非线性动力学和一个环在垂直杆上的稳定前行开始

A torsional pendulum involves the oscillation of a rigid body in which the restoring force is provided by the torsion in the string from which the rigid body is suspended. Ideally, the string should be massless; practically, its mass is much smaller than the rigid body's mass and is neglected.
As long as the rigid body's angular displacement is small, its oscillation can be modeled as a linear angular oscillation. The amplitude of the oscillation is an angle. The role of mass is played...

An object undergoing circular motion, like a race car, is accelerating because it is changing the direction of its velocity. This centrally directed acceleration is called centripetal acceleration. This acceleration acts along the radius of the curved path (thus is also referred to as radial acceleration).
Any acceleration must be produced by some force. Therefore, any force or combination of forces can cause centripetal acceleration. A few examples include the tension in the rope on a...

Precession can be demonstrated effectively through a spinning top. If a spinning top is placed on a flat surface near the surface of the Earth at a vertical angle and is not spinning, it will fall over due to the force of gravity producing a torque acting on its center of mass. However, if the top is spinning on its axis, it precesses about the vertical direction, rather than topple over due to this torque. Precessional motion is a combination of a steady circular motion of the axis and the...

Suppose a car moves on flat ground and turns to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. For this, a minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway. Let's now consider banked curves, where the slope of the road helps in negotiating the curve. The greater the angle of the curve, the faster one can take the curve. It is common for race tracks for...

Consider a scenario where a circular shaft is subject to torque that remains within the boundaries of Hooke's Law, avoiding any permanent deformation. So, the formula for shearing strain is revisited. This formula is multiplied by the modulus of rigidity, and then Hooke's Law for the shearing stress and strain is applied. As a result, the equation for shearing stress in a shaft can be derived.

One of the distinctive characteristics of circular shafts is their ability to maintain their cross-sectional integrity under torsion. In other words, each cross-section continues to exist as a flat, unaltered entity, simply rotating like a solid, rigid slab. To understand the distribution of shearing stress within such a shaft, consider a cylindrical section inside this circular shaft. This section has a length of L and a radius of R, with one end fixed. The radius of the cylindrical section is...