Find the Angular velocity and period of oscillations of a solid sphere of mass m and radius R about a point on its surface book of Marion,Stephen&Thornton, Jerry (

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Find the Angular velocity and period of oscillations of a solid sphere of mass m and radius R about a point on its surface book of Marion,Stephen&Thornton, Jerry (Classical Dynamics of Particles and Systems)
by v (units: vibrations per unit time or Hertz, Hz,
cumes a will be
for brevity, although "angular frequency" is to be understood.
1
3.2 SIMPLE HARMONIC OSCILLATOR
10
EXAMPLE 3.1
Find the angular velocity and period of oscillation of a solid sphere of mass m
and radius Rabout a point on its surface. See Figure 3-1.
Solution. Let the rotational inertia of the sphere be Iabout the pivot point. In ele-
mentary physics we learn that the value of the rotational inertia about an axis
through the sphere's center is 2/5 mR. If we use the parallel-axis theorem, the rota-
tional inertia about the pivot point on the surface is 2/5 mR + mR = 7/5 mR.
The equilibrium position of the sphere occurs when the center of mass (center of
sphere) is hanging directly below the pivot point. The gravitational force F= mg
pulls the sphere back towards the equilibrium position as the sphere swings back
and forth with angle 0. The torque on the sphere is N= la, where a - ö is the an-
gular acceleration. The torque is also N = R x F, with N= RFsin 0 = Rmg sin 8.
For small oscillations, we have N- Rmg 0. We must have 1ö = -Rmg 0 for the
equation of motion in this case, because as 8 increases, ở is negative. We need to
solve the equation of motion for 6.
Rmg
This equation is similar to Equation 3.5 and has solutions for the angular fre-
quency and period from Equations 3.14 and 3.15,
Rmg
Rmg
5g
VTR
7
and
= 27\
1= 27 Rmg
- 27N 5g
Rmg
Pivot
R
FIGURE 3-1 Example 3.1. The physical pendulum (sphere).
104
3/ OSCILLATIONS
Note that the mass m does not enter. Only the distance R to the center of
mass determines the oscillation frequency.
3.3 Harmonic Oscillations in Two Dimensions
We next consider the motion of a particle that is allowed two degrees of freedom.
We take the restoring force to be proportional to the distance of the particle from
a force center located at the origin and to be directed toward the origin:
F = -kr
(3.16)
which can be resolved in polar coordinates into the components
F, = - kr cos 8 = - kx
F, = - kr sin e = – ky S
(3.17)
%3D
The equations of motion are
Transcribed Image Text:by v (units: vibrations per unit time or Hertz, Hz, cumes a will be for brevity, although "angular frequency" is to be understood. 1 3.2 SIMPLE HARMONIC OSCILLATOR 10 EXAMPLE 3.1 Find the angular velocity and period of oscillation of a solid sphere of mass m and radius Rabout a point on its surface. See Figure 3-1. Solution. Let the rotational inertia of the sphere be Iabout the pivot point. In ele- mentary physics we learn that the value of the rotational inertia about an axis through the sphere's center is 2/5 mR. If we use the parallel-axis theorem, the rota- tional inertia about the pivot point on the surface is 2/5 mR + mR = 7/5 mR. The equilibrium position of the sphere occurs when the center of mass (center of sphere) is hanging directly below the pivot point. The gravitational force F= mg pulls the sphere back towards the equilibrium position as the sphere swings back and forth with angle 0. The torque on the sphere is N= la, where a - ö is the an- gular acceleration. The torque is also N = R x F, with N= RFsin 0 = Rmg sin 8. For small oscillations, we have N- Rmg 0. We must have 1ö = -Rmg 0 for the equation of motion in this case, because as 8 increases, ở is negative. We need to solve the equation of motion for 6. Rmg This equation is similar to Equation 3.5 and has solutions for the angular fre- quency and period from Equations 3.14 and 3.15, Rmg Rmg 5g VTR 7 and = 27\ 1= 27 Rmg - 27N 5g Rmg Pivot R FIGURE 3-1 Example 3.1. The physical pendulum (sphere). 104 3/ OSCILLATIONS Note that the mass m does not enter. Only the distance R to the center of mass determines the oscillation frequency. 3.3 Harmonic Oscillations in Two Dimensions We next consider the motion of a particle that is allowed two degrees of freedom. We take the restoring force to be proportional to the distance of the particle from a force center located at the origin and to be directed toward the origin: F = -kr (3.16) which can be resolved in polar coordinates into the components F, = - kr cos 8 = - kx F, = - kr sin e = – ky S (3.17) %3D The equations of motion are
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