Centripetal force (from Latin centrum 'center' and petere 'to seek') is the force that makes a body follow a curved path. The direction of the centripetal force is always orthogonal to the motion of the body and towards the fixed point of the instantaneous center of curvature of the path. Isaac Newton coined the term, describing it as "a force by which bodies are drawn or impelled, or in any way tend, towards a point as to a centre". In Newtonian mechanics, gravity provides the centripetal force causing astronomical orbits. One common example involving centripetal force is the case in which a body moves with uniform speed along a circular path. The centripetal force is directed at right angles to the motion and also along the radius towards the centre of the circular path. The mathematical description was derived in 1659 by the Dutch physicist Christiaan Huygens.
Formula From the kinematics of curved motion it is known that an object moving at tangential speed v along a path with radius of curvature r accelerates toward the center of curvature at a rate
a c = lim Δ t → 0 Δ v Δ t , a c = v 2 r {\displaystyle {\textbf {a}}_{c}=\lim _{\Delta t\to 0}{\frac {\Delta {\textbf {v}}}{\Delta t}},\quad a_{c}={\frac {v^{2}}{r}}}
Here, a c {\displaystyle a_{c}} is the centripetal acceleration and Δ v {\displaystyle \Delta {\textbf {v}}} is the difference between the velocity vectors at t + Δ t {\displaystyle t+\Delta {t}} and t {\displaystyle t} . By Newton's second law, the cause of acceleration is a net force acting on the object, which is proportional to its mass m and its acceleration. The force, usually referred to as a centripetal force, has a magnitude
F c = m a c = m v 2 r {\displaystyle F_{c}=ma_{c}=m{\frac {v^{2}}{r}}}
and is, like centripetal acceleration, directed toward the center of curvature of the object's trajectory.
Derivation
The centripetal acceleration can be inferred from the diagram of the velocity vectors at two instances. In the case of uniform circular motion the velocities have constant magnitude. Because each one is perpendicular to its respective position vector, simple vector subtraction implies two similar isosceles triangles with congruent angles – one comprising a base of Δ v {\displaystyle \Delta {\textbf {v}}} and a leg length of v {\displaystyle v} , and the other a base of Δ r {\displaystyle \Delta {\textbf {r}}} (position vector difference) and a leg length of r {\displaystyle r}
| Δ v | v = | Δ r | r {\displaystyle {\frac {|\Delta {\textbf {v}}|}{v}}={\frac {|\Delta {\textbf {r}}|}{r}}}
| Δ v | = v r | Δ r | {\displaystyle |\Delta {\textbf {v}}|={\frac {v}{r}}|\Delta {\textbf {r}}|}
Therefore, | Δ v | {\displaystyle |\Delta {\textbf {v}}|} can be substituted with v r | Δ r | {\displaystyle {\frac {v}{r}}|\Delta {\textbf {r}}|} :
a c = lim Δ t → 0 | Δ v | Δ t = v r lim Δ t → 0 | Δ r | Δ t = v 2 r {\displaystyle a_{c}=\lim _{\Delta t\to 0}{\frac {|\Delta {\textbf {v}}|}{\Delta t}}={\frac {v}{r}}\lim _{\Delta t\to 0}{\frac {|\Delta {\textbf {r}}|}{\Delta t}}={\frac {v^{2}}{r}}}
The direction of the force is toward the center of the circle in which the object is moving, or the osculating circle (the circle that best fits the local path of the object, if the path is not circular). The speed in the formula is squared, so twice the speed needs four times the force, at a given radius. This force is also sometimes written in terms of the angular velocity ω of the object about the center of the circle, related to the tangential velocity by the formula
v = ω r {\displaystyle v=\omega r}
so that
F c = m r ω 2 . {\displaystyle F_{c}=mr\omega ^{2}\,.}
Expressed using the orbital period T for one revolution of the circle,
ω = 2 π T {\displaystyle \omega ={\frac {2\pi }{T}}}
the equation becomes
F c = m r ( 2 π T ) 2 . {\displaystyle F_{c}=mr\left({\frac {2\pi }{T}}\right)^{2}.}
In particle accelerators, velocity can be very high (close to the speed of light in vacuum) so the same rest mass now exerts greater inertia (relativistic mass) thereby requiring greater force for the same centripetal acceleration, so the equation becomes:
F c = γ m v 2 r {\displaystyle F_{c}={\frac {\gamma mv^{2}}{r}}}
where
γ = 1 1 − v 2 c 2 {\displaystyle \gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}
is the Lorentz factor. Thus the centripetal force is given by:
F c = γ m v ω {\displaystyle F_{c}=\gamma mv\omega }
which is the rate of change of relativistic momentum γ m v {\displaystyle \gamma mv} .
Sources
In the case of an object that is swinging around on the end of a rope in a horizontal plane, the centripetal force on the object is supplied by the tension of the rope. The rope example is an example involving a 'pull' force. The centripetal force can also be supplied as a 'push' force, such as in the case where the normal reaction of a wall supplies the centripetal force for a wall of death or a Rotor rider. Newton's idea of a centripetal force corresponds to what is nowadays referred to as a central force. When a satellite is in orbit around a planet, gravity is considered to be a centripetal force even though in the case of eccentric orbits, the gravitational force is directed towards the focus, and not towards the instantaneous center of curvature. Another example of centripetal force arises in the helix that is traced out when a charged particle moves in a uniform magnetic field in the absence of other external forces. In this case, the magnetic force is the centripetal force that acts towards the helix axis.
Analysis of several cases Below are three examples of increasing complexity, with derivations of the formulas governing velocity and acceleration.
Uniform circular motion
Uniform circular motion refers to the case of constant rate of rotation. Here are two approaches to describing this case.
Calculus derivation In two dimensions, the position vector r {\displaystyle {\textbf {r}}} , which has magnitude (length) r {\displaystyle r} and directed at an angle θ {\displaystyle \theta } above the x-axis, can be expressed in Cartesian coordinates using the unit vectors x ^ {\displaystyle {\hat {\mathbf {x} }}} and y ^ {\displaystyle {\hat {\mathbf {y} }}} :
r = r cos ( θ ) x ^ + r sin ( θ ) y ^ . {\displaystyle {\textbf {r}}=r\cos(\theta ){\hat {\mathbf {x} }}+r\sin(\theta ){\hat {\mathbf {y} }}.}
The assumption of uniform circular motion requires three things:
The object moves only on a circle. The radius of the circle r {\displaystyle r} does not change in time. The object moves with constant angular velocity ω {\displaystyle \omega } around the circle. Therefore, θ = ω t {\displaystyle \theta =\omega t} where t {\displaystyle t} is time. The velocity v {\displaystyle {\textbf {v}}} and acceleration a {\displaystyle {\textbf {a}}} of the motion are the first and second derivatives of position with respect to time:
r = r cos ( ω t ) x ^ + r sin ( ω t ) y ^ , {\displaystyle {\textbf {r}}=r\cos(\omega t){\hat {\mathbf {x} }}+r\sin(\omega t){\hat {\mathbf {y} }},}
v = r ˙ = − r ω sin ( ω t ) x ^ + r ω cos ( ω t ) y ^ , {\displaystyle {\textbf {v}}={\dot {\textbf {r}}}=-r\omega \sin(\omega t){\hat {\mathbf {x} }}+r\omega \cos(\omega t){\hat {\mathbf {y} }},}
a = r ¨ = − ω 2 ( r cos ( ω t ) x ^ + r sin ( ω t ) y ^ ) . {\displaystyle {\textbf {a}}={\ddot {\textbf {r}}}=-\omega ^{2}(r\cos(\omega t){\hat {\mathbf {x} }}+r\sin(\omega t){\hat {\mathbf {y} }}).}
The term in parentheses is the original expression of r {\displaystyle {\textbf {r}}} in Cartesian coordinates. Consequently,
a = − ω 2 r . {\displaystyle {\textbf {a}}=-\omega ^{2}{\textbf {r}}.}
The negative sign shows that the acceleration is pointed towards the center of the circle (opposite the radius), hence it is called "centripetal" (i.e. "center-seeking"). While objects naturally follow a straight path (due to inertia), this centripetal acceleration describes the circular motion path caused by a centripetal force.
Derivation using vectors
The image at right shows the vector relationships for uniform circular motion. The rotation itself is represented by the angular velocity vector Ω, which is normal to the plane of the orbit (using the right-hand rule) and has magnitude given by:
| Ω | = d θ d t = ω , {\displaystyle |\mathbf {\Omega } |={\frac {\mathrm {d} \theta }{\mathrm {d} t}}=\omega \ ,}
with θ the angular position at time t. In this subsection, dθ/dt is assumed constant, independent of time. The distance traveled dℓ of the particle in time dt along the circular path is
d ℓ = Ω × r ( t ) d t , {\displaystyle \mathrm {d} {\boldsymbol {\ell }}=\mathbf {\Omega } \times \mathbf {r} (t)\mathrm {d} t\ ,}
which, by properties of the vector cross product, has magnitude rdθ and is in the direction tangent to the circular path. Consequently,
d r d t = lim Δ t → 0 r ( t + Δ t ) − r ( t ) Δ t = d ℓ d t . {\displaystyle {\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}=\lim _{{\Delta }t\to 0}{\frac {\mathbf {r} (t+{\Delta }t)-\mathbf {r} (t)}{{\Delta }t}}={\frac {\mathrm {d} {\boldsymbol {\ell }}}{\mathrm {d} t}}\ .}
In other words,
v = d e f d r d t = d ℓ d t = Ω × r ( t ) . {\displaystyle \mathbf {v} \ {\stackrel {\mathrm {def} }{=}}\ {\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}={\frac {\mathrm {d} \mathbf {\boldsymbol {\ell }} }{\mathrm {d} t}}=\mathbf {\Omega } \times \mathbf {r} (t)\ .}
Differentiating with respect to time,
a = d e f d v d t = Ω × d r ( t ) d t = Ω × [ Ω × r ( t ) ] . {\displaystyle \mathbf {a} \ {\stackrel {\mathrm {def} }{=}}\ {\frac {\mathrm {d} \mathbf {v} }{d\mathrm {t} }}=\mathbf {\Omega } \times {\frac {\mathrm {d} \mathbf {r} (t)}{\mathrm {d} t}}=\mathbf {\Omega } \times \left[\mathbf {\Omega } \times \mathbf {r} (t)\right]\ .}
Lagrange's formula states:
a × ( b × c ) = b ( a ⋅ c ) − c ( a ⋅ b ) . {\displaystyle \mathbf {a} \times \left(\mathbf {b} \times \mathbf {c} \right)=\mathbf {b} \left(\mathbf {a} \cdot \mathbf {c} \right)-\mathbf {c} \left(\mathbf {a} \cdot \mathbf {b} \right)\ .}
Applying Lagrange's formula with the observation that Ω • r(t) = 0 at all times,
a = − | Ω | 2 r ( t ) . {\displaystyle \mathbf {a} =-{|\mathbf {\Omega |} }^{2}\mathbf {r} (t)\ .}
In words, the acceleration is pointing directly opposite to the radial displacement r at all times, and has a magnitude:
| a | = | r ( t ) | ( d θ d t ) 2 = r ω 2 {\displaystyle |\mathbf {a} |=|\mathbf {r} (t)|\left({\frac {\mathrm {d} \theta }{\mathrm {d} t}}\right)^{2}=r{\omega }^{2}}
where vertical bars |...| denote the vector magnitude, which in the case of r(t) is simply the radius r of the path. This result agrees with the previous section, though the notation is slightly different. When the rate of rotation is made constant in the analysis of nonuniform circular motion, that analysis agrees with this one. A merit of the vector approach is that it is manifestly independent of any coordinate system.
Example: The banked turn
The upper panel in the image at right shows a ball in circular motion on a banked curve. The curve is banked at an angle θ from the horizontal, and the surface of the road is considered to be slippery. The objective is to find what angle the bank must have so the ball does not slide off the road. Intuition tells us that, on a flat curve with no banking at all, the ball will simply slide off the road; while with a very steep banking, the ball will slide to the center unless it travels the curve rapidly. Apart from any acceleration that might occur in the direction of the path, the lower panel of the image above indicates the forces on the ball. There are two forces; one is the force of gravity vertically downward through the center of mass of the ball mg, where m is the mass of the ball and g is the gravitational acceleration; the second is the upward normal force exerted by the road at a right angle to the road surface man. The centripetal force demanded by the curved motion is also shown above. This centripetal force is not a third force applied to the ball, but rather must be provided by the net force on the ball resulting from vector addition of the normal force and the force of gravity. The resultant or net force on the ball found by vector addition of the normal force exerted by the road and vertical force due to gravity must equal the centripetal force dictated by the need to travel a circular path. The curved motion is maintained so long as this net force provides the centripetal force requisite to the motion. The horizontal net force on the ball is the horizontal component of the force from the road, which has magnitude |Fh| = m|an| sin θ. The vertical component of the force from the road must counteract the gravitational force: |Fv| = m|an| cos θ = m|g|, which implies |an| = |g| / cos θ. Substituting into the above formula for |Fh| yields a horizontal force to be:
| F h | = m | g | sin θ cos θ = m | g | tan θ . {\displaystyle |\mathbf {F} _{\mathrm {h} }|=m|\mathbf {g} |{\frac {\sin \theta }{\cos \theta }}=m|\mathbf {g} |\tan \theta \,.}
On the other hand, at velocity |v| on a circular path of radius r, kinematics says that the force needed to turn the ball continuously into the turn is the radially inward centripetal force Fc of magnitude:
| F c | = m | a c | = m | v | 2 r . {\displaystyle |\mathbf {F} _{\mathrm {c} }|=m|\mathbf {a} _{\mathrm {c} }|={\frac {m|\mathbf {v} |^{2}}{r}}\,.}
Consequently, the ball is in a stable path when the angle of the road is set to satisfy the condition:
m | g | tan θ = m | v | 2 r , {\displaystyle m|\mathbf {g} |\tan \theta ={\frac {m|\mathbf {v} |^{2}}{r}}\,,}
or,
tan θ = | v | 2 | g | r . {\displaystyle \tan \theta ={\frac {|\mathbf {v} |^{2}}{|\mathbf {g} |r}}\,.}
As the angle of bank θ approaches 90°, the tangent function approaches infinity, allowing larger values for |v|2/r. In words, this equation states that for greater speeds (bigger |v|) the road must be banked more steeply (a larger value for θ), and for sharper turns (smaller r) the road also must be banked more steeply, which accords with intuition. When the angle θ does not satisfy the above condition, the horizontal component of force exerted by the road does not provide the correct centripetal force, and an additional frictional force tangential to the road surface is called upon to provide the difference. If friction cannot do this (that is, the coefficient of friction is exceeded), the ball slides to a different radius where the balance can be realized. These ideas apply to air flight as well. See the FAA pilot's manual.
Nonuniform circular motion
As a generalization of the uniform circular motion case, suppose the angular rate of rotation is not constant. The acceleration now has a tangential component, as shown the image at right. This case is used to demonstrate a derivation strategy based on a polar coordinate system. Let r(t) be a vector that describes the position of a point mass as a function of time. Since we are assuming circular motion, let r(t) = R·ur, where R is a constant (the radius of the circle) and ur is the unit vector pointing from the origin to the point mass. The direction of ur is described by θ, the angle between the x-axis and the unit vector, measured counterclockwise from the x-axis. The other unit vector for polar coordinates, uθ is perpendicular to ur and points in the direction of increasing θ. These polar unit vectors can be expressed in terms of Cartesian unit vectors in the x and y directions, denoted i ^ {\displaystyle {\hat {\mathbf {i} }}} and j ^ {\displaystyle {\hat {\mathbf {j} }}} respectively:
u r = cos θ i ^ + sin θ j ^ {\displaystyle \mathbf {u} _{r}=\cos \theta \ {\hat {\mathbf {i} }}+\sin \theta \ {\hat {\mathbf {j} }}} and u θ = − sin θ i ^ + cos θ j ^ . {\displaystyle \mathbf {u} _{\theta }=-\sin \theta \ {\hat {\mathbf {i} }}+\cos \theta \ {\hat {\mathbf {j} }}.}
One can differentiate to find velocity:
v = r d u r d t = r d d t ( cos θ i ^ + sin θ j ^ ) = r d θ d t d d θ ( cos θ i ^ + sin θ j ^ ) = r d θ d t ( − sin θ i ^ + cos θ j ^ ) = r d θ d t u θ = ω r u θ {\displaystyle {\begin{aligned}\mathbf {v} &=r{\frac {d\mathbf {u} _{r}}{dt}}\\&=r{\frac {d}{dt}}\left(\cos \theta \ {\hat {\mathbf {i} }}+\sin \theta \ {\hat {\mathbf {j} }}\right)\\&=r{\frac {d\theta }{dt}}{\frac {d}{d\theta }}\left(\cos \theta \ {\hat {\mathbf {i} }}+\sin \theta \ {\hat {\mathbf {j} }}\right)\\&=r{\frac {d\theta }{dt}}\left(-\sin \theta \ {\hat {\mathbf {i} }}+\cos \theta \ {\hat {\mathbf {j} }}\right)\\&=r{\frac {d\theta }{dt}}\mathbf {u} _{\theta }\\&=\om