Chapter 8- Rotational Kinematics Angular Variables Kinematic Equations

Chapter 8- Rotational Kinematics Angular Variables Kinematic Equations Chapter 9- Rotational Dynamics Torque Center of Gravity Newton s 2 nd Law- Angular Rotational Work & Energy Angular Momentum

Angular Momentum (chapter 9) There are 2 types of pure unmixed motion: Translational - linear motion Rotational - motion involving a rotation or revolution around a fixed chosen axis. It is useful to think about how BOTH types of motion work together on a system.

9.4 Newton s Second Law for Rotational Motion About a Fixed Axis τ = ma T r r ( ) α τ = mr 2

9.4 Newton s Second Law for Rotational Motion About a Fixed Axis This is Newton s Second Law for Rotational Motion

Consider Newton s second law for the inertia of rotation to be patterned after the law for translation (linear motion). ΣF T = m a T

F = 20 N a = 4 m/s 2 F = 20 N R = 0.5 m α = 4 rad/s 2 Linear Inertia, m m = 20 N 4 m/s 2 = 5 kg Rotational Inertia, I τ (20 N)(0.5 m) I = = = 2.5 kg m α 4 m/s 2 2

9.4 Newton s Second Law for Rotational Motion About a Fixed Axis τ = ( mr 2 ) α

Rotational Kinetic Energy and Inertia Just like massive bodies tend to resist changes in their motion ( AKA - "Inertia"). Rotating bodies also tend to resist changes in their motion. We call this ROTATIONAL INERTIA. Let s take a look at how rotational inertia relates to Kinetic Energy. K = 1 2 mv 2, v t = rω K = 1 2 m(rω)2 K = 1 2 mr2 ω 2 I = mr 2 K rot = 1 2 Iω 2

Consider a tiny mass m: v = ωr KE = ½mv 2 KE = ½m(ωR) 2 ω m 1 m m 4 m 3 KE = ½(mR 2 )ω 2 axis m 2 Sum to find KE total: KE = ½(ΣmR 2 )ω 2 (½ω 2 same for all m ) Object rotating at constant ω. Rotational Inertia Defined: I = ΣmR 2

First: I = ΣmR 2 I = (3 kg)(1 m) 2 + (2 kg)(3 m) 2 + (1 kg)(2 m) 2 2 kg 3 m 2 m 3 kg 1 m 1 kg ω I = 25 kg m 2 ω = 600 rpm = 62.8 rad/s K = ½Iω 2 = ½(25 kg m 2 )(62.8 rad/s) 2 K = 49,300 J

Continuous Masses The equation I =Σmr 2 works fine for point masses. But what about more continuous masses like disks, rods, or spheres where the mass is extended over a volume or area? In this case, calculus is needed. This suggests that we will take small discrete amounts of mass and add them up over a set of limits.

Rotational Dynamics; Rotational Inertia The quantity is called the rotational inertia of an object. The distribution of mass matters here these two objects have the same mass, but the one on the left has a greater rotational inertia, as so much of its mass is far from the axis of rotation.

Rotational Dynamics; Rotational Inertia The rotational inertia of an object depends not only on its mass distribution but also the location of the axis of rotation compare (f) and (g), for example.

R I = 0.270 kg m 2 I = mr 2 Hoop R I = ½mR 2 Disk I = 0.135 kg m 2

Check for Understanding

Check for Understanding

For many problems involving rotation, there is an analogy to be drawn from linear motion. m A resultant force F produces negative acceleration a for a mass m. I R 4 kg A resultant torque τ produces angular acceleration α of disk with rotational inertia I.

Displacement: a = αr α = a R

If you are to solve for a linear parameter, you must convert all angular terms to linear terms: α = a R If you are to solve for an angular parameter, you must convert all linear terms to angular terms: a = αr

How many revolutions will the disk undergo before the rotation is stopped? τ = Iα F R 4 kg ω ω ο = 50 rad/s R = 0.20 m F = 40 N FR = (½mR 2 )α 0 2αθ = ω 2 f - ω 2 o α = 100 rad/s 2 θ = 12.5 rad = 1.99 rev

Apply Newton s 2nd law to rotating disk: τ = Iα TR = (½MR 2 )α T = ½MRα a T = ½MR( ) ; R but a = αr; α = and T = ½Ma Apply Newton s 2nd law to falling mass: mg - T = ma mg - ½Ma T = ma (2 kg)(9.8 m/s 2 ) - ½(6 kg) a = (2 kg) a 19.6 N - (3 kg) a = (2 kg) a a = 3.92 m/s 2 a R R = 50 cm M 6 kg a =? R = 50 cm +a 6 kg 2 kg T T mg 2 kg

Work = Fs = FRθ Work = τθ τ = FR θ s F Work Power= = t τθ t ω = θ t s = Rθ F Power = τ ω Power = Torque x average angular velocity

Rotational Work and Kinetic Energy

Rotational Work and Kinetic Energy Rotational Translational Work: W = τθ W = Fd = Fx Kinetic Energy: KE = ½ I ω 2 KE = ½ mv 2 Power: P = τω P = Fv Work-Energy Theorem: W = ΔKE = ½ I ω f 2 - ½ I ω o 2 W = ½ mv f 2 - ½ mv o 2 The kinetic energy of a rolling body (without slipping) relative to an axis through the contact point is the sum of the rotational kinetic energy about an axis through the center of mass and the translational kinetic energy of the center of mass. KE = ½ I CM ω 2 + ½ mv CM 2 total = rotational + translational KE KE + KE

Check for Understanding From W = τθ, the units of rotational work are a. watt b. N m c. kg rad/s 2 d. N rad Answer: b

Work = τθ = FR θ s 20 m θ = = = 50 rad R 0.4 m F = mg = (2 kg)(9.8 m/s 2 ); F = 19.6 N Work = (19.6 N)(0.4 m)(50 rad) s θ 2 kg 6 kg F F=mg s = 20 m Work = 392 J Work Power = = t 392 J 4s Power = 98 W

Now consider a ball rolling without slipping. The angular velocity ω about the point P is same as ω for disk, so we write: R ω P v Or

ω R P v Total Kinetic Energy of a Rolling Object:

Check for Understanding A bowling ball rolls without slipping on a flat surface. The ball has a. rotational kinetic energy b. translational kinetic energy c. both rotational and translational kinetic energies Answer: c

R = 50 cm 6 kg 2 kg h = 10 m 2.5v 2 = 196 m 2 /s 2 v = 8.85 m/s

Recall for linear motion that the work done is equal to the change in linear kinetic energy: Using angular analogies, we find the rotational work is equal to the change in rotational kinetic energy:

What work is needed to stop wheel rotating? F R ω ω ο = 60 rad/s R = 0.30 m Work = ΔΚ r 4 kg F = 40 N First find I for wheel: I = mr 2 = (4 kg)(0.3 m) 2 = 0.36 kg m 2 0 Work = -½Iω ο 2 Work = -½(0.36 kg m 2 )(60 rad/s) 2 Work = -648 J

Total energy: E = ½mv2 + ½Iω2

Total energy: E = ½mv2 + ½Iω2

Rotational Kinetic Energy The kinetic energy of a rotating object is given by By substituting the rotational quantities, we find that the rotational kinetic energy can be written: A object that has both translational and rotational motion also has both translational and rotational kinetic energy:

Total Kinetic Energy An object that has both translational and rotational motion also has both translational and rotational kinetic energy:

Kinetic Energy When using conservation of energy, both rotational and translational kinetic energy must be taken into account. All these objects have the same potential energy at the top, but the time it takes them to get down the incline depends on how much rotational inertia they have.

Check for Understanding

8.4 Check for Understanding

Two kinds of energy: K T = ½mv 2 K r = ½Iω 2 ω v ω v Total energy: E = ½mv 2 + ½Iω 2 ω = v R Disk: E = ¾mv 2 Hoop: E = mv 2

The total energy is still conserved for systems in rotation and translation. Remember to always consider rotational energy as well as translational. Begin: (U + K t + K R ) o = End: (U + K t + K R ) f

mgh o = ½mv 2 + ½Iω 2 Hoop: I = mr 2 20 m mgh o = ½mv 2 + ½mv 2 ; mgh o = mv 2 Hoop: v = 14.0 m/s Disk: I = ½mR 2 ; mgh o = ½mv 2 + ½Iω 2 v = 16.2 m/s

Angular Momentum Momentum resulting from an object moving in linear motion is called linear momentum. Momentum resulting from the rotation (or spin) of an object is called angular momentum.

Consider a particle m moving with velocity v in a circle of radius r. Define angular momentum L: L = mvr Substituting v= ωr, gives: L = m(ωr) r = mr 2 ω For extended rotating body: L = (Σmr 2 ) ω v = ωr ω axis m 1 m m 4 m 3 m 2 Object rotating at constant ω. Since I = Σmr 2, we have: L = Iω Angular Momentum

Angular Momentum and Its Conservation In analogy with linear momentum (p = mv), we can define angular momentum L: We can then write the total torque as being the rate of change of angular momentum. Recall that F = mδv t = ma τ = IΔω Similarly t = Iα

Conservation of Angular Momentum Angular momentum is important because it obeys a conservation law, as does linear momentum. The total angular momentum of a closed system stays the same.

Angular Momentum and Its Conservation Therefore, systems that can change their rotational inertia through internal forces will also change their rate of rotation:

Check for Understanding 1. The units of angular momentum are a) N m b) kg m/s 2 c) kg m 2 /s d) J m Answer: c

Calculating angular momentum

l = 2 m m = 4 kg For rod: I = 1/12 m l 2 = 1/12 (4 kg)(2 m) 2 I = 1.33 kg m 2 L = Iω = (1.33 kg m 2 )(31.4 rad/s) L = 41.8 kg m 2 /s

Recall for linear motion the linear impulse is equal to the change in linear momentum: Using angular analogies, we find angular impulse to be equal to the change in angular momentum:

I = mr 2 = (2 kg)(0.4 m) 2 I = 0.32 kg m 2 Applied torque τ = FR Δ t = 0.002 s ω R F 2 kg ω ο = 0 rad/s R = 0.40 m F = 200 N Impulse = change in angular momentum 0 τ Δt = Iω f Ιω o FR Δt = Iω f ω f = FRΔt I = (200N)(0.40m)(0.002s) 0.32kg m 2 ω f = 0.5 rad/s

In the absence of external torque the rotational momentum of a system is conserved (constant). 0 I f ω f Ι ο ω o = τ Δt I f ω f = Ι ο ω o I o = 2 kg m 2 ; ω ο = 600 rpm I f = 6 kg m 2 ; ω ο =? ω f = 200 rpm

Quantity Linear Rotational Displacement Displacement x Radians θ Inertia Mass (kg) I (kg m 2 ) Force Newtons N Torque N m Velocity v m/s ω Rad/s Acceleration a m/s 2 α Rad/s 2 Momentum mv (kg m/s) Iω (kg m 2 rad/s)

Linear Motion F = ma Rotational Motion τ = Iα K = ½mv 2 K = ½Iω 2 Work = Fx Work = τθ Power = Fv Power = Iω Fx = ½mv f 2 - ½mv o 2 τθ = ½Iω f 2 - ½Iω o 2

I = ΣmR2