Centripetal Force

4 Key excerpts on "Centripetal Force"

• 

  eBook - ePub

  Engineering Science

  • W. Bolton(Author)
  • 2015(Publication Date)
  • Routledge

    (Publisher)

  r gives the acceleration an object must experience, at right angles to its direction of motion, if it is to move in a circular path. The Centripetal Force necessary for this acceleration is thus:

  According to Newton’s third law, to every action there is an opposite and equal reaction. In this case the reaction to the Centripetal Force is called the centrifugal force and acts in an outwards direction on the pivot C around which the motion is occurring.

  Example An object of mass 0.5 kg is whirled round in a horizontal circle of radius 0.8 m on the end of a rope. What is the tension in the rope when the object rotates at 4 rev/s?

  F = mω 2 r = 0.5 X (2π X 4)2 × 0.8 = 253 N

  Example Calculate the force acting on a bearing which is carrying a crankshaft with an out- of-balance load of 0.10 kg at a radius of 100 mm and rotating at 50 rev/s.

  Centripetal Force = mω 2 r = 0.10 X (2π X 50)2 X 0.100 = 987 N. The force acting on the bearing is the reaction force, i.e. the centrifugal force, and is thus 987 N radially outwards.

  Example

  An object of mass 3 kg is attached to the end of a rope and whirled round in a vertical circle of radius 1 m. What are the maximum and minimum values of the tension in the rope when it rotates at 3 rev/s?

  The Centripetal Force F = mω 2 r = 3 X (2π X 3)2 X 1 = 1066 N. At the top of the path, the Centripetal Force is provided by the tension in the rope plus the weight since they are acting in the same direction. Hence, T + 3 X 9.8 = 1066 and so tension T = 1037 N. At the bottom of the path, the tension and the weight act in opposite directions and so T - 3 X 9.8 = 1066 and the tension T = 1095 N. Thus the maximum tension is 1095 N and the minimum tension 1037 N.

  23.3 Cornering

  Figure 23.3 Cornering on the flat

  Consider a vehicle of mass m rounding a horizontal corner of radius r (Figure 23.3 ). Since the reactive force R is mg at right angles to the plane, the maximum frictional force F = μR = μmg . Hence, when the Centripetal Force exceeds this frictional force, skidding will occur. Thus the maximum speed is given when mv 2 /r = μmg

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  eBook - ePub

  Biomechanics of Human Motion

  Applications in the Martial Arts, Second Edition

  • Emeric Arus, Ph.D.(Authors)
  • 2017(Publication Date)
  • CRC Press

    (Publisher)

  We can observe this force when we are sitting in a car. When the car is turning to the left, our body moves to the right. This motion to the right is the centrifugal force. In martial arts, an example of the centrifugal force manifestation is the aikido technique named “entering throw—negative” execution (Irimi-nage). Here, the executor (attacker) (Shite) guides the opponent in a rotary fashion. The defender (Uke) feels that the guiding force throws him away from the closeness of his attacker. Besides the attacker’s guiding force, the defender also feels the centrifugal force. More description about the centripetal and centrifugal forces can be found under Part IV. Pseudocentrifugal force appears when a rotating reference frame is used for analysis. The true frame acceleration is substituted by a pseudocentrifugal force that is exerted on all objects, and directed away from the axis of rotation. Figure 9.1 is seen from the top of the head of the karateka. As you can see the tangential force is perpendicular to the radial force. The drawing shows that the technique starts with the elbow bent. However, for the correct calculation of the angular velocity and the radial force, the arm should be completely extended. r represents the radius of the circle and also the total length of the arm. 9.2 CENTER OF MASS As we know already that every object/matter is comprised of mass. This amount of mass has a volume and density. The amount of mass also has its center, what we call the center of mass (CoM). This CoM represents the average position of the mass distribution of an object and it is considered that the mass of the body lies within the CoG. The linear and/or rotational movements and their positions of the rigid bodies can be analyzed by finding out the CoG positions of the body in case. There are different methods to find out the CoG of a rigid body. One method is the reaction board method. We will not describe these methods in this book

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  eBook - ePub

  Unearthing Fermi's Geophysics

  • Gino C. Segrè, John D. Stack(Authors)
  • 2022(Publication Date)
  • University of Chicago Press

    (Publisher)

  Eq. (2.9) from here on.

  Centrifugal Effects The Earth is rotating, so a position fixed with respect to the Earth is in a rotating coordinate system and centrifugal effects must be taken into account. Recall that in elementary physics a mass μ , which is rotating at angular frequency ω experiences a “centrifugal force.” For a mass located at a distance of r ⟘ from the axis of rotation, the centrifugal force is

  and points away from the axis of rotation. The centrifugal force can be derived from a centrifugal potential, given by

  Denoting the centrifugal force on a mass μ as

  Fcent

  , we have

  Returning to the case of the Earth’s rotation and using spherical coordinates, r ⟘ is just the distance from the Earth’s axis and is given by r ⟘ = r sin θ . The centrifugal potential becomes

  and ω = 2

  π/Tsid

  . Adding

  Vcent

  to our expression for

  Vg

  from Eq. (2.9) and dropping all angular dependence from Legendre functions with l > 2,, the total potential is

  The local acceleration due to gravity and centrifugal effects is defined as

  Evaluating g using Eq. (2.15) at the equator, the spherically symmetric term in

  Vtot

  gives ~ 980 cm/ s2 . The J 2 term is ~ 1.6 cm/ s2 . The centrifugal term is ~ 3.4 cm/ s2 . We see that the spherically symmetric term is dominant, while the J 2 and centrifugal terms are much smaller.

  It is important to note that in a rotating coordinate system, Newton’s second law is modified in two ways, commonly known as centrifugal and Coriolis forces. The addition of

  Vcent

  to

  Vg

  does account for centrifugal effects, but it leaves out the Coriolis force. The latter comes into play only when the body being considered is moving relative to the Earth, e.g., a part of the atmosphere which is in motion as in a storm, discussed further in Sec. (6). For a body of mass μ

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  eBook - ePub

  Basic Engineering Mechanics Explained, Volume 1

  Principles and Static Forces

  • Gregory Pastoll, Gregory Pastoll(Authors)
  • 2019(Publication Date)
  • Gregory Pastoll

    (Publisher)

  One might think that the displacement of the CG by so small an amount was insignificant. However, while rotating, the wheel is now behaving as if it were a point mass of 15.45 kg situated 0.233 mm from the axis of rotation. At a vehicle speed of 120 km/h, the centrifugal force on this wheel would be approximately 36 N, rotating at approximately 16 times per second, sufficient to cause a noticeable wobble. Centrifugal force is explained in a subsequent chapter, in volume 3 of this series.

  Centroids of areas and volumes

  The centroid of a plane area is a point on its surface that corresponds with the CG which that surface would have, if it were a thin plate, of uniform thickness and density. (This type of thin plate is often referred to as a ‘lamina’. This word comes directly from Latin, meaning a thin plate of marble or metal.)

  The centroid of a volume is that point within the volume that corresponds with the CG of a solid object with uniform density, and the same shape as that volume. Standard expressions for the location of the centroids of various regular geometric areas and volumes

  (derived by methods similar to that demonstrated at the end of this chapter. )

  Example

  The object illustrated here is turned from solid brass, density 8400 kg/m3 , and can be thought of as a hollow hemisphere adjoining a hollow cylinder.

  Determine the location of its CG above point ‘O’.

  Since all the constituent ‘parts’ of this assembly are of the same material, their masses are proportional to their volumes. It will thus not be necessary to determine their masses, as we can work with volumes instead. The equation for determining Ȳ becomes : Ȳ = (Σȳv)/(Σv)

  Consider the assembly to be made up of four parts:

  A. Solid hemisphere, r = 90 mm,

  B. Removed solid hemisphere, r = 50 mm, hence negative volume

  C. Solid cylinder, r = 90 mm and h = 150 mm

  D. Removed solid cylinder, r = 50 mm and h = 150 mm, hence negative volume

  Determine the volumes of the respective parts, and insert these values into the standard table. Note: the heights above the bases of the two solid hemispheres are added to the height of the cylinder, to give the ȳ dimensions relative to the base of the assembly.

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