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Home , Acceleration, Angular acceleration, Angular displacement, Angular momentum, Angular velocity, Displacement (geometry)

Rotational & Angular Rotational Motion Every quantity that we have studied with translational motion has a rotational counterpart

TRANSLATIONAL ROTATIONAL x v a m I Momentum p L F  Angular

s Arclength   (from axis of r )

Measured in – all angular quantities will be measured in radians (NOT degrees) Angular Position

• In translational motion, x = 3 position is represented by a x , such as x. 0 5 linear p/2

• In rotational motion, r position is represented  p 0 by an , such as , and a radius, r. angular 3p/2 Displacement

Dx = 4 • Linear displacement x is represented by the 0 5 vector Dx. linear p/2

• Angular displacement is D  60 represented by D, which is not a vector, but p 0 behaves like one for small values. angular 3p/2 Angular Displacement

Compare to D풙 = 풙 − 풙풐 D    0

Which direction is positive (by convention)? Positive – Counterclockwise Negative - Clockwise EXAMPLE: (a) What is the angular position, , if we go around a circle two ? s 2(2p푟) Ans.  = = = 4π r r

(b) Say you go a quarter more, what is D? 9휋 휋 Ans. D =    = − 4π = o 2 2 (c) What is the arclength covered in total between Probls. (a) and (b) if the radius of the circle is 3 m?

휋 9휋 27휋 Ans. s = r = (3 m)(4p + ) =3m( ) = m 2 2 2 P.O.D. 1: A Boy on a horse is hunting a goose around a strange world of radius 25 m. The angular separation between the hunter/hunted is a constant 5. What is the angular (in m) between the boy/horse and the goose? v and velocity s T

• The same rotates with an D r avg. angular velocity given by vT D훉  = D퐭 • Linear (Tangential) and angular are related by the equation v = r Angular Velocity Compare to ∆풙 풗 = D 풂풗품 ∆풕 avg  Dt Units – rad/s, rev/s Direction – Same as displacement (positive is counterclockwise, negative is clockwise) EXAMPLE: A station ring of radius 500 m spins twice in 12 minutes.

(a) Find its angular velocity (in rads/s). (b) Find the linear (tangential) velocity (in m/s) of a point on the outer edge of the ring.

Ans. (a) Twice means two revolutions = 2(2p) = 4 p radians 휃 4휋  = = 푠푒푐 = 0.017 rad/sec 푡 12 min × 60 Τ푚𝑖푛 (b) tangential velocity, vT = r = 0.017 (500 m) = 8.73 m/s P.O.D. 2: A figure skater spins through five revolutions in a of 2.4 s.

(a) Find her angular velocity (in rads/s). (b) Find the linear (tangential) velocity (in m/s) of her foot if it is 0.3 meters away from her body. v Acceleration s T

• The same particle rotates with an D r avg. angular acceleration given by vT D = D퐭 • Linear (Tangential) and angular are related by the equation a = r Angular Acceleration

D Compare to ∆풗 풂 =   풂풗품 ∆풕 avg Dt • Angular Acceleration is how the angular velocity changes with time Units – rad/s2 Centripetal Acceleration A is swinging back and forth. At the bottom of the swing the force of is pulling it downwards but it doesn’t fall down. This means there must be a force pulling upwards to balance it out. Fc This is the . Since F = ma, the center-seeking acceleration is called centripetal Fg acceleration and is given by: v2 a    2r c r Sample Problem A compact disk rotates about an axis according to the formula: (t) = t2 – 6

(a) What is the linear speed of a point 20 cm from the center at t = 5 s? (b) What is the linear acceleration at 0.5 m at t = 5 s? Ans. (a) To find the linear speed, use the relationship, where (5) we obtained from the (t) = t2 – 6  (5)= 52 –6=19 v = r = (19 rad/s)(0.20 m) = 3.8 m/s. (b) To find the linear acceleration, use the relationship 2 2 2 ac =  r = (19 rad/s) (0.20 m) = 72.2 m/s P.O.D. 3: Diana Prince starts rotating with an angular velocity of 5 rad/s, where the negative indicates a clockwise rotation, to transform into Wonder Woman. After 5 she has completed the transformation and her angular velocity is 2 rad/s.

(a) Find Wonder Woman’s angular acceleration (in rad/s2). (b) Find the centripetal acceleration (in m/s2) of her indestructible bracelets (at  = 2 rad/s) if they are 0.8 m away from her body when she is spinning. Constant Angular Acceleration • Our equations have angular equivalents • Just as with their linear counterparts, these only for constant acceleration First Kinematic Equation

2 • x = vot + ½ at () – Substitute angle for position. – Substitute angular velocity for linear velocity. – Substitute angular acceleration for linear acceleration. 2  = ot + ½ t (angular form) Kinematic Equation

• v = vo + at (linear form) – Substitute angular velocity for linear velocity. – Substitute angular acceleration for linear acceleration.

 = o + t (angular form) Sample Problem

A bicycle starts from rest and for 10.0 s has a constant linear acceleration of 0.8 m/s2 to the right. During this period, the tires do not slip. The radius of the tires is 0.50 m. At the end of the 10.0 s interval what is the angle through which each wheel has rotated? Ans. The angular acceleration can be found from the formula 풎 풂 ퟎ.ퟖ ퟐ a = r  휶 =  휶 = 풔 = 1.6 rad/s2 풓 ퟎ.ퟓퟎ 풎 The angular acceleration should be negative because the tire spins clockwise. To find the angular displacement: 2 2 2 (t) = ot + ½t = 0(10 s) + ½(1.6 rad/s )(10 s) = 80 rad P. O. D. 4: an extreme diver rotates at an angular velocity of +3.75 rad/s while doing his first set of flips. When he does his second set of flips, he accelerates and reaches a greater angular velocity of +4.4 rads/s. If his angular acceleration is 1.7 rad/s2... (a) How long does this flipping part of the dive take (in s)? (b) Find his angular displacement after the time found in (a). ANGULAR MOMENTUM! Angular Momentum Angular momentum depends on linear momentum and the distance from a particular point. It is a vector quantity with symbol L. If r and v are  then the of angular momentum w/ resp. to point Q is given by L = rp = mvr. In this case L points out of the page. If the mass were moving in the opposite direction, L would point into the page. The SI unit for angular momentum is the kgm2 / s. (It has no special v name.) Angular momentum is a . A torque is needed to change L, just a force is r m needed to change p. Anything spinning has angular has angular Q momentum. The more it has, the harder it is to stop it from spinning. Angular Momentum: General Definition If r and v are not  then the angle between these two vectors must be taken into account. The general definition of angular momentum is given by a vector : L = r  p This formula works regardless of the angle. From cross products, the magnitude of the angular momentum of m relative to point Q is: L = r p sin = m v r. In this case, by the right-hand rule, L points out of the page. If the mass were moving in the opposite direction, L would point into the page. v  r m

Q of Inertia vs. Angular Momentum Any moving body has inertia. (It wants to keep moving at constant v.) The more inertia a body has, the harder it is to change its . Rotating bodies possess a rotational inertia called the , I. The more rotational inertia a body has, the harder it is change its rotation. For a single point-like mass w/ respect to a given point Q, I = mr 2. For a system, I = the sum of each mass m times its respective distance from the point of interest. r

m2 Q m1 2 r1 I = mr r2 Q

2 2 2 I = mi ri = m1 r1 + m2r2 Moment of Inertia of various Moment of Inertia Example Two merry-go-rounds have the same mass and are spinning with the same angular velocity. One is solid wood (a disc), and the other is a metal ring. Which has a bigger moment of inertia relative to its ?

r r  

m m

Ans. I is independent of the angular speed. Since their and radii are the same, the ring has a greater moment of inertia. This is because more of its mass is farther from the axis of rotation. Since I is bigger for the ring, it would more difficult to increase or decrease its angular speed. Torque & Angular Acceleration

nd ’s 2 Law, as you know, is Fnet = ma

nd The 2 Law has a rotational analog: net = I A force is required for a body to undergo acceleration. A “turning force” (a torque) is required for a body to undergo angular acceleration.

The bigger a body’s mass, the more force is required to accelerate it. Similarly, the bigger a body’s rotational inertia, the more torque is required to accelerate it angularly. Both m and I are measures of a body’s inertia (resistance to change in motion). Example: The torque of an Electric Saw Motor The motor in an electric saw brings the circular blade up to the rated angular speed of 80.0 rev/s in 240.0 rev. One type of blade has a moment of inertia of 1.41 x 10-3 kgm2. What net torque (assumed constant) must the motor apply to the blade? Ans. First we need to convert our values into rad/s for calculation purposes.

   o t 240 rev x 2p = 1508 rads ? 80 revs/s x 2p = 503 rad/s 0 rad/s ? 2 2 We can find the angular acceleration from  = o + 2 2− 2 (503 rad/s)2−02 Solving for :  = o = = 83.89 rad/s2 2 2(1508 rad/s)  = I  = 1.41 x 10-3 kgm2 83.89 rad/s2 = 0.118 Nm PROBLEM 6: A Chinese of mass 0.025 kg and radius 0.03 m is thrown by Bruce Lee at his adversary. The Chinese star is thrown from rest. If its final angular velocity is 15 revs/s after 3 sec, (a) find the angular acceleration of the Chinese star. (b) Find the torque of the Chinese star (Assume the Chinese star is hoop-shaped). Linear Momentum vs. Angular Momentum If a acts on an object, it must accelerate, which means its momentum must change. Similarly, if a net torque acts on a body, it undergoes angular acceleration, which means its angular momentum changes. Recall, angular momentum’s magnitude is given by

L = mvr (if v and r are )

v So, if a net torque is applied, angular velocity must r m change, which changes angular momentum.

Proof:  net = r  Fnet = r m a Dv DL = r  m = t t So net torque is the of change of angular momentum, just as net force is the rate of change of linear momentum. continued on next slide Linear & Angular Momentum (cont.) Here is yet another pair of similar equations, one linear, one rotational. From the formula v = r , we get

2 L = mv r = m  (  r) r = m  r  = I   This is very much like p = mv, and this is one reason I is defined the way it is. In terms of magnitudes, linear momentum is inertia times speed, and angular momentum is rotational inertia times angular speed.

L = I  p = m v Comparison: Linear & Angular Momentum

Linear Momentum, p Angular Momentum, L • Tendency for a mass to continue • Tendency for a mass to continue moving in a straight line. rotating. • Parallel to v. • Perpendicular to both v and r. • A conserved, vector quantity. • A conserved, vector quantity. • Magnitude is inertia (mass) • Magnitude is rotational inertia times speed. times angular speed. • Net force required to change it. • Net torque required to change it. • The greater the mass, the greater • The greater the moment of the force needed to change inertia, the greater the torque momentum. needed to change angular momentum. Example: Spinning Ice Skater Suppose Mr. Stickman is sitting on a stool that swivels holding a pair of dumbbells. His axis of rotation is vertical. With the far from that axis, his moment of inertia is 600 kgm2 and he is spinning at an angular velocity of 20 rad/s. When he pulls his arms in as he’s spinning, the weights are closer to the axis, so his moment of inertia gets to 400 kgm2. What will be his angular velocity at this moment? Ans.

I  = L = I  2 2 600 kgm (20 rad/s) = 400 kgm 

12,000 = 400 30 rad/s =  PROBLEM 7: An artificial (m = 1500 kg) is placed into an elliptical about the . Telemetry data indicate that its point of closes approach

(called the perigee) is rp = 8.37 x 106 m from the center of the earth, while its point of greatest distance (called the apogee) is 6 rA = 25.1 x 10 m from the center of the earth. The speed of the satellite at the perigee is vp = 8450 m/s.

(a) Find its speed vA at the apogee. (b)Find its angular momentum at any point in its orbit. Rotational Kinetic • A particle in a rotating object has rotational :

2 where v r Ki = ½ mivi , i = i (tangential velocity) The whole rotating object has a rotational kinetic energy given by:

1 22 KR K i m i r i  ii2

112 2 2 KR m i r i  I 22i Rotational Kinetic Energy example: A thin walled hollow cylinder (mass

= mh, radius = rh) and a solid cylinder ( mass = ms, radius = rs) start from rest at the of an incline. Both cylinders start at the same vertical height ho. All heights are measured relative to an arbitrarily chosen zero level that passes through the center of mass of a cylinder when it is at the bottom of the incline. Ignoring energy losses due to retarding , determine which cylinder has the greatest translational speed upon reaching the bottom. Rotational Kinetic Energy example (cont.): Ans. At the top of the incline the cylinder have only gravitational . At the bottom of the incline this energy has converted into translational kinetic and rotational kinetic energy.

Ein = Eout GPEin = TKEout + RKEout 2 2 mgh = ½mvf + ½If 퐯 The angular velocity  can be related to the linear velocity v by  = 퐟 f f 퐫 Substituting the given values and for the angular velocity: 2 퐯퐟 2 mhgho = ½mhvf + ½I( ) 퐫풉 For the hollow cylinder, the moment of inertia is given by: I = mr2 2 2 vf 2 Substituting: mhgho = ½mhvf + ½(mhrh )( ) rℎ 2 2 퐯퐟 2 Simplifying: gho = ½vf + ½rh ( ) 퐫풉 퐯 ퟐ 2 2 풇 gho = ½vf + ½rh  ퟐ 풓풉 2 ퟐ gho = ½vf + ½퐯풇 2 gho = vf

퐠퐡퐨= vf Rotational Kinetic Energy example (cont.): Ans. At the top of the incline the cylinder have only energy. At the bottom of the incline this energy has converted into translational kinetic and rotational kinetic energy.

Ein = Eout GPEin = TKEout + RKEout 2 2 mgh = ½mvf + ½If 퐯 The angular velocity  can be related to the linear velocity v by  = 퐟 f f 퐫 Substituting the given values and for the angular velocity: 2 퐯퐟 2 msgho = ½msvf + ½I( ) 퐫풔 For the solid cylinder, the moment of inertia is given by: I = ½mr2 2 2 vf 2 Substituting: msgho = ½msvf + ½(½msrs )( ) r푠 2 2 퐯퐟 2 Simplifying: gho = ½vf + ½ ½rs ( ) 퐫풔 퐯 ퟐ 2 2 풇 gho = ½vf + ¼rs  ퟐ 풓풔 2 ퟐ gho = ½vf + ¼퐯풇 2 gho = ¾vf ퟒ 퐠퐡 = v ퟑ 퐨 f PROBLEM 8: