Chapter 10 Rotation. Engineering Physics: Lecture 10 Topics Rotational Kinematics – Analogy with one-dimensional kinematics Kinetic energy of a rotating.

Chapter 10 Rotation

Engineering Physics: Lecture 10 Topics Rotational Kinematics – Analogy with one-dimensional kinematics Kinetic energy of a rotating system – Moment of inertia – Discrete particles – Continuous solid objects Parallel axis theorem

Rotation Introduction Up until now we have gracefully avoided dealing with the rotation of objects. – We have studied objects that slide, not roll. – We have assumed pulleys are without mass. Rotation is extremely important, however, and we need to understand it! Most of the equations we will develop are simply rotational analogues of ones we have already learned when studying linear kinematics and dynamics.

The Rotational Variables A rigid body is a body that can rotate with all its parts locked together and without any change in its shape. A fixed axis means that the rotation occurs about an axis that does not move. Figure skater Sasha Cohen in motion of pure rotation about a vertical axis. (Elsa/Getty Images, Inc.)

The Rotational Variables

The Rotational Variables: Angular Position Here s is the length of a circular arc that extends from the x axis (the zero angular position) to the reference line, and r is the radius of the circle. An angle defined in this way is measured in radians (rad).

The Rotational Variables: Angular Displacement If a body rotates about the rotation axis as in, changing the angular position of the reference line from  1 to  2, the body undergoes an angular displacement  given by An angular displacement in the counterclockwise direction is positive, and one in the clockwise direction is negative.

The Rotational Variables: Angular Velocity Suppose that our rotating body is at angular position  1 at time t 1 and at angular position  2 at time t 2. Then the average angular velocity of the body in the time interval t from t 1 to t 2 to be. The instantaneous angular velocity  is the limit of the ratio as  t approaches zero.

Relation Between Linear and Angular Quantities In Cartesian coordinates, we say velocity dx/dt = v. – x = vt In polar coordinates, angular velocity d  /dt = . –  =  t –  has units of radians/second. Displacement s = vt. but s = r  = r  t, so: r v x y s  t v =  r

Question Rotations Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds. – Klyde’s angular velocity is: (a) (a) the same as Bonnie’s (b) (b) twice Bonnie’s (c) (c) half Bonnie’s

Answer Rotations The angular velocity  of any point on a solid object rotating about a fixed axis is the same. – Both Bonnie & Klyde go around once (2  radians) every two seconds.  (Their “linear” speed v will be different since v =  r).

Period and Frequency Recall that 1 revolution = 2  radians – frequency (f) = revolutions / second (a) – angular velocity (  ) = radians / second (b) By combining (a) and (b) –  = 2  f Realize that: – period (T) = seconds / revolution – So T = 1 / f = 2  /  R v s  = 2  / T = 2  f 

Summary x = R cos(  )  = R cos(  t)  y = R sin(  )  = R sin(  t)  = arctan (y/x)  =  t s = v t s = R  = R  t v =  R R v s tt (x,y)

Are Angular Quantities Vectors?

The Rotational Variables: Angular Acceleration If the angular velocity of a rotating body is not constant, then the body has an angular acceleration. If  2 and  1 be its angular velocities at times t 2 and t 1, respectively, then the average angular acceleration of the rotating body in the interval from t 1 to t 2 is defined as The instantaneous angular acceleration , is the limit of this quantity as  t approaches zero. These relations hold for every particle of that body. The unit of angular acceleration is (rad/s 2 ).

Aside: Polar Unit Vectors i j k r  We are familiar with the Cartesian unit vectors: i j k Now introduce “polar unit-vectors” r and  : – r – r points in radial direction –  –  points in tangential direction R x y i j  r^^^ ^ ^ ^ (counter clockwise)

Centripetal Acceleration UCM results in acceleration: – Magnitude:a = v 2 / R r – Direction:- r (toward center of circle) R a  ^

Derivation: We know that and v =  R Substituting for v we find that:  a =  2 R

Problem Uniform Circular Motion A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ? (a) 500 m (b) 1000 m (c) 2000 m

Solution 2km

Example: Propeller Tip The propeller on a stunt plane spins with frequency f = 3500 rpm. The length of each propeller blade is L = 80cm. What centripetal acceleration does a point at the tip of a propeller blade feel? f L a what is a here?

Example: First calculate the angular velocity of the propeller: – so 3500 rpm means  = 367 s -1 Now calculate the acceleration. – a =  2 R = (367s -1 ) 2 x (0.8m) = 1.1 x 10 5 m/s 2 = 11,000 g ar – direction of a points at the propeller hub (-r ). ^

Example: Newton & the Moon What is the acceleration of the Moon due to its motion around the Earth? What we know (Newton knew this also): – T = 27.3 days = 2.36 x 10 6 s (period ~ 1 month) – R = 3.84 x 10 8 m(distance to moon) – R E = 6.35 x 10 6 m(radius of earth) RRERE

Moon... Calculate angular velocity: So  = 2.66 x 10 -6 s -1. Now calculate the acceleration. – a =  2 R = 0.00272 m/s 2 = 0.000278 g ar – direction of a points at the center of the Earth (-r ). ^

Moon... So we find that a moon / g = 0.000278 Newton noticed that R E 2 / R 2 = 0.000273 This inspired him to propose that F Mm  1 / R 2 – (more on gravity later) RRERE a moon g

Space Shuttle Centripetal Acceleration The Space Shuttle is in Low Earth Orbit (LEO) about 300 km above the surface. The period of the orbit is about 91 min. What is the acceleration of an astronaut in the Shuttle in the reference frame of the Earth? (The radius of the Earth is 6.4 x 10 6 m.) (a) 0 m/s 2 (b) 8.9 m/s 2 (c) 9.8 m/s 2

Space Shuttle Centripetal Acceleration First calculate the angular frequency  : Realize that: RORO 300 km R O = R E + 300 km = 6.4 x 10 6 m + 0.3 x 10 6 m = 6.7 x 10 6 m RERE

Space Shuttle Centripetal Acceleration Now calculate the acceleration: a =  2 R a = (0.00115 s -1 ) 2 x 6.7 x 10 6 m a = 8.9 m/s 2

Relating Linear and Angular Acceleration: Differentiating the linear velocity with respect to time -with r held constant-leads to : a t =dv/dt= d(rω)dt = rdω/dt Note that dv/dt =a t (tangential acceleration) represents only the part of the linear acceleration that is responsible for changes in the magnitude v of the linear velocity. Like v, that part of the linear acceleration is tangent to the path of the point in question. Also, the radial part of the acceleration is the centripetal acceleration given by:

Rotational Quantities: Rotation about a fixed axis: – Consider a disk rotating about an axis through its center: First, recall what we learned about Uniform Circular Motion: (Analogous to )  

Rotational Kinematics Equations Now suppose  can change as a function of time: We define the angular acceleration:    Consider the case when  is constant.  We can integrate this to find  and  as a function of time: constant

Rotational Kinematics Equations (α = const.) Recall also that for a point at a distance R away from the axis of rotation: – x =  r – v =  r And taking the derivative of this we find: – a =  r   r v x 

Summary (with comparison to 1-D kinematics) AngularLinear And for a point at a distance r from the rotation axis: x =  r  v =  r  a =  r

Rotation with Constant Angular Acceleration Just as in the basic equations for constant linear acceleration, the basic equations for constant angular acceleration can be derived in a similar manner. The constant angular acceleration equations are similar to the constant linear acceleration equations.

A ball is spinning about an axis that passes through its center with a constant angular acceleration of  rad/s 2. During a time interval from t 1 to t 2, the angular displacement of the ball is  radians. At time t 2, the angular velocity of the ball is 2  rad/s. What is the ball’s angular velocity at time t 1 ? a) 6.28 rad/s b) 3.14 rad/s c) 2.22 rad/s d) 1.00 rad/s e) zero rad/s

The original Ferris wheel had a radius of 38 m and completed a full revolution (2  radians) every two minutes when operating at its maximum speed. If the wheel were uniformly slowed from its maximum speed to a stop in 35 seconds, what would be the magnitude of the instantaneous tangential speed at the outer rim of the wheel 15 seconds after it begins its deceleration? a) 0.295 m/s b) 1.12 m/s c) 1.50 m/s d) 1.77 m/s e) 2.03 m/s

Sample problem, Angular Velocity and Acceleration

Example: Wheel And Rope A wheel with radius R = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4 m/s 2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2  radians)a R

Wheel And Rope... Use a =  R to find  :  = a / R = 4 m/s 2 / 0.4 m = 10 rad/s 2 l Now use the equations we derived above just as you would use the kinematic equations from the beginning of the semester. = 0 + 0(10) + (10)(10) 2 = 500 rad a R 

Sample problem Consider an induction roller coaster (which can be accelerated by magnetic forces even on a horizontal track). Each passenger is to leave the loading point with acceleration g along the horizontal track. That first section of track forms a circular arc (Fig. 10-10), so that the passenger also experiences a centripetal acceleration. As the passenger accelerates along the arc, the magnitude of this centripetal acceleration increases alarmingly. When the magnitude a of the net acceleration reaches 4g at some point P and angle  P along the arc, the passenger moves in a straight line, along a tangent to the arc. (a)What angle  P should the arc subtend so that a is 4g at point P? Calculations: Substituting  o =0, and  o =0, and we find: But which gives: This leads us to a total acceleration: Substituting for a r, and solving for q lead to: When a reaches the design value of 4g, angle  is the angle  P. Substituting a =4g,  =  P, and a t = g, we find:

Sample problem, cont. (b) What is the magnitude a of the passenger’s net acceleration at point P and after point P? Reasoning: At P, a has the design value of 4g. Just after P is reached, the passenger moves in a straight line and no longer has centripetal acceleration. Thus, the passenger has only the acceleration magnitude g along the track. Hence, a =4g at P and a =g after P. (Answer) Roller-coaster headache can occur when a passenger’s head undergoes an abrupt change in acceleration, with the acceleration magnitude large before or after the change. The reason is that the change can cause the brain to move relative to the skull, tearing the veins that bridge the brain and skull. Our design to increase the acceleration from g to 4g along the path to P might harm the passenger, but the abrupt change in acceleration as the passenger passes through point P is more likely to cause roller-coaster headache.

Rotation & Kinetic Energy Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods). The kinetic energy of this system will be the sum of the kinetic energy of each piece: rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3 

Rotation & Kinetic Energy... So: but v i =  r i rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3  vv4vv4 vv1vv1 vv3vv3 vv2vv2 which we write as: moment of inertia Define the moment of inertia about the rotation axis I has units of kg m 2.

Rotation & Kinetic Energy... Point Particle Rotating System The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle Rotating System v is “linear” velocity m is the mass.  is angular velocity I is the moment of inertia about the rotation axis.

Moment of Inertia Notice that the moment of inertia I depends on the distribution of mass in the system. – The further the mass is from the rotation axis, the bigger the moment of inertia. For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass). We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics! l So where

Calculating Moment of Inertia We have shown that for N discrete point masses distributed about a fixed axis, the moment of inertia is: where r is the distance from the mass to the axis of rotation. Example: Calculate the moment of inertia of four point masses (m) on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square: mm mm L

Calculating Moment of Inertia... The squared distance from each point mass to the axis is: mm mm L r L/2 so I = 2mL 2 Using the Pythagorean Theorem

Calculating Moment of Inertia... Now calculate I for the same object about an axis through the center, parallel to the plane (as shown): mm mm L r I = mL 2

Calculating Moment of Inertia... Finally, calculate I for the same object about an axis along one side (as shown): mm mm L r I = 2mL 2

Calculating Moment of Inertia... For a single object, I clearly depends on the rotation axis!! L I = 2mL 2 I = mL 2 mm mm I = 2mL 2

Question: Moment of Inertia 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 I a, I b, and I c respectively. – Which of the following is correct: (a) (a) I a > I b > I c (b) (b) I a > I c > I b (c) (c) I b > I a > I c a b c

Lecture 17, Act 2 Moment of Inertia a b c l Label masses and lengths: m m m L L l Calculate moments of inerta: So (b) is correct: I a > I c > I b

Calculating Moment of Inertia... For a discrete collection of point masses we found: For a continuous solid object we have to add up the mr 2 contribution for every infinitesimal mass element dm. – We have to do an integral to find I : r dm

Moments of Inertia Some examples of I for solid objects: Thin hoop (or cylinder) of mass M and radius R, about an axis through its center, perpendicular to the plane of the hoop. R Thin hoop of mass M and radius R, about an axis through a diameter. R Hoop

Moments of Inertia... Some examples of I for solid objects: Solid sphere of mass M and radius R, about an axis through its center. R R Solid disk or cylinder of mass M and radius R, about a perpendicular axis through its center. Sphere and disk

Question: Moment of Inertia Two spheres have the same radius and equal masses. One is made of solid aluminum, and the other is made from a hollow shell of gold. – Which one has the biggest moment of inertia about an axis through its center? same mass & radius solid hollow (a) solid aluminum(b) hollow gold(c) same

Question: Moment of Inertia Moment of inertia depends on mass (same for both) and distance from axis squared, which is bigger for the shell since its mass is located farther from the center. – The spherical shell (gold) will have a bigger moment of inertia. same mass & radius I SOLID < I SHELL solid hollow

Moments of Inertia... Some examples of I for solid objects. Thin rod of mass M and length L, about a perpendicular axis through its center. L Thin rod of mass M and length L, about a perpendicular axis through its end. L Rod

A hollow cylinder is rotating about an axis that passes through the center of both ends. The radius of the cylinder is r. At what angular speed  must the this cylinder rotate to have the same total kinetic energy that it would have if it were moving horizontally with a speed v without rotation? a) b) c) d) e)

Parallel Axis Theorem Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass, I CM, is known. The moment of inertia about an axis parallel to this axis but a distance h away is given by: I PARALLEL = I CM + Mh 2 So if we know I CM, it is easy to calculate the moment of inertia about a parallel axis.

Parallel Axis Theorem: Example Consider a thin uniform rod of mass M and length L. Figure out the moment of inertia about an axis through the end of the rod. I PARALLEL = I CM + MD 2 L D=L/2 M x CM We know So which agrees with the result on a previous slide. I CM I END

Connection with CM motion The kinetic energy of a system of particles is: K REL K CM l For a solid object rotating about its center of mass, we now see that the first term becomes: Substituting but

Connection with CM motion... So for a solid object which rotates about its center or mass and whose CM is moving:  V CM We will use this formula more in coming lectures.

Recap of lecture Rotational Kinematics – Analogy with one-dimensional kinematics Kinetic energy of a rotating system – Moment of inertia – Discrete particles – Continuous solid objects Parallel axis theorem

Chapter 10 Rotation. Engineering Physics: Lecture 10 Topics Rotational Kinematics – Analogy with one-dimensional kinematics Kinetic energy of a rotating.

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Chapter 10 Rotation. Engineering Physics: Lecture 10 Topics Rotational Kinematics – Analogy with one-dimensional kinematics Kinetic energy of a rotating.

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Presentation on theme: "Chapter 10 Rotation. Engineering Physics: Lecture 10 Topics Rotational Kinematics – Analogy with one-dimensional kinematics Kinetic energy of a rotating."— Presentation transcript:

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