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Circular motion

From Wikipedia, the free encyclopedia
Object movement along a circular path
"Radial motion" redirects here and is not to be confused withradial velocity orrotational speed.
Part of a series on
Classical mechanics
F=dpdt{\displaystyle {\textbf {F}}={\frac {d\mathbf {p} }{dt}}}

Inphysics,circular motion ismovement of an object along thecircumference of acircle orrotation along acircular arc. It can be uniform, with a constantrate of rotation and constanttangential speed, or non-uniform with a changing rate of rotation. Therotation around a fixed axis of a three-dimensional body involves the circular motion of its parts. The equations of motion describe the movement of thecenter of mass of a body, which remains at a constant distance from theaxis of rotation. In circular motion, the distance between the body and a fixed point on its surface remains the same, i.e., the body is assumedrigid.

Examples of circular motion include: special satellite orbits around the Earth (circular orbits), aceiling fan's blades rotating around a hub, a stone that is tied to a rope and is being swung in circles, a car turning through a curve in arace track, an electron moving perpendicular to a uniformmagnetic field, and agear turning inside a mechanism.

Since the object'svelocity vector is constantly changing direction, the moving object is undergoingacceleration by acentripetal force in the direction of the center of rotation. Without this acceleration, the object would move in a straight line, according toNewton's laws of motion.

Uniform circular motion

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Figure 1: Velocityv and accelerationa in uniform circular motion at angular rateω; the speed is constant, but the velocity is always tangential to the orbit; the acceleration has constant magnitude, but always points toward the center of rotation.
Figure 2: The velocity vectors at timet and timet +dt are moved from the orbit on the left to new positions where their tails coincide, on the right. Because the velocity is fixed in magnitude atv =rω, the velocity vectors also sweep out a circular path at angular rateω. Asdt → 0, the acceleration vectora becomes perpendicular tov, which means it points toward the center of the orbit in the circle on the left. Angleωdt is the very small angle between the two velocities and tends to zero asdt → 0.
Figure 3: (Left) Ball in a circular motion – rope provides centripetal force to keep the ball in a circle (Right) Rope is cut and the ball continues in a straight line with the velocity at the time of cutting the rope, in accord with Newton's law of inertia, because centripetal force is no longer there.

Inphysics,uniform circular motion describes the motion of a body traversing acircular path at a constantspeed. Since the body describes circular motion, itsdistance from the axis of rotation remains constant at all times. Though the body's speed is constant, itsvelocity is not constant: velocity, avector quantity, depends on both the body's speed and its direction of travel. This changing velocity indicates the presence of an acceleration; thiscentripetal acceleration is of constant magnitude and directed at all times toward the axis of rotation. This acceleration is, in turn, produced by acentripetal force which is also constant in magnitude and directed toward the axis of rotation.

In the case ofrotation around a fixed axis of arigid body that is not negligibly small compared to the radius of the path, each particle of the body describes a uniform circular motion with the same angular velocity, but with velocity and acceleration varying with the position with respect to the axis.

Formula

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Figure 1: Vector relationships for uniform circular motion; vectorΩ representing the rotation is normal to the plane of the orbit.

For motion in a circle ofradiusr, the circumference of the circle isC = 2πr. If the period for one rotation isT, the angular rate of rotation, also known asangular velocity,ω is:ω=2πT=2πf=dθdt{\displaystyle \omega ={\frac {2\pi }{T}}=2\pi f={\frac {d\theta }{dt}}} and the units are radians/second.

The speed of the object traveling the circle is:v=2πrT=ωr{\displaystyle v={\frac {2\pi r}{T}}=\omega r}

The angleθ swept out in a timet is:θ=2πtT=ωt{\displaystyle \theta =2\pi {\frac {t}{T}}=\omega t}

Theangular acceleration,α, of the particle is:α=dωdt{\displaystyle \alpha ={\frac {d\omega }{dt}}}

In the case of uniform circular motion,α will be zero.

The acceleration due to change in the direction is:ac=v2r=ω2r{\displaystyle a_{c}={\frac {v^{2}}{r}}=\omega ^{2}r}

Thecentripetal andcentrifugal force can also be found using acceleration:Fc=p˙=m˙=0mac=mv2r{\displaystyle F_{c}={\dot {p}}\mathrel {\overset {{\dot {m}}=0}{=}} ma_{c}={\frac {mv^{2}}{r}}}

The vector relationships are shown in Figure 1. The axis of rotation is shown as a vectorω perpendicular to the plane of the orbit and with a magnitudeω =dθ /dt. The direction ofω is chosen using theright-hand rule. With this convention for depicting rotation, the velocity is given by a vectorcross product asv=ω×r,{\displaystyle \mathbf {v} ={\boldsymbol {\omega }}\times \mathbf {r} ,}which is a vector perpendicular to bothω andr(t), tangential to the orbit, and of magnitudeωr. Likewise, the acceleration is given bya=ω×v=ω×(ω×r),{\displaystyle \mathbf {a} ={\boldsymbol {\omega }}\times \mathbf {v} ={\boldsymbol {\omega }}\times \left({\boldsymbol {\omega }}\times \mathbf {r} \right),}which is a vector perpendicular to bothω andv(t) of magnitudeω |v| =ω2r and directed exactly opposite tor(t).[1]

In the simplest case the speed, mass, and radius are constant.

Consider a body of one kilogram, moving in a circle ofradius one metre, with anangular velocity of oneradian persecond.

In polar coordinates

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See also:Velocity § Polar coordinates
Figure 4: Polar coordinates for circular trajectory. On the left is a unit circle showing the changesdu^R{\displaystyle \mathbf {d{\hat {\mathbf {u} }}_{R}} } anddu^θ{\displaystyle \mathbf {d{\hat {\mathbf {u} }}_{\theta }} } in the unit vectorsu^R{\displaystyle \mathbf {{\hat {\mathbf {u} }}_{R}} } andu^θ{\displaystyle \mathbf {{\hat {\mathbf {u} }}_{\theta }} } for a small incrementdθ{\displaystyle d\theta } in angleθ{\displaystyle \theta }.

During circular motion, the body moves on a curve that can be described in thepolar coordinate system as a fixed distanceR from the center of the orbit taken as the origin, oriented at an angleθ(t) from some reference direction. See Figure 4. The displacementvectorr{\displaystyle \mathbf {r} } is the radial vector from the origin to the particle location:r(t)=Ru^R(t),{\displaystyle \mathbf {r} (t)=R{\hat {\mathbf {u} }}_{R}(t)\,,}whereu^R(t){\displaystyle {\hat {\mathbf {u} }}_{R}(t)} is theunit vector parallel to the radius vector at timet and pointing away from the origin. It is convenient to introduce the unit vectororthogonal tou^R(t){\displaystyle {\hat {\mathbf {u} }}_{R}(t)} as well, namelyu^θ(t){\displaystyle {\hat {\mathbf {u} }}_{\theta }(t)}. It is customary to orientu^θ(t){\displaystyle {\hat {\mathbf {u} }}_{\theta }(t)} to point in the direction of travel along the orbit.

The velocity is the time derivative of the displacement:v(t)=ddtr(t)=dRdtu^R(t)+Rdu^Rdt.{\displaystyle \mathbf {v} (t)={\frac {d}{dt}}\mathbf {r} (t)={\frac {dR}{dt}}{\hat {\mathbf {u} }}_{R}(t)+R{\frac {d{\hat {\mathbf {u} }}_{R}}{dt}}\,.}

Because the radius of the circle is constant, the radial component of the velocity is zero. The unit vectoru^R(t){\displaystyle {\hat {\mathbf {u} }}_{R}(t)} has a time-invariant magnitude of unity, so as time varies its tip always lies on a circle of unit radius, with an angleθ the same as the angle ofr(t){\displaystyle \mathbf {r} (t)}. If the particle displacement rotates through an angledθ in timedt, so doesu^R(t){\displaystyle {\hat {\mathbf {u} }}_{R}(t)}, describing an arc on the unit circle of magnitudedθ. See the unit circle at the left of Figure 4. Hence:du^Rdt=dθdtu^θ(t),{\displaystyle {\frac {d{\hat {\mathbf {u} }}_{R}}{dt}}={\frac {d\theta }{dt}}{\hat {\mathbf {u} }}_{\theta }(t)\,,}where the direction of the change must be perpendicular tou^R(t){\displaystyle {\hat {\mathbf {u} }}_{R}(t)} (or, in other words, alongu^θ(t){\displaystyle {\hat {\mathbf {u} }}_{\theta }(t)}) because any changedu^R(t){\displaystyle d{\hat {\mathbf {u} }}_{R}(t)} in the direction ofu^R(t){\displaystyle {\hat {\mathbf {u} }}_{R}(t)} would change the size ofu^R(t){\displaystyle {\hat {\mathbf {u} }}_{R}(t)}. The sign is positive because an increase indθ implies the object andu^R(t){\displaystyle {\hat {\mathbf {u} }}_{R}(t)} have moved in the direction ofu^θ(t){\displaystyle {\hat {\mathbf {u} }}_{\theta }(t)}.Hence the velocity becomes:v(t)=ddtr(t)=Rdu^Rdt=Rdθdtu^θ(t)=Rωu^θ(t).{\displaystyle \mathbf {v} (t)={\frac {d}{dt}}\mathbf {r} (t)=R{\frac {d{\hat {\mathbf {u} }}_{R}}{dt}}=R{\frac {d\theta }{dt}}{\hat {\mathbf {u} }}_{\theta }(t)=R\omega {\hat {\mathbf {u} }}_{\theta }(t)\,.}

The acceleration of the body can also be broken into radial and tangential components. The acceleration is the time derivative of the velocity:a(t)=ddtv(t)=ddt(Rωu^θ(t))=R(dωdtu^θ(t)+ωdu^θdt).{\displaystyle {\begin{aligned}\mathbf {a} (t)&={\frac {d}{dt}}\mathbf {v} (t)={\frac {d}{dt}}\left(R\omega {\hat {\mathbf {u} }}_{\theta }(t)\right)\\&=R\left({\frac {d\omega }{dt}}{\hat {\mathbf {u} }}_{\theta }(t)+\omega {\frac {d{\hat {\mathbf {u} }}_{\theta }}{dt}}\right)\,.\end{aligned}}}

The time derivative ofu^θ(t){\displaystyle {\hat {\mathbf {u} }}_{\theta }(t)} is found the same way as foru^R(t){\displaystyle {\hat {\mathbf {u} }}_{R}(t)}. Again,u^θ(t){\displaystyle {\hat {\mathbf {u} }}_{\theta }(t)} is a unit vector and its tip traces a unit circle with an angle that isπ/2 +θ. Hence, an increase in angledθ byr(t){\displaystyle \mathbf {r} (t)} impliesu^θ(t){\displaystyle {\hat {\mathbf {u} }}_{\theta }(t)} traces an arc of magnitudedθ, and asu^θ(t){\displaystyle {\hat {\mathbf {u} }}_{\theta }(t)} is orthogonal tou^R(t){\displaystyle {\hat {\mathbf {u} }}_{R}(t)}, we have:du^θdt=−dθdtu^R(t)=−ωu^R(t),{\displaystyle {\frac {d{\hat {\mathbf {u} }}_{\theta }}{dt}}=-{\frac {d\theta }{dt}}{\hat {\mathbf {u} }}_{R}(t)=-\omega {\hat {\mathbf {u} }}_{R}(t)\,,}where a negative sign is necessary to keepu^θ(t){\displaystyle {\hat {\mathbf {u} }}_{\theta }(t)} orthogonal tou^R(t){\displaystyle {\hat {\mathbf {u} }}_{R}(t)}. (Otherwise, the angle betweenu^θ(t){\displaystyle {\hat {\mathbf {u} }}_{\theta }(t)} andu^R(t){\displaystyle {\hat {\mathbf {u} }}_{R}(t)} woulddecrease with an increase indθ.) See the unit circle at the left of Figure 4. Consequently, the acceleration is:a(t)=R(dωdtu^θ(t)+ωdu^θdt)=Rdωdtu^θ(t)−ω2Ru^R(t).{\displaystyle {\begin{aligned}\mathbf {a} (t)&=R\left({\frac {d\omega }{dt}}{\hat {\mathbf {u} }}_{\theta }(t)+\omega {\frac {d{\hat {\mathbf {u} }}_{\theta }}{dt}}\right)\\&=R{\frac {d\omega }{dt}}{\hat {\mathbf {u} }}_{\theta }(t)-\omega ^{2}R{\hat {\mathbf {u} }}_{R}(t)\,.\end{aligned}}}

Thecentripetal acceleration is the radial component, which is directed radially inward:aR(t)=−ω2Ru^R(t),{\displaystyle \mathbf {a} _{R}(t)=-\omega ^{2}R{\hat {\mathbf {u} }}_{R}(t)\,,}while the tangential component changes themagnitude of the velocity:aθ(t)=Rdωdtu^θ(t)=dRωdtu^θ(t)=d|v(t)|dtu^θ(t).{\displaystyle \mathbf {a} _{\theta }(t)=R{\frac {d\omega }{dt}}{\hat {\mathbf {u} }}_{\theta }(t)={\frac {dR\omega }{dt}}{\hat {\mathbf {u} }}_{\theta }(t)={\frac {d\left|\mathbf {v} (t)\right|}{dt}}{\hat {\mathbf {u} }}_{\theta }(t)\,.}

Using complex numbers

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Circular motion can be described usingcomplex numbers. Let thex axis be the real axis and they{\displaystyle y} axis be the imaginary axis. The position of the body can then be given asz{\displaystyle z}, a complex "vector":z=x+iy=R(cos⁡[θ(t)]+isin⁡[θ(t)])=Reiθ(t),{\displaystyle z=x+iy=R\left(\cos[\theta (t)]+i\sin[\theta (t)]\right)=Re^{i\theta (t)}\,,}wherei is theimaginary unit, andθ(t){\displaystyle \theta (t)} is the argument of the complex number as a function of time,t.

Since the radius is constant:R˙=R¨=0,{\displaystyle {\dot {R}}={\ddot {R}}=0\,,}where adot indicates differentiation in respect of time.

With this notation, the velocity becomes:v=z˙=ddt(Reiθ[t])=Rddt(eiθ[t])=Reiθ(t)ddt(iθ[t])=iRθ˙(t)eiθ(t)=iωReiθ(t)=iωz{\displaystyle v={\dot {z}}={\frac {d}{dt}}\left(Re^{i\theta [t]}\right)=R{\frac {d}{dt}}\left(e^{i\theta [t]}\right)=Re^{i\theta (t)}{\frac {d}{dt}}\left(i\theta [t]\right)=iR{\dot {\theta }}(t)e^{i\theta (t)}=i\omega Re^{i\theta (t)}=i\omega z}and the acceleration becomes:a=v˙=iω˙z+iωz˙=(iω˙−ω2)z=(iω˙−ω2)Reiθ(t)=−ω2Reiθ(t)+ω˙eiπ2Reiθ(t).{\displaystyle {\begin{aligned}a&={\dot {v}}=i{\dot {\omega }}z+i\omega {\dot {z}}=\left(i{\dot {\omega }}-\omega ^{2}\right)z\\&=\left(i{\dot {\omega }}-\omega ^{2}\right)Re^{i\theta (t)}\\&=-\omega ^{2}Re^{i\theta (t)}+{\dot {\omega }}e^{i{\frac {\pi }{2}}}Re^{i\theta (t)}\,.\end{aligned}}}

The first term is opposite in direction to the displacement vector and the second is perpendicular to it, just like the earlier results shown before.

Velocity

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Figure 1 illustrates velocity and acceleration vectors for uniform motion at four different points in the orbit. Because the velocityv is tangent to the circular path, no two velocities point in the same direction. Although the object has a constantspeed, itsdirection is always changing. This change in velocity is caused by an accelerationa, whose magnitude is (like that of the velocity) held constant, but whose direction also is always changing. Theacceleration points radially inwards (centripetally) and is perpendicular to the velocity. This acceleration is known as centripetal acceleration.

For a path of radiusr, when an angleθ is swept out, the distance traveled on theperiphery of the orbit iss =rθ. Therefore, the speed of travel around the orbit isv=rdθdt=rω,{\displaystyle v=r{\frac {d\theta }{dt}}=r\omega ,}where the angular rate of rotation isω. (By rearrangement,ω =v/r.) Thus,v is a constant, and the velocity vectorv also rotates with constant magnitudev, at the same angular rateω.

Relativistic circular motion

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In this case, the three-acceleration vector is perpendicular to the three-velocity vector,u⋅a=0.{\displaystyle \mathbf {u} \cdot \mathbf {a} =0.}and the square of proper acceleration, expressed as a scalar invariant, the same in all reference frames,α2=γ4a2+γ6(u⋅a)2/c2,{\displaystyle \alpha ^{2}=\gamma ^{4}a^{2}+\gamma ^{6}\left(\mathbf {u} \cdot \mathbf {a} \right)^{2}/c^{2},}becomes the expression for circular motion,α2=γ4a2.{\displaystyle \alpha ^{2}=\gamma ^{4}a^{2}.}or, taking the positive square root and using the three-acceleration, we arrive at the proper acceleration for circular motion:α=γ2v2r.{\displaystyle \alpha =\gamma ^{2}{\frac {v^{2}}{r}}.}

Acceleration

[edit]
Main article:Acceleration

The left-hand circle in Figure 2 is the orbit showing the velocity vectors at two adjacent times. On the right, these two velocities are moved so their tails coincide. Because speed is constant, the velocity vectors on the right sweep out a circle as time advances. For a swept angledθ =ωdt the change inv is a vector at right angles tov and of magnitudevdθ, which in turn means that the magnitude of the acceleration is given byac=vdθdt=vω=v2r{\displaystyle a_{c}=v{\frac {d\theta }{dt}}=v\omega ={\frac {v^{2}}{r}}}

Centripetal acceleration for some values of radius and magnitude of velocity
|v|
r
1 m/s
3.6 km/h
2.2 mph		2 m/s
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4.5 mph		5 m/s
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36 km/h
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180 km/h
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360 km/h
220 mph
SlowwalkBicycleCity carAerobatics
10 cm
3.9 in		Laboratory
centrifuge		10 m/s2
1.0g		40 m/s2
4.1g		250 m/s2
25g		1.0 km/s2
100g		4.0 km/s2
410g		25 km/s2
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0.20g		8.0 m/s2
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20g		800 m/s2
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1 m
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carousel		1.0 m/s2
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10g		400 m/s2
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2 m
6.6 ft		500 mm/s2
0.051g		2.0 m/s2
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5.1g		200 m/s2
20g		1.3 km/s2
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0.020g		800 mm/s2
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0.51g		20 m/s2
2.0g		80 m/s2
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33 ft		Roller-coaster
vertical loop		100 mm/s2
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0.041g		2.5 m/s2
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66 ft		50 mm/s2
0.0051g		200 mm/s2
0.020g		1.3 m/s2
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2g		130 m/s2
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0.0020g		80 mm/s2
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0.051g		2.0 m/s2
0.20g		8.0 m/s2
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5.1g		200 m/s2
20g
100 m
330 ft		Freeway
on-ramp		10 mm/s2
0.0010g		40 mm/s2
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0.025g		1.0 m/s2
0.10g		4.0 m/s2
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2.5g		100 m/s2
10g
200 m
660 ft		5.0 mm/s2
0.00051g		20 mm/s2
0.0020g		130 m/s2
0.013g		500 mm/s2
0.051g		2.0 m/s2
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0.082g		5.0 m/s2
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2.0g
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3300 ft		High-speed
railway		1.0 mm/s2
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0.0025g		100 mm/s2
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0.041g		2.5 m/s2
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1.0g

Non-uniform circular motion

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Velocity and acceleration in non-uniform circular motion.

Innon-uniform circular motion, an object moves in a circular path with varyingspeed. Since the speed is changing, there istangential acceleration in addition to normal acceleration.

The net acceleration is directed towards the interior of the circle (but does not pass through its center).

The net acceleration may be resolved into two components: tangential acceleration and centripetal acceleration. Unlike tangential acceleration, centripetal acceleration is present in both uniform and non-uniform circular motion.

This diagram shows the normal force (n) pointing in other directions rather than opposite to the weight force.

In non-uniform circular motion, thenormal force does not always point to the opposite direction ofweight.

Here, 'n' is the normal force.

The normal force is actually the sum of the radial and tangential forces. The component of weight force is responsible for the tangential force (when we neglect friction). The centripetal force is due to the change in the direction of velocity.

The normal force and weight may also point in the same direction. Both forces can point downwards, yet the object will remain in a circular path without falling down.

The normal force can point downwards.

The normal forcecan point downwards. Considering that the object is a person sitting inside a plane moving in a circle, the two forces (weight and normal force) will point down only when the plane reaches the top of the circle. The reason for this is that the normal force is the sum of the tangential force and centripetal force. The tangential force is zero at the top (as no work is performed when the motion is perpendicular to the direction of force). Since weight is perpendicular to the direction of motion of the object at the top of the circle and the centripetal force points downwards, the normal force will point down as well.

From a logical standpoint, a person travelling in that plane will be upside down at the top of the circle. At that moment, the person's seat is actually pushing down on the person, which is the normal force.

The reason why an object does not fall down when subjected to only downward forces is a simple one. Once an object is thrown into the air, there is only the downward gravitational force that acts on the object. That does not mean that once an object is thrown into the air, it will fall instantly. Thevelocity of the object keeps it up in the air. The first ofNewton's laws of motion states that an object'sinertia keeps it in motion; since the object in the air has a velocity, it will tend to keep moving in that direction.

A varyingangular speed for an object moving in a circular path can also be achieved if the rotating body does not have a homogeneous mass distribution.[2]

One can deduce the formulae of speed, acceleration and jerk, assuming that all the variables depend ont{\displaystyle t}:r=RuR{\displaystyle \mathbf {r} =R\mathbf {u} _{R}}u˙R=ωuθ{\displaystyle {\dot {\mathbf {u} }}_{R}=\omega \mathbf {u} _{\theta }}u˙θ=−ωuR{\displaystyle {\dot {\mathbf {u} }}_{\theta }=-\omega \mathbf {u} _{R}}v=ddtr=r˙=R˙uR+Rωuθ{\displaystyle \mathbf {v} ={\frac {d}{dt}}\mathbf {r} ={\dot {\mathbf {r} }}={\dot {R}}\mathbf {u} _{R}+R\omega \mathbf {u} _{\theta }}a=ddtv=v˙=R¨uR+(R˙ωuθ+R˙ωuθ)+Rω˙uθ−Rω2uR{\displaystyle \mathbf {a} ={\frac {d}{dt}}\mathbf {v} ={\dot {\mathbf {v} }}={\ddot {R}}\mathbf {u} _{R}+\left({\dot {R}}\omega \mathbf {u} _{\theta }+{\dot {R}}\omega \mathbf {u} _{\theta }\right)+R{\dot {\omega }}\mathbf {u} _{\theta }-R\omega ^{2}\mathbf {u} _{R}}

j=ddta=a˙=R¨˙uR+R¨ωuθ+(2R¨ωuθ+2R˙ω˙uθ−2R˙ω2uR)+R˙ω˙uθ+Rω¨uθ−Rω˙ωuR−R˙ω2uR−R2ω˙ωuR−Rω3uθ{\displaystyle \mathbf {j} ={\frac {d}{dt}}\mathbf {a} ={\dot {\mathbf {a} }}={\dot {\ddot {R}}}\mathbf {u} _{R}+{\ddot {R}}\omega \mathbf {u} _{\theta }+\left(2{\ddot {R}}\omega \mathbf {u} _{\theta }+2{\dot {R}}{\dot {\omega }}\mathbf {u} _{\theta }-2{\dot {R}}\omega ^{2}\mathbf {u} _{R}\right)+{\dot {R}}{\dot {\omega }}\mathbf {u} _{\theta }+R{\ddot {\omega }}\mathbf {u} _{\theta }-R{\dot {\omega }}\omega \mathbf {u} _{R}-{\dot {R}}\omega ^{2}\mathbf {u} _{R}-R2{\dot {\omega }}\omega \mathbf {u} _{R}-R\omega ^{3}\mathbf {u} _{\theta }}

j=(R¨˙−3R˙ω2−3Rω˙ω)uR+(3R¨ω+3R˙ω˙+Rω¨−Rω3)uθ{\displaystyle \mathbf {j} =\left({\dot {\ddot {R}}}-3{\dot {R}}\omega ^{2}-3R{\dot {\omega }}\omega \right)\mathbf {u} _{R}+\left(3{\ddot {R}}\omega +3{\dot {R}}{\dot {\omega }}+R{\ddot {\omega }}-R\omega ^{3}\right)\mathbf {u} _{\theta }}

Further transformations may involvecurvature=c=1R,ω=vR=vc{\displaystyle curvature=c={\frac {1}{R}},\omega ={\frac {v}{R}}=vc} and their corresponding derivatives:R˙=−c˙c2R¨=2(c˙)2c3−c¨c2ω˙=v˙R−R˙vR2=v˙c+vc˙.{\displaystyle {\begin{aligned}{\dot {R}}&=-{\frac {\dot {c}}{c^{2}}}\\{\ddot {R}}&={\frac {2\left({\dot {c}}\right)^{2}}{c^{3}}}-{\frac {\ddot {c}}{c^{2}}}\\{\dot {\omega }}&={\frac {{\dot {v}}R-{\dot {R}}v}{R^{2}}}={\dot {v}}c+v{\dot {c}}.\\\end{aligned}}}

Applications

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Solving applications dealing with non-uniform circular motion involves force analysis. With a uniform circular motion, the only force acting upon an object traveling in a circle is the centripetal force. In a non-uniform circular motion, there are additional forces acting on the object due to a non-zero tangential acceleration. Although there are additional forces acting upon the object, the sum of all the forces acting on the object will have to be equal to the centripetal force.Fnet=ma=mar=mv2r=Fc{\displaystyle {\begin{aligned}F_{\text{net}}&=ma\\&=ma_{r}\\&={\frac {mv^{2}}{r}}\\&=F_{c}\end{aligned}}}

Radial acceleration is used when calculating the total force. Tangential acceleration is not used in calculating total force because it is not responsible for keeping the object in a circular path. The only acceleration responsible for keeping an object moving in a circle is the radial acceleration. Since the sum of all forces is the centripetal force, drawing centripetal force into a free body diagram is not necessary and usually not recommended.

UsingFnet=Fc{\displaystyle F_{\text{net}}=F_{c}}, we can draw free body diagrams to list all the forces acting on an object and then set it equal toFc{\displaystyle F_{c}}. Afterward, we can solve for whatever is unknown (this can be mass, velocity, radius of curvature, coefficient of friction, normal force, etc.). For example, the visual above showing an object at the top of a semicircle would be expressed asFc=n+mg{\displaystyle F_{c}=n+mg}.

In a uniform circular motion, the total acceleration of an object in a circular path is equal to the radial acceleration. Due to the presence of tangential acceleration in a non uniform circular motion, that does not hold true any more. To find the total acceleration of an object in a non uniform circular, find the vector sum of the tangential acceleration and the radial acceleration.ar2+at2=a{\displaystyle {\sqrt {a_{r}^{2}+a_{t}^{2}}}=a}

Radial acceleration is still equal tov2r{\textstyle {\frac {v^{2}}{r}}}. Tangential acceleration is simply the derivative of the speed at any given point:at=dvdt{\textstyle a_{t}={\frac {dv}{dt}}}. This root sum of squares of separate radial and tangential accelerations is only correct for circular motion; for general motion within a plane with polar coordinates(r,θ){\displaystyle (r,\theta )}, the Coriolis termac=2(drdt)(dθdt){\textstyle a_{c}=2\left({\frac {dr}{dt}}\right)\left({\frac {d\theta }{dt}}\right)} should be added toat{\displaystyle a_{t}}, whereas radial acceleration then becomesar=−v2r+d2rdt2{\textstyle a_{r}={\frac {-v^{2}}{r}}+{\frac {d^{2}r}{dt^{2}}}}.

See also

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References

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  1. ^Knudsen, Jens M.; Hjorth, Poul G. (2000).Elements of Newtonian mechanics: including nonlinear dynamics (3 ed.). Springer. p. 96.ISBN 3-540-67652-X.
  2. ^Gomez, R W; Hernandez-Gomez, J J; Marquina, V (25 July 2012)."A jumping cylinder on an inclined plane".Eur. J. Phys.33 (5). IOP:1359–1365.arXiv:1204.0600.Bibcode:2012EJPh...33.1359G.doi:10.1088/0143-0807/33/5/1359.S2CID 55442794. Retrieved25 April 2016.

External links

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Linear/translational quantitiesAngular/rotational quantities
Dimensions1LL2Dimensions1θθ2
Ttime:t
s		absement:A
m s		Ttime:t
s
1distance:d,position:r,s,x,displacement
m		area:A
m2		1angle:θ,angular displacement:θ
rad		solid angle:Ω
rad2, sr
T−1frequency:f
s−1,Hz		speed:v,velocity:v
m s−1		kinematic viscosity:ν,
specific angular momentum: h
m2 s−1		T−1frequency:f,rotational speed:n,rotational velocity:n
s−1,Hz		angular speed:ω,angular velocity:ω
rad s−1
T−2acceleration:a
m s−2		T−2rotational acceleration
s−2		angular acceleration:α
rad s−2
T−3jerk:j
m s−3		T−3angular jerk:ζ
rad s−3
Mmass:m
kg		weighted position:M ⟨x⟩ = ∑mxmoment of inertia: I
kg m2		ML
MT−1Mass flow rate:m˙{\displaystyle {\dot {m}}}
kg s−1		momentum:p,impulse:J
kg m s−1,N s		action:𝒮,actergy:ℵ
kg m2 s−1,J s		MLT−1angular momentum:L,angular impulse:ΔL
kg m rad s−1
MT−2force:F,weight:Fg
kg m s−2,N		energy:E,work:W,Lagrangian:L
kg m2 s−2,J		MLT−2torque:τ,moment:M
kg m rad s−2,N m
MT−3yank:Y
kg m s−3, N s−1		power:P
kg m2 s−3, W		MLT−3rotatum:P
kg m rad s−3, N m s−1
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