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If the bicycle in the preceding example had been on its wheels instead of upside-down, it would first have accelerated along the ground and then come to a stop. This connection between circular motion and linear motion needs to be explored. For example, it would be useful to know how linear and angular acceleration are related. In circular motion, linear acceleration is tangent to the circle at the point of interest, as seen in [link] . Thus, linear acceleration is called tangential acceleration     a t size 12{a rSub { size 8{t} } } {} .

In the figure, a semicircle is drawn, with its radius r, shown here as a line segment. The anti-clockwise motion of the circle is shown with an arrow on the path of the circle. Tangential velocity vector, v, of the point, which is on the meeting point of radius with the circle, is shown as a green arrow and the linear acceleration, a-t is shown as a yellow arrow in the same direction along v.
In circular motion, linear acceleration a size 12{a} {} , occurs as the magnitude of the velocity changes: a size 12{a} {} is tangent to the motion. In the context of circular motion, linear acceleration is also called tangential acceleration a t size 12{a rSub { size 8{t} } } {} .

Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction. We know from Uniform Circular Motion and Gravitation that in circular motion centripetal acceleration, a c size 12{a rSub { size 8{t} } } {} , refers to changes in the direction of the velocity but not its magnitude. An object undergoing circular motion experiences centripetal acceleration, as seen in [link] . Thus, a t size 12{a rSub { size 8{t} } } {} and a c size 12{a rSub { size 8{t} } } {} are perpendicular and independent of one another. Tangential acceleration a t size 12{a rSub { size 8{t} } } {} is directly related to the angular acceleration α size 12{α} {} and is linked to an increase or decrease in the velocity, but not its direction.

In the figure, a semicircle is drawn, with its radius r, shown here as a line segment. The anti-clockwise motion of the circle is shown with an arrow on the path of the circle. Tangential velocity vector, v, of the point, which is on the meeting point of radius with the circle, is shown as a green arrow and the linear acceleration, a sub t is shown as a yellow arrow in the same direction along v. The centripetal acceleration, a sub c, is also shown as a yellow arrow drawn perpendicular to a sub t, toward the direction of the center of the circle. A label in the figures states a sub t affects magnitude and a sub c affects direction.
Centripetal acceleration a c size 12{a rSub { size 8{t} } } {} occurs as the direction of velocity changes; it is perpendicular to the circular motion. Centripetal and tangential acceleration are thus perpendicular to each other.

Now we can find the exact relationship between linear acceleration a t size 12{a rSub { size 8{t} } } {} and angular acceleration α size 12{α} {} . Because linear acceleration is proportional to a change in the magnitude of the velocity, it is defined (as it was in One-Dimensional Kinematics ) to be

a t = Δ v Δ t . size 12{a rSub { size 8{t} } = { {Δv} over {Δt} } "."} {}

For circular motion, note that v = size 12{v=rω} {} , so that

a t = Δ Δ t . size 12{a rSub { size 8{t} } = { {Δ left (rω right )} over {Δt} } "."} {}

The radius r size 12{r} {} is constant for circular motion, and so Δ ( ) = r ( Δ ω ) size 12{Δ \( rω \) =r \( Δω \) } {} . Thus,

a t = r Δ ω Δ t . size 12{a rSub { size 8{t} } =r { {Δω} over {Δt} } "."} {}

By definition, α = Δ ω Δ t size 12{α= { {Δω} over {Δt} } } {} . Thus,

a t = , size 12{a rSub { size 8{t} } =rα} {}

or

α = a t r . size 12{α= { {a rSub { size 8{t} } } over {r} } } {}

These equations mean that linear acceleration and angular acceleration are directly proportional. The greater the angular acceleration is, the larger the linear (tangential) acceleration is, and vice versa. For example, the greater the angular acceleration of a car’s drive wheels, the greater the acceleration of the car. The radius also matters. For example, the smaller a wheel, the smaller its linear acceleration for a given angular acceleration α size 12{α} {} .

Calculating the angular acceleration of a motorcycle wheel

A powerful motorcycle can accelerate from 0 to 30.0 m/s (about 108 km/h) in 4.20 s. What is the angular acceleration of its 0.320-m-radius wheels? (See [link] .)

The figure shows the right side view of a man riding a motorcycle hence, depicting linear acceleration a of the motorcycle pointing toward the front of the bike as a horizontal arrow and the angular acceleration alpha of its wheels, shown here as curved arrows along the front of both the wheels pointing downward.
The linear acceleration of a motorcycle is accompanied by an angular acceleration of its wheels.

Strategy

We are given information about the linear velocities of the motorcycle. Thus, we can find its linear acceleration a t size 12{a rSub { size 8{t} } } {} . Then, the expression α = a t r size 12{a rSub { size 8{t} } =rα,`````α= { {a rSub { size 8{t} } } over {r} } } {} can be used to find the angular acceleration.

Solution

The linear acceleration is

a t = Δ v Δ t = 30.0 m/s 4.20 s = 7.14 m/s 2 . alignl { stack { size 12{a rSub { size8{t} } = { {Δv} over {Δt} } } {} # `````= { {"30" "." 0" m/s"} over {4 "." "20 s"} } {} #`````=7 "." "14"" m/s" rSup { size 8{2} "."} {} } } {}

We also know the radius of the wheels. Entering the values for a t size 12{a rSub { size 8{t} } } {} and r size 12{r} {} into α = a t r size 12{a rSub { size 8{t} } =rα,`````α= { {a rSub { size 8{t} } } over {r} } } {} , we get

α = a t r = 7.14 m/s 2 0.320 m = 22.3 rad/s 2 . alignl { stack { size 12{α= { {a rSub { size 8{t} } } over {r} } } {} #```= { {7 "." "14"" m/s" rSup { size 8{2} } } over {0 "." "320 m"} } {} # " "="22" "." "3 rad/s" rSup { size 8{2} } {}} } {}

Discussion

Units of radians are dimensionless and appear in any relationship between angular and linear quantities.

So far, we have defined three rotational quantities— θ ω size 12{θ,ω} {} , and α size 12{α} {} . These quantities are analogous to the translational quantities x v size 12{x,v} {} , and a size 12{a} {} . [link] displays rotational quantities, the analogous translational quantities, and the relationships between them.

Rotational and translational quantities
Rotational Translational Relationship
θ size 12{θ} {} x size 12{x} {} θ = x r size 12{θ= { {x} over {r} } } {}
ω size 12{ω} {} v size 12{v} {} ω = v r size 12{ω= { {v} over {r} } } {}
α size 12{α} {} a size 12{a} {} α = a t r size 12{α= { {a rSub { size 8{t} } } over {r} } } {}

Angular acceleration is a vector, having both magnitude and direction. How do we denote its magnitude and direction? Illustrate with an example.

The magnitude of angular acceleration is α size 12{α} {} and its most common units are rad/s 2 size 12{"rad/s" rSup { size 8{2} } } {} . The direction of angular acceleration along a fixed axis is denoted by a + or a – sign, just as the direction of linear acceleration in one dimension is denoted by a + or a – sign. For example, consider a gymnast doing a forward flip. Her angular momentum would be parallel to the mat and to her left. The magnitude of her angular acceleration would be proportional to her angular velocity (spin rate) and her moment of inertia about her spin axis.

Section summary

  • Uniform circular motion is the motion with a constant angular velocity ω = Δ θ Δ t size 12{ω= { {Δθ} over {Δt} } } {} .
  • In non-uniform circular motion, the velocity changes with time and the rate of change of angular velocity (i.e. angular acceleration) is α = Δ ω Δ t size 12{α= { {Δω} over {Δt} } } {} .
  • Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction, given as a t = Δ v Δ t size 12{a rSub { size 8{t} } = { {Δv} over {Δt} } } {} .
  • For circular motion, note that v = size 12{v=rω} {} , so that
    a t = Δ Δ t . size 12{a rSub { size 8{t} } = { {Δ left (rω right )} over {Δt} } } {}
  • The radius r is constant for circular motion, and so Δ = r Δ ω size 12{Δ left (rω right )=rΔω} {} . Thus,
    a t = r Δ ω Δ t . size 12{a rSub { size 8{t} } =r { {Δω} over {Δt} } } {}
  • By definition, Δ ω / Δ t = α size 12{ {Δω} slash {Δt=α} } {} . Thus,
    a t = size 12{a rSub { size 8{t} } =rα} {}

    or

    α = a t r . size 12{α= { {a rSub { size 8{t} } } over {r} } } {}

Conceptual questions

Analogies exist between rotational and translational physical quantities. Identify the rotational term analogous to each of the following: acceleration, force, mass, work, translational kinetic energy, linear momentum, impulse.

Suppose a piece of food is on the edge of a rotating microwave oven plate. Does it experience nonzero tangential acceleration, centripetal acceleration, or both when: (a) The plate starts to spin? (b) The plate rotates at constant angular velocity? (c) The plate slows to a halt?

Problems&Exercises

At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second?

ω = 0 . 737 rev/s size 12{ω= {underline {0 "." "737 rev/s"}} } {}

Integrated Concepts

An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min. (a) What is its angular acceleration in rad/s 2 size 12{"rad/s" rSup { size 8{2} } } {} ? (b) What is the tangential acceleration of a point 9.50 cm from the axis of rotation? (c) What is the radial acceleration in m/s 2 size 12{"m/s" rSup { size 8{2} } } {} and multiples of g size 12{gs} {} of this point at full rpm?

You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial force of 20.0 N. (a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (b) How many turns will the stone make before coming to rest?

(a) 0 . 26 rad/s 2 size 12{ - 0 "." "26 rad/s" rSup { size 8{2} } } {}

(b) 27 rev size 12{"27"`"rev"} {}

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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or infinite solutions?
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Unit 8 - rotational motion. OpenStax CNX. Feb 22, 2016 Download for free at https://legacy.cnx.org/content/col11970/1.1
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