Centripetal Acceleration and Centripetal Force

5 Key excerpts on "Centripetal Acceleration and Centripetal Force"

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  Mechanical Design for the Stage

  • Alan Hendrickson(Author)
  • 2012(Publication Date)
  • Routledge

    (Publisher)

  Figure 7.4c ). Mathematically, centripetal acceleration equals

  Figure 7.4 Illustration of the terms involved in: a. Angular and linear displacement, b. Angular and linear speed and acceleration, c. Tangential and centripetal acceleration. (In both b. and c. rotational speed and acceleration are indicated in an easily understood way. The actual vectors for rotational velocity and acceleration would be shown by arrows along the axis of rotation, straight out of the page.)

  a

  c e n t r i p e t a l

  =

  v 2

  r

  = r

  ω 2

  Newton’s second law states that force is required to accelerate mass, and so a centripetal force can be defined.

  F

  c e n t r i p e t a l

  = m

  a

  c e n t r i p e t a l

  = m r

  ω 2

  If you were to tie a weight to a string and spin it around above your head, the tension in the line would equal Fcentripetal . The cohesive forces within the steel of a shaft supply Fcentripetal , keeping the shaft from flying apart as it spins (though there are limits to the speed it can withstand). And a free-standing prop set on a turntable could slide outward, or away from the axis of rotation, if the turntable spins fast enough so that Fcentripetal exceeds Fstatic friction

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  Biomechanics of Human Motion

  Applications in the Martial Arts, Second Edition

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

    (Publisher)

  centripetal force or center-seeking force.

  The radial acceleration involves change of direction continuing perpendicular to the direction of motion. Here are the equations for radial acceleration and radial (centripetal) force. Radial acceleration a rad = v 2 /r = ω2 r and radial force F rad = m ⋅ v 2 /r = m ⋅ r ⋅ ω2 . Here, v is the magnitude of the tangential circular velocity or speed and r is the length of the radius of rotation.

  Let us take an example and calculate the radial force. A karateka strikes toward the opponent’s face with a technique back-fist strike (Uraken-uchi) as shown in Figure 9.1 . The karateka weighs 70 kg, which we do not count at this time; his total arm mass is 3.45 kg; the velocity is 11 m/s; and the total arm length (with the fist closed) is 0.69 m. F rad = m ⋅ v 2 /r = 3.45 kg × 112 / 0.69 m = 605 N.

  FIGURE 9.1. Radial and tangential forces.

  Centripetal force coexists with another force named centrifugal force. The word comes from the Latin word “centrum” (center) and “fugere” to (flee). The centrifugal force has two slightly different manifestation forms:

  • Reactive centrifugal force that occurs in reaction to a centripetal acceleration on a mass. This force is equal in magnitude to the centripetal force and is directed from the center of rotation. 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)

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  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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  Science and Mathematics for Engineering

  • John Bird(Author)
  • 2019(Publication Date)
  • Routledge

    (Publisher)

  Chapter 18 Acceleration Why it is important to understand: Acceleration Acceleration may be defined as a ‘change in velocity’. This change can be in the magnitude (speed) of the velocity or the direction of the velocity. In daily life we use acceleration as a term for the speeding up of objects and decelerating for the slowing down of objects. If there is a change in the velocity, whether it is slowing down or speeding up, or changing its direction, we say that the object is accelerating. If an object is moving at constant speed in a circular motion – such as a satellite orbiting the earth – it is said to be accelerating because change in direction of motion means its velocity is changing even if speed may be constant. This is called centripetal (directed towards the centre) acceleration. On the other hand, if the direction of motion of the object is not changing but its speed is, this is called tangential acceleration. If the direction of acceleration is in the same direction as that of velocity then the object is said to be speeding up or accelerating. If the acceleration and velocity are in opposite directions then the object is said to be slowing down or decelerating. An example of constant acceleration is the effect of the gravity of earth on an object in free fall. Measurement of the acceleration of a vehicle enables an evaluation of the overall vehicle performance and response. Detection of rapid negative acceleration of a vehicle is used to detect vehicle collision and deploy airbags. The measurement of acceleration is also used to measure seismic activity, inclination and machine vibration

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  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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