Physics 2210 Fall 2015

smartPhysics 14 Rotational kinematics

11/02/2015

Exam 3: smartPhysics units 10-13 Midterm Exam 3: Day: Fri Nov. 06, 2015 Time: regular class time Section 01 12:55 – 1:45 pm Section 10 02:00 – 02:50 pm

Location: FMAB Practice problems have been posted on Canvas and on the class web page http://www.physics.utah.edu/~woolf/2210_Jui/rev3.pdf

𝒓

It is typically convenient to choose the rotation axis to be the z-axis. Here we have chosen z to point out of this page. The symbol indicates an arrow pointing out of the page. (And this symbol indicates an arrow into the page) The angular position of the body can be determined by the position 𝑟 of an ARBITRARY reference point (which we will mark with a peg )

Rotation About a fixed axis

𝑥 𝑧

𝑦 In the position shown the reference point lies on the x-axis, which we generally choose to be the ZERO (i.e. ORIGIN) of rotation about the z-axis. And we generally pick the counter-clock-wise (CCW) direction to be the positive rotational sense The angular position is then the (azimuthal) angle 𝜃 of the position vector 𝑟 of the reference point measured CCW from the +x axis. smartPhysics uses 𝜃 for the angular position and so we will too…

𝜃

𝑥 𝑧

𝑦

As the object rotates about the axel (rotation axis) , the reference point (remember this is an arbitrary fixed point in the body) traces out a circle

𝜃 = −𝜋

18 rad

𝜃 = +5𝜋18

rad

𝜃 = +11𝜋18 rad

𝜃 = +17𝜋18

rad

𝜃 = +23𝜋18 rad

𝜃 = +29𝜋18 rad

Definition of radians Radians is a natural measure of angle: (90 degrees for a right angle was a Babylonian tradition that tried to match ONE degree to ONE day of the Sun orb Iting around the Earth) Given the wedge colored in light blue subtended by radii separated by angle 𝜃: The arc length S is then given by

𝑠 = 𝑟𝜃 If 𝜃 = 1 rad then 𝑠 = 𝑟 For a reference point on a rotating object moving at constant speed in a circle, then

𝑣 ≡𝑑𝑑𝑑𝑑

=𝑑 𝑟𝜃𝑑𝑑

= 𝑟𝑑𝜃𝑑𝑑

= 𝑟𝑟

𝒓

Choose the rotation axis to be the z-axis, pointing out of the page. ARBITRARY reference point, marked by at position 𝑟, and distance 𝑟 from axis. Angular position given by angle 𝜃 of 𝑟 measured CCW from the +x axis.

Rotation About a fixed axis

𝑥 𝑧

𝑦

Definitions Average Angular Velocity:

𝑟� ≡∆𝜃∆𝑑

Instantaneous Angular Velocity:

ω ≡𝑑𝜃𝑑𝑑

≡ lim∆𝑡→0

∆𝜃∆𝑑

Average Angular Acceleration:

𝛼� ≡∆𝑟∆𝑑

Instantaneous Angular Acceleration:

𝛼 ≡𝑑2𝜃𝑑𝑑2

≡𝑑𝑟𝑑𝑑

≡ lim∆𝑡→0

∆𝑟∆𝑑

𝜃

Comparison of Rotational Kinematics to Motion in a Straight Line

Case 1 of 2

Linear motion Rotational motion

𝑣 = constant 𝑟 = constant

𝑥 = 𝑥0 + 𝑣𝑑 𝜃 = 𝜃0 + 𝑟𝑑

Constant Angular Velocity Rotation Compared to Constant Velocity Motion in a Straight Line

�𝑑𝑑

Comparison of Kinematics: Case 2 of 2

�𝑑𝑑

�𝑑𝑑

Constant Angular Acceleration Rotation compared to Constant Acceleration Motion in a Straight Line

Poll 11-02-01

A wheel which is initially at rest starts to turn with a constant angular acceleration. After 4 seconds it has made 4 complete revolutions.

How many revolutions has it made after 8 seconds?

A. 8 B. 12 C. 16

Example 14.1 (1 of 3) To start a lawn mower, you must pull on a rope wound around the perimeter of a flywheel. After you pull the rope for 0.93 s, the flywheel is rotating at 5.1 revolutions per second, at which point the rope disengages. This attempt at starting the mower does not work, however, and the flywheel slows, coming to rest 0.24 s after the disengagement. Assume constant acceleration during both spin up and spin down. (a) Determine the average angular acceleration during the 0.93-s spin-up and again during the 0.24-s spin-down. (b) What is the maximum angular speed reached by the flywheel? (c) Determine the ratio of the number of revolutions made during spin-up to the number made during spin-down. (%i1) /* Here we started with omega=0 and after 0.93s of constant angular acceleration, we reached omega=5.1 rev/s */

omega1: 0;

(%o1) 0

(%i2) /* convert 5.1 rev/s to rad/s by multiplying by 1=2*pi rad/rev */

omega2: 5.1*2*%pi, numer;

(%o2) 32.04424506661588

(%i3) alpha: (omega_f - omega_i)/Dt;

omega_f - omega_i

(%o3) -----------------

Dt ... continued

Example 14.1 (2 of 3) … After you pull the rope for 0.93 s, the flywheel is rotating at 5.1 revolutions per second … flywheel slows, coming to rest 0.24 s after the disengagement. Assume constant acceleration during both spin up and spin down. (a) Determine the average angular acceleration during the 0.93-s spin-up and again during the 0.24-s spin-down. (b) What is the maximum angular speed reached by the flywheel? (c) Determine the ratio of the number of revolutions made during spin-up to the number made during spin-down. (%i4) alpha1: alpha, omega_i=omega1, omega_f=omega2, Dt=0.93;

(%o4) 34.45617749098482

Answer: (a) spin-up alpha = 34.5 rad/s^2 (%i5) /* spin down goes from omega2 to omega3=0 */ omega3: 0; (%o5) 0 (%i6) alpha2: alpha, omega_i=omega2, omega_f=omega3, Dt=0.24; (%o6) - 133.5176877775662 Answer: (a) spin-down alpha = -134 rad/s^2 Answer (b) Max ang speed = omega2 = 32.0 rad/s

Example 14.1 (3 of 3) … After you pull the rope for 0.93 s, the flywheel is rotating at 5.1 revolutions per second … flywheel slows, coming to rest 0.24 s after the disengagement. Assume constant acceleration during both spin up and spin down. (c) Determine the ratio of the number of revolutions made during spin-up to the number made during spin-down. (%i9) kill(alpha);

(%o9) done

(%i14) /* rotation at constant angular acceleration is given by */

Dtheta: omega_i*t + alpha/2*t^2;

(%i14)

2

alpha t

(%o14) -------- + omega_i t

2

(%i15) Dtheta1: Dtheta, omega_i=omega1, alpha=alpha1, t=0.93;

(%o15) 14.90057395597638

(%i16) Dtheta2: Dtheta, omega_i=omega2, alpha=alpha2, t=0.24;

(%o16) 3.845309407993905

(%i17) R: Dtheta1/Dtheta2;

(%o17) 3.875000000000001

Answer (c) Spin-up angle to spin-down angle ratio is R = 3.875

v

sθ

y

xr

P

O

Relating Linear and Rotational Quantities

Relationship between Linear and Rotational quantities depends on the distance 𝑟 of the point in question from the rotation axis: • Distance traveled by a point (at distance

𝑟 from the axis through angular displacement ∆𝜃 (measured in radians):

∆𝑑 = 𝑟∆𝜃 • Speed 𝑣 of point when body is rotating at

angular velocity 𝑟: 𝑣 = 𝑟𝑟

• Tangential Acceleration 𝑎𝑡 of point:

𝑎𝑡 =𝑑𝑣𝑑𝑑

= 𝑟𝑑𝑟𝑑𝑑

= 𝑟𝛼 • Centripetal acceleration 𝑎𝑐 of point:

𝑎𝑐 =𝑣2

𝑟= 𝑟𝑟2

ta

caa

x

y

P

O

𝑟

Any point in the body is executing circular motion about the rotation axis.

Unit 14 Main Points 1/2

LIMITED APPLICABILITY WARNING: The Main Points for unit 14 in smartPhysics gives similar formulae but specifically for a point on the rim of a disk of radius 𝑹

Rotation Dynamics: KE of particle A particle of mass 𝑚 (attached by a rigid, massless rod to the z-axis) is confined to rotate in a circle of radius 𝑟 (measured perpendicularly from the z-axis), on plane perpendicular to the z axis. Its angular velocity is 𝑟 (CCW around the z-axis is the positive sense) What is its kinetic energy?

Its “linear” speed is 𝑣 = 𝑟𝑟, and 𝐾 = 12𝑚𝑣2 = 1

2𝑚 𝑟𝑟 2 = 1

2(𝑚𝑟2)𝑟2

𝑧

𝑟

𝑟

𝑣

𝑚

Quantity Linear Angular

position 𝑥 𝜃

velocity 𝑣 𝑟

Acceleration a 𝛼

Kinetic Energy 12𝑚𝑣2

12𝐼𝑟2

We define the angular/rotational equivalent of (inertial )mass by:

𝐾 ≡12 𝐼𝑟

2

𝐼 ≡ Moment of Inertia

For a particle at radius 𝑟 (measured perpendicularly) from axis of rotation,

𝐼 = 𝑚𝑟2

Moment of Inertia of collection of particles We define the angular/rotational equivalent of (inertial )mass by:

𝐾 ≡12𝐼𝑟2

𝐼 ≡ Moment of Inertia

For a particle at radius 𝑟 (measured perpendicularly) from axis of rotation,

𝐼 = 𝑚𝑟2

𝑧

𝑟1

𝑟

𝑣1 𝑚1

𝑟2 𝑣2 𝑚2

𝑚3

𝑟3 𝑣3

For a collection of particles (say they are joined together by rigid, massless rods) with mass 𝑚1,𝑚2,⋯ ,𝑚𝑁, at radii (measured perpendicularly from the rotation axis): 𝑟1, 𝑟2,⋯ , 𝑟𝑁 , then its total kinetic energy under angular velocity 𝑟 (about rot. axis) is

𝐾 = �12𝑚𝑖𝑣𝑖2

𝑁

𝑖=1= �

12𝑚𝑖 𝑟𝑖𝑟 2

𝑁

𝑖=1=

12 � 𝑚𝑖𝑟𝑖2

𝑁

𝑖=1𝑟2

So the moment of inertia of the composite body is

𝐼 ≡� 𝑚𝑖𝑟𝑖2𝑁

𝑖=1= 𝑀

1𝑀� 𝑚𝑖𝑟𝑖2

𝑁

𝑖=1= 𝑀 𝑟2

where 𝑀 = ∑ 𝑚𝑖𝑖 is the total mass, and 𝑟2 = 1𝑀∑ 𝑚𝑖𝑟𝑖2𝑁𝑖=1 is the weighted average 𝑟2

Unit 14 Main Points 2/2

Poll 11-02-02

A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is Ia, Ib, and Ic respectively.

Which of the following orderings is correct?

A. Ia > Ib > Ic B. Ia > Ic > Ib C. Ib > Ia > Ic

Poll 11-02-03

In both cases shown below a hula hoop with mass M and radius R is spun with the same angular velocity about a vertical axis through its center. In Case 1 the plane of the hoop is parallel to the floor and in Case 2 it is perpendicular. In which case does the spinning hoop have the most kinetic energy?

A. Case 1 B. Case 2 C. Same

Example 14.2 (1 of 4) A uniform rod of total mass 𝑀 = 2.5 kg and length ℓ=1.2 m is rotating at a constant 720 rpm about a perpendicular axle at one end. (a) Find its angular velocity in rad/s. (b) Find its angular displacement (in rad.) over a time interval of 2.5 seconds. (c) Find the force exerted on the rod by the axle. (d) Find the kinetic energy of the rod. (e) repeat (c) and (d) for a rod rotating about a perpendicular axle through its center

Solution: (a) (constant) angular velocity is given by

𝑟 =∆𝜃∆𝑑

=720 ∙ (2𝜋 rad)

60 s= 24𝜋

rads

= 75.4rad

s

(b) We have constant angular velocity, so

𝜃 = 𝜃0 + 𝑟𝑑, ∆𝜃 = 𝑟∆𝑑 = 75.4rad

s 2.5 s = 188.5 rad

(c) This part requires us to look at the motion of the center-of-mass. The CM is located at a distance 𝑅𝐶𝑀 = ℓ 2⁄ from the axis. So we can treat (for the purpose of an external force) the rod as a point particle located at its CM, executing uniform circular motion at a radius of 𝑟 = 𝑅𝐶𝑀 = ℓ 2⁄ : At constant angular velocity: 𝑣 = 𝑟𝑟 is constant, so the tangential acceleration is 𝑎𝑡 = 𝑟𝛼 = 0. But the centripetal acceleration (directed radially inward) is

𝑎𝑐 = 𝑟𝑟2 = ℓ 2⁄ 𝑟2 The only external force on the rod is the force by the axle:

𝐹 = 𝑀𝑎𝑐 = 𝑀 ℓ 2⁄ 𝑟2 = 2.5 kg 0.60 m 75.4 rad s⁄ 2 = 8.53 × 103 N

𝑧

𝑀

ℓ

Solution: (d) Consider a small piece of the rod of length 𝑑𝑥. It has mass 𝑑𝑚 = 𝜇𝑑𝑥 where 𝜇 = 𝑀/ℓ is the linear mass density. This piece is at 𝑟 = 𝑥 from the axle and has kinetic energy

𝑑𝐾 = 12 𝑑𝑚 𝑣2 = 1

2 𝑑𝑚 𝑟𝑟 2

𝑧

𝑀 𝑟

ℓ

𝑑𝑥 𝑂 𝑥

𝑧 𝑟 = 𝑥

The total kinetic energy of the rod is then obtained by summing/integrating over the pieces:

𝐾 = �12 𝑑𝑚 𝑟𝑟 2 =

12 �𝑟2𝑑𝑚 𝑟 2 ≡

12 𝐼𝑟 2

This formula for rotational kinetic energy has the same form linear KE: 𝐾 = 12𝑚𝑣2 where we

defined the “moment of inertia” 𝐼, as the weighted sum/integral of the 𝑟2 from the axle:

𝐼 ≡ �𝑟2𝑑𝑚 , or, 𝐼 ≡�𝑚𝑖𝑟𝑖2𝑖

The word “moment” refers to different measures of how samples are distributed over a distribution. In this instance, it is the 2nd radial moment of the mass. The zeroth moment of the mass is just the total mass of the body. But it depends on where you choose the center

Uniform rod 𝑀 = 2.5 kg , ℓ = 1.2 m, 720 rpm about a perpendicular axle at one end. (a) Ans.: 𝑟 = 75.4 rad/s. (b) ∆𝜃 = 188.5 rad. in 2.5 s. (c) Ans.: 𝐹 = 𝑀𝑎𝑐 = 8.53 × 103 N (d) KE=?. (e) repeat (c) and (d) for perpendicular axle through center of rod

Units: kg⋅m2

Example 14.2 (2 of 4)

Uniform rod 𝑀 = 2.5 kg , ℓ = 1.2 m, 720 rpm about a perpendicular axle at one end. (a) Ans.: 𝑟 = 75.4 rad/s. (b) ∆𝜃 = 188.5 rad. in 2.5 s. (c) Ans.: 𝐹 = 𝑀𝑎𝑐 = 8.53 × 103 N (d) KE=3.41 × 103 J (e) repeat (c) and (d) for perpendicular axle through center of rod

(d) Continued: For this rod, the moment of inertia 𝐼 is given by:

𝐼 ≡ �𝑟2𝑑𝑚 = � 𝑥2 ∙ 𝜇𝑑𝑥ℓ

0= 𝜇� 𝑥2𝑑𝑥

ℓ

0= 𝜇

𝑥3

3ℓ0

=13 𝜇ℓ

3

𝑧

𝑀 𝑟

ℓ

𝑑𝑥 𝑂 𝑥

𝑧 𝑟 = 𝑥

Recall the linear mass density is given by 𝜇 = 𝑀/ℓ. So the moment inertia of a rod (mass M, length ℓ ) about ONE END is

𝐼𝐸𝑁𝐸 =13𝑀ℓ

2

The total KE is given by

𝐾 =12 𝐼𝐸𝑁𝐸𝑟

2 =16𝑀ℓ

2𝑟2 =2.5 kg 1.2 m 2 75.4 rad/s 2

6 = 3.41 × 103 J

Part (e) on Next Page

Note it is ℓ SQUARED

Example 14.2 (3 of 4)

In calculating the moment of inertia 𝐼 (about its CM) the integration limits have changed:

𝐼 ≡ �𝑟2𝑑𝑚 = � 𝑥2 ∙ 𝜇𝑑𝑥+ℓ 2⁄

−ℓ 2⁄= 𝜇� 𝑥2𝑑𝑥

ℓ

0= 𝜇

𝑥3

3

ℓ2−ℓ2

=13 𝜇

ℓ2

3

−−ℓ2

3

=13 𝜇ℓ

3 18 −

−18 =

13 𝜇ℓ

3 ∙14

Again: 𝜇 = 𝑀/ℓ. So the moment inertia of a rod (mass M, length ℓ ) about its CM is

𝐼𝐶𝑀 =1

12𝑀ℓ2

The total KE is given by

𝐾 =12 𝐼𝐶𝑀𝑟

2 =1

24𝑀ℓ2𝑟2 =

2.5 kg 1.2 m 2 75.4 rad/s 2

24 = 853 J

Uniform rod 𝑀 = 2.5 kg , ℓ = 1.2 m, 720 rpm about a perpendicular axle at one end. (a) Ans.: 𝑟 = 75.4 rad/s. (b) ∆𝜃 = 188.5 rad. in 2.5 s. (c) Ans.: 𝐹 = 𝑀𝑎𝑐 = 8.53 × 103 N (d) KE=?. (e) repeat (c) and (d) for perpendicular axle through center of rod

(e) In this case: the rod is rotating about its CM: So the CM is at 𝑟 = 𝑅𝐶𝑀 = 0

The centripetal acceleration of the CM is therefore: 𝑎𝑐 = 𝑟𝑟2 = 0 → 𝐹 = 0

𝑧

𝑀 ℓ/2

(e) 𝑟

ℓ

𝑑𝑥 𝑂 𝑥 𝑧 𝑟 = 𝑥

Note it is ℓ SQUARED

Example 14.2 (4 of 4)