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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 { size 8{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.

Questions & Answers

show that the set of all natural number form semi group under the composition of addition
Nikhil Reply
what is the meaning
Dominic
explain and give four Example hyperbolic function
Lukman Reply
_3_2_1
felecia
⅗ ⅔½
felecia
_½+⅔-¾
felecia
The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
SABAL Reply
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
Pawel
ok
Ifeanyi
on number 2 question How did you got 2x +2
Ifeanyi
combine like terms. x + x + 2 is same as 2x + 2
Pawel
x*x=2
felecia
2+2x=
felecia
×/×+9+6/1
Debbie
Q2 x+(x+2)+(x+4)=60 3x+6=60 3x+6-6=60-6 3x=54 3x/3=54/3 x=18 :. The numbers are 18,20 and 22
Naagmenkoma
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
mariel Reply
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
Pawel
how do I set up the problem?
Harshika Reply
what is a solution set?
Harshika
find the subring of gaussian integers?
Rofiqul
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Shirley Reply
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Mark
I need quadratic equation link to Alpa Beta
Abdullahi Reply
find the value of 2x=32
Felix Reply
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
Beyan
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Mark
16
Makan
x=16
Makan
use the y -intercept and slope to sketch the graph of the equation y=6x
Only Reply
how do we prove the quadratic formular
Seidu Reply
please help me prove quadratic formula
Darius
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
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Seidu
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Opoku
what is math number
Tric Reply
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
can you teacch how to solve that🙏
Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
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Source:  OpenStax, Introduction to applied math and physics. OpenStax CNX. Oct 04, 2012 Download for free at http://cnx.org/content/col11426/1.3
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