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Learning objectives

By the end of this section, you will be able to:

  • Describe uniform circular motion.
  • Explain nonuniform circular motion.
  • Calculate angular acceleration of an object.
  • Observe the link between linear and angular acceleration.

Uniform Circular Motion and Gravitation discussed only uniform circular motion, which is motion in a circle at constant speed and, hence, constant angular velocity. Recall that angular velocity ω size 12{ω} {} was defined as the time rate of change of angle θ size 12{θ} {} :

ω = Δ θ Δ t , size 12{ω= { {Δθ} over {Δt} } ","} {}

where θ size 12{θ} {} is the angle of rotation as seen in [link] . The relationship between angular velocity ω size 12{ω} {} and linear velocity v size 12{v} {} was also defined in Rotation Angle and Angular Velocity as

v = size 12{v=rω} {}

or

ω = v r , size 12{ω= { {v} over {r} } } {}

where r size 12{r} {} is the radius of curvature, also seen in [link] . According to the sign convention, the counter clockwise direction is considered as positive direction and clockwise direction as negative

The given figure shows counterclockwise circular motion with a horizontal line, depicting radius r, drawn from the center of the circle to the right side on its circumference and another line is drawn in such a manner that it makes an acute angle delta theta with the horizontal line. Tangential velocity vectors are indicated at the end of the two lines. At the bottom right side of the figure, the formula for angular velocity is given as v upon r.
This figure shows uniform circular motion and some of its defined quantities.

Angular velocity is not constant when a skater pulls in her arms, when a child starts up a merry-go-round from rest, or when a computer's hard disk slows to a halt when switched off. In all these cases, there is an angular acceleration    , in which ω size 12{ω} {} changes. The faster the change occurs, the greater the angular acceleration. Angular acceleration α size 12{α} {} is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows:

α = Δ ω Δ t , size 12{α= { {Δω} over {Δt} } ","} {}

where Δ ω size 12{Δω} {} is the change in angular velocity    and Δ t size 12{Δt} {} is the change in time. The units of angular acceleration are rad/s /s size 12{ left ("rad/s" right )"/s"} {} , or rad/s 2 size 12{"rad/s" rSup { size 8{2} } } {} . If ω size 12{ω} {} increases, then α size 12{α} {} is positive. If ω size 12{ω} {} decreases, then α size 12{α} {} is negative.

Calculating the angular acceleration and deceleration of a bike wheel

Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the angular acceleration in rad/s 2 size 12{"rad/s" rSup { size 8{2} } } {} . (b) If she now slams on the brakes, causing an angular acceleration of 87.3 rad/s 2 size 12{"-87" "." 3`"rad/s" rSup { size 8{2} } } {} , how long does it take the wheel to stop?

Strategy for (a)

The angular acceleration can be found directly from its definition in α = Δ ω Δ t size 12{α= { {Δω} over {Δt} } } {} because the final angular velocity and time are given. We see that Δ ω size 12{Δω} {} is 250 rpm and Δ t size 12{Δt} {} is 5.00 s.

Solution for (a)

Entering known information into the definition of angular acceleration, we get

α = Δ ω Δ t = 250 rpm 5.00 s . alignl { stack { size 12{α= { {Δω} over {Δt} } } {} #size 12{ {}= { {"250"" rpm"} over {5 "." "00 s"} } "."} {} } } {}

Because Δ ω size 12{Δω} {} is in revolutions per minute (rpm) and we want the standard units of rad/s 2 size 12{"rad/s" rSup { size 8{2} } } {} for angular acceleration, we need to convert Δ ω size 12{Δω} {} from rpm to rad/s:

Δ ω = 250 rev min 2π rad rev 1 min 60 sec = 26.2 rad s . alignl { stack { size 12{Δω="250" { {"rev"} over {"min"} } cdot { {2π" rad"} over {"60" "." "0 s"} } } {} #size 12{ {}="26" "." 2 { {"rad"} over {"s"} } } {} } } {}

Entering this quantity into the expression for α size 12{α} {} , we get

α = Δ ω Δ t = 26.2 rad/s 5.00 s = 5.24  rad/s 2 . alignl { stack { size 12{α= { {Δω} over {Δt} } } {} #size 12{ {}= { {"26" "." 2" rad/s"} over {5 "." "00"" s"} } "." } {} # size 12{ {}=5 "." "24"" rad/s" rSup { size 8{2} } } {}} } {}

Strategy for (b)

In this part, we know the angular acceleration and the initial angular velocity. We can find the stoppage time by using the definition of angular acceleration and solving for Δ t size 12{Δt} {} , yielding

Δ t = Δ ω α . size 12{Δt= { {Δω} over {α} } "."} {}

Solution for (b)

Here the angular velocity decreases from 26.2 rad/s size 12{"26" "." 2`"rad/s"} {} (250 rpm) to zero, so that Δ ω size 12{Δω} {} is 26.2 rad/s , and α size 12{α} {} is given to be 87.3 rad/s 2 size 12{"-87" "." 3`"rad/s" rSup { size 8{2} } } {} . Thus,

Δ t = 26.2 rad/s 87.3 rad/s 2 = 0.300 s. alignl { stack { size 12{Δt= { { - "26" "." 2`"rad/s"} over { - "87" "." 3`"rad/s" rSup { size 8{2} } } } } {} #=0 "." "300"`"s" "." {} } } {}

Discussion

Note that the angular acceleration as the girl spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero. In both cases, the relationships are analogous to what happens with linear motion. For example, there is a large deceleration when you crash into a brick wall—the velocity change is large in a short time interval.

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Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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