How to calculate the radial force on the shaft
Publish: 2021-05-26 21:22:04
1.
The solution is as follows:
If Z is used to represent the number of teeth of the gear, then: the circumference of the dividing circle= π D = ZP, that is, d = ZP/ π Make p/ π= m. Then d = MZ where. It's called molus. Because the pitch P of two gears must be equal, so the molus is equal
For the convenience of gear design and machining, the molus value has been standardized. The higher the molus is, the higher the height and thickness of the gear teeth are, and the greater the load they bear. Under the same conditions, the larger the molus is, the larger the gear isextended data
Properties of circular force:
1. In any case, the resultant force of the moment center is zero (that is, the moment center is a fixed point, which should have the conditions to balance the circular force)
The circular force can be divided into several parts or couples without changing the effect on the figure The translation theorem is not completely applicable to circular force2. It has been installed many times, but it has never been calculated. The torque transmitted by the motor through the tail shaft should be determined according to the actual size of the motor encoder. This is not determined by the motor, but by the size of the encoder.
3. Bending radius generally depends on the material, space, design capacity requirements and other factors, so it is considered in many aspects.
4. The horizontal disc rotates uniformly around the vertical central axis, a small block of wood is placed on the disc and rotates with the disc, and the block remains stationary relative to the disc, as shown in the figure; A. The greater the mass of the block, the easier it is to slide on the disc
C. The greater the distance between the block and the rotating shaft, the easier it is to slide on the disc
D. the smaller the rotation period of the disc, The more difficult it is for the wooden block to slide on the disk
homework helps users with physics 2017-10-20
do you want to get the secret script of fast score raising? A. when the wood block moves in a uniform circular motion, the friction force in the motion provides centripetal force, so its direction is perpendicular to the linear velocity direction, so a is wrong
B. the critical state of relative sliding of wood block on the turntable is: mg μ= m ω 2R, the mass is eliminated, so B is wrong
C. The greater the distance from the block to the shaft, the greater the centripetal force required, and the easier it is to slide, so C is correct
D. when relative sliding occurs, the maximum static friction provides centripetal force, and the value is: mg μ= m ω Therefore, the smaller the period, i.e. the angular velocity ω As shown in the figure, there is a horizontal disc which can rotate around the vertical central axis, on which a spring with stiffness coefficient K is placed. One end of the spring is fixed on the shaft o, and the other end is connected with a small block a (which can be regarded as a particle) with mass M. the dynamic friction coefficient between the block and the disc is 0 μ， Open
as shown in Figure 2, a disk rotates uniformly around the vertical axis passing through the center in the horizontal plane, and a small object on the disk is stationary relative to the disk and moves with the disk. As for the centripetal force on the object, the following statement is correct: a
the horizontal disk rotates uniformly around the central axis, and... Two small objects a and B are placed on the disk, which rotate with the disk, Their masses are M1 and M2 respectively, and the distances from the rotating axis are R1 and R2 respectively, and R1 > R2. The friction coefficients between the two objects and the disc are the same, and the static friction forces they are subjected to are F1 and F2 respectively. The following statement is correct:
as shown in the figure, a disc can rotate around a vertical axis passing through the center of the disc and perpendicular to the surface of the disc, The wood block moves with the disc, then () A. the wood block is subject to the friction of the disc, the direction
the object with a mass of 1.0kg is placed on a horizontal disc which can rotate around the vertical axis, and the object is connected with the rotating axis by a light spring. The maximum static friction between the object and the disc is 0.1 times of the gravity, the stiffness coefficient of the spring is 600N / m, and the original length is 4cm,
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problem analysis
when the wood block moves in a uniform circular motion on the horizontal turntable, the static friction provides centripetal force. Since the static friction has a maximum value, it is necessary to keep relatively static ring rotation, There is a maximum value of rotation speed or period.
comments from famous teachers
examination site of this question:
centripetal force.
examination site comments:
C. The greater the distance between the block and the rotating shaft, the easier it is to slide on the disc
D. the smaller the rotation period of the disc, The more difficult it is for the wooden block to slide on the disk
homework helps users with physics 2017-10-20
do you want to get the secret script of fast score raising? A. when the wood block moves in a uniform circular motion, the friction force in the motion provides centripetal force, so its direction is perpendicular to the linear velocity direction, so a is wrong
B. the critical state of relative sliding of wood block on the turntable is: mg μ= m ω 2R, the mass is eliminated, so B is wrong
C. The greater the distance from the block to the shaft, the greater the centripetal force required, and the easier it is to slide, so C is correct
D. when relative sliding occurs, the maximum static friction provides centripetal force, and the value is: mg μ= m ω Therefore, the smaller the period, i.e. the angular velocity ω As shown in the figure, there is a horizontal disc which can rotate around the vertical central axis, on which a spring with stiffness coefficient K is placed. One end of the spring is fixed on the shaft o, and the other end is connected with a small block a (which can be regarded as a particle) with mass M. the dynamic friction coefficient between the block and the disc is 0 μ， Open
as shown in Figure 2, a disk rotates uniformly around the vertical axis passing through the center in the horizontal plane, and a small object on the disk is stationary relative to the disk and moves with the disk. As for the centripetal force on the object, the following statement is correct: a
the horizontal disk rotates uniformly around the central axis, and... Two small objects a and B are placed on the disk, which rotate with the disk, Their masses are M1 and M2 respectively, and the distances from the rotating axis are R1 and R2 respectively, and R1 > R2. The friction coefficients between the two objects and the disc are the same, and the static friction forces they are subjected to are F1 and F2 respectively. The following statement is correct:
as shown in the figure, a disc can rotate around a vertical axis passing through the center of the disc and perpendicular to the surface of the disc, The wood block moves with the disc, then () A. the wood block is subject to the friction of the disc, the direction
the object with a mass of 1.0kg is placed on a horizontal disc which can rotate around the vertical axis, and the object is connected with the rotating axis by a light spring. The maximum static friction between the object and the disc is 0.1 times of the gravity, the stiffness coefficient of the spring is 600N / m, and the original length is 4cm,
ask the operator for more questions
problem analysis
when the wood block moves in a uniform circular motion on the horizontal turntable, the static friction provides centripetal force. Since the static friction has a maximum value, it is necessary to keep relatively static ring rotation, There is a maximum value of rotation speed or period.
comments from famous teachers
examination site of this question:
centripetal force.
examination site comments:
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