Angular torque

In the case of a polyatomic molecule, rotation can occur in three dimensions about the molecular center of mass. Any possible mode of rotation can be expressed as projections on the three mutually perpendicular axes, x, y, and z hence, three moments of inertia are necessar y to give the resistance to angular acceleration by any torque (twisting force) in a , y, and z space. In the MM3 output file, they are denoted IX, lY, and IZ and are given in the nonstandard units of grams square centimeters. 

The relationship between viscosity, angular velocity, and torque for a Newtonian fluid in a concentric cylinder viscometer is given by the Margules equation (eq. 26) (21,146), where M is the torque on the inner cylinder, h the length of the inner cylinder, Q the relative angular velocity of the cylinder in radians per second, T the radius of the inner cylinder wall, the radius of the outer cylinder wall, and an instmment constant. 

Therefore, the viscosity can be determined from the torque and angular velocity. However, the viscosity is usually calculated from the shear rate and shear stress, which can be obtained from the Margules equation. The shear rate is given by equation 27, where r is any given radius. 

In addition to non-Newtonian flow, the main correction necessary for concentric cylinder measurements is that on account of end effects. Because the inner cylinder is not infinitely long, there is drag on the ends as well as on the face of the cylinder. The correction appears as an addition, to the length, b. The correction is best deterrnined by measuring the angular velocity and torque at several values of b, that is, at various depths of immersion. The data are plotted as M/Q vs b, and extrapolation is made to a value of at M/H = 0. The quantity (/i + h ) is substituted for b in the various equations. 

Specific Commercial Rotational Viscometers. Information on selected commercial rotational viscometers can be found ia Table 7. The ATS RheoSystems Stresstech rheometer is an iastmment that combines controlled stress as well as controlled strain (shear rate) and oscillatory measurements. It has a torque range of 10 to 50 mN-m, an angular velocity range of 0 to 300 rad/s, and a frequency range of seven decades. Operation and temperature programming (—30 to 150°C higher temperatures optional) are computer controlled. 

The rate of change of energy transfer (ft-lb//sec) is the product of the torque and the angular velocity (lu)... 

There are three relevant bending moments caused by a gear coupling when transmitting torque with angular or parallel misalignment ... 

A flywheel of moment of inertia / sits in bearings that produee a frietional moment of C times the angular veloeity uj t) of the shaft as shown in Figure 2.7. Find the differential equation relating the applied torque T t) and the angular veloeity uj t). 

Figure 2.8 shows a reduetion gearbox being driven by a motor that develops a torque T tn(t). It has a gear reduetion ratio of and the moments of inertia on the motor and output shafts are and /q, and the respeetive damping eoeffieients Cm and Cq. Find the differential equation relating the motor torque CmfO and the output angular position 6a t). 

Consider a nucleus with magnetic moment pi in a magnetic field Ho- According to classical mechanics the rate of change of the angular momentum G is the torque T. 

A force is required to rotate an object. The response to the force depends not only the size of the force, but also on the manlier in which the force is applied. A bicyclist must push down on a pedal to cause the sprocket to rotate. But if the shaft to which the pedal is attached is vertical, no rotation results. The greatest response occurs when the bicyclist pushes down on the pedal when the shaft is horizontal. The concept of torque is used to describe rotational motion. The bicyclist pushing down on the pedal when the shaft is vertical produces zero torque. The maxiniuiii torque is produced when the shaft is horizontal. In this case the torque is the product of the force and the length of the shaft. For any other position between vertical and horizontal, the torque is the product of the force, shaft length, and sine of the angle made by the shaft and direction of the force. A force develops power when linear motion is involved and a torque develops power when rotational motion is involved. The power developed by a force is the product of force and linear velocity (P = Fv) and the power developed by a torque is the product of torque and angular velocity (P = Tw). 

Locked-rotor (static) torque, starting, or breakaway The minimum torque that a motor will develop at rest for all angular positions of the rotor, with rated voltage applied at rated frequency. 

A uniform steel bar 3 ft long and 2.5 in. in diameter is subjected to a torque of 800 ft-lb. What will be the maximum torsional stress and the angular deflection between the two ends ... 

Fig. 3-14 Shaft with diameter d and length L under torque T undergoing angular deformation.

The angular rotation of the shaft is caused by torque is given by ... 

Moment of inertia It is the ratio of torque applied to a rigid body free to rotate about a given axis to the angular acceleration thus produced about that axis. 

The torque acting on the fluid dr is equal to the rate of change of angular momentum with time, as it goes through the pumps or ... 

Fig. 1.1. Time-dependence of the components of angular momentum J, (Markovian process) and the torque M, (white noise) in the impact approximation.

Two other features of the classical mechanics are illustrated in Fig. 3. The first is that the torque responsible for the angular deflection at low k comes from points close to r = 0 in the scaled form of Eq. (2), around which the quartic term can be ignored. Thus... 

This is an equation of rotation of an elementary mass around the y-axis. Here r can be treated as the moment of inertia of the unit mass and dco/dt is the angular acceleration. The product gx characterizes the torque with respect to the point 0. Multiplying Equation (3.49) by dm and performing integration over the pendulum mass, we obtain... 

If there is no force acting on the particle, the torque is zero. Consequently, the rate of change of the angular momentum is zero and the angular momentum is conserved. 

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