radially inward. For equilibrium, the net torque about the axle must be zero.
Let's denote the angle between the vertical and the line connecting the axle and the point mass as θ.
In the first equilibrium position, the point mass is directly below the axle (θ = 0). In this case, the gravitational force and the tension force balance each other, resulting in a net torque of zero.
In the second equilibrium position, the point mass is at the highest point of the circle (θ = π). Here again, the gravitational force and the tension force balance each other, resulting in a net torque of zero.
Now, to identify the stable equilibrium position, we need to analyze small oscillations around each equilibrium position. For small oscillations, we can use the small angle approximation sin(θ) ≈ θ.
When the wheel is slightly displaced from the first equilibrium position, a restoring torque is produced due to the tension force. This torque acts to return the wheel to its original position, thus making it a stable equilibrium.
However, when the wheel is slightly displaced from the second equilibrium position, the restoring torque acts in the opposite direction of the displacement, making it an unstable equilibrium.
Therefore, the first equilibrium position (θ = 0) is a stable equilibrium, while the second equilibrium position (θ = π) is an unstable equilibrium.
(b) To calculate the period for small angle oscillations about the stable equilibrium, we can use the formula for the period (T) of a simple pendulum:
T = 2π√(I/mgd)
where I is the moment of inertia of the wheel, m is the mass of the point mass, g is the acceleration due to gravity, and d is the distance from the center of mass to the point mass.
The moment of inertia of a hoop about its axis of rotation is I = MR^2, where M is the mass of the hoop and R is the radius.
Plugging in the given values:
M = 1.0 kg
R = 0.6 m
m = 0.005 kg
g ≈ 9.8 m/s^2 (acceleration due to gravity)
We also need to determine the value of d, which is the distance between the center of mass of the system and the point mass. For a hoop, the center of mass is at the center of the hoop, so d = R.
Substituting these values into the formula, we have:
T = 2π√((MR^2)/(mgR))
Simplifying further:
T = 2π√((R^2)/(g))
Evaluating this expression numerically, we get:
T ≈ 2π√((0.6^2)/(9.8))
T ≈ 0.774 s
Therefore, the period for small angle oscillations of the wheel about the stable equilibrium is approximately 0.774 seconds.