Rotational Kinematics and Energy

Rotational Motion Up until now we have been looking at the kinematics and dynamics of translational motion – that is, motion without rotation. Now we will widen our view of the natural world to include objects that both rotate and translate. We will develop descriptions (equations) that describe rotational motion Now we can look at motion of bicycle wheels, roundabouts and divers.

II.Rotation with constant angular acceleration II. Rotation with constant angular acceleration III. Relation between linear and angular variables - Position, speed, acceleration I.Rotational variables - Angular position, displacement, velocity, acceleration - Angular position, displacement, velocity, acceleration Kinetic energy of rotation IV. Kinetic energy of rotation V. Rotational inertia VI. Torque VII. Newton’s second law for rotation VIII. Work and rotational kinetic energy

Rotational kinematics In the kinematics of rotation we encounter new kinematic quantitiesIn the kinematics of rotation we encounter new kinematic quantities –Angular displacement  –Angular speed  –Angular acceleration  –Rotational Inertia I –Torque  All these quantities are defined relative to an axis of rotationAll these quantities are defined relative to an axis of rotation

Angular displacement Measured in radians or degreesMeasured in radians or degrees There is no dimensionThere is no dimension  =  f -  i Axis of rotation  ii ff

Angular displacement and arc length Arc length depends on the distance it is measured away from the axis of rotationArc length depends on the distance it is measured away from the axis of rotation Axis of rotation Q spsp sqsq P r ii ff

Angular Speed Angular speed is the rate of change of angular positionAngular speed is the rate of change of angular position We can also define the instantaneous angular speedWe can also define the instantaneous angular speed

Average angular velocity and tangential speed Recall that speed is distance divided by time elapsedRecall that speed is distance divided by time elapsed Tangential speed is arc length divided by time elapsedTangential speed is arc length divided by time elapsed And because we can writeAnd because we can write

Average Angular Acceleration Rate of change of angular velocityRate of change of angular velocity Instantaneous angular accelerationInstantaneous angular acceleration

Angular acceleration and tangential acceleration We can find a link between tangential acceleration a t and angular acceleration αWe can find a link between tangential acceleration a t and angular acceleration α SoSo

Centripetal acceleration We have thatWe have that But we also know thatBut we also know that So we can also saySo we can also say

Example: Rotation A dryer rotates at 120 rpm. What distance do your clothes travel during one half hour of drying time in a 70 cm diameter dryer? What angle is swept out?A dryer rotates at 120 rpm. What distance do your clothes travel during one half hour of drying time in a 70 cm diameter dryer? What angle is swept out?

Example: Rotation A dryer rotates at 120 rpm. What distance do your clothes travel during one half hour of drying time in a 70 cm diameter dryer? What angle is swept out?A dryer rotates at 120 rpm. What distance do your clothes travel during one half hour of drying time in a 70 cm diameter dryer? What angle is swept out? Angle:  t = 120 r/min x 0.5 x 60 min = 120x2  rad) /min x 60 min/h x 0.5 h = 2.3 x 10 4 rAngle:  t = 120 r/min x 0.5 x 60 min = 120x2  rad) /min x 60 min/h x 0.5 h = 2.3 x 10 4 r –Distance: s =  r and  =  /  t so s =  tr –s = 120x2  rad)/min x 60 min/h x 0.5 h x 0.35 m = 1.3 km = 1.3 km

Rotational motion with constant angular acceleration We will consider cases where  is constantWe will consider cases where  is constant Definitions of rotational and translational quantities look similarDefinitions of rotational and translational quantities look similar The kinematic equations describing rotational motion also look similarThe kinematic equations describing rotational motion also look similar Each of the translational kinematic equations has a rotational analogueEach of the translational kinematic equations has a rotational analogue

Rotational and Translational Kinematic Equations

Constant  motion What is the angular acceleration of a car’s wheels (radius 25 cm) when a car accelerates from 2 m/s to 5 m/s in 8 seconds?

Constant  motion What is the angular acceleration of a car’s wheels (radius 25 cm) when a car accelerates from 2 m/s to 5 m/s in 8 seconds?

A rotating wheel requires 3.00 s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.00-s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel?

Example: Centripetal Acceleration A 1000 kg car goes around a bend that has a radius of 100 m, travelling at 50 km/h. What is the centripetal force? What keeps the car on the bend? [What keeps the skater in the arc?]A 1000 kg car goes around a bend that has a radius of 100 m, travelling at 50 km/h. What is the centripetal force? What keeps the car on the bend? [What keeps the skater in the arc?]

Example: Centripetal Acceleration A 1000 kg car goes around a bend that has a radius of 100 m, travelling at 50 km/h. What is the centripetal force? What keeps the car on the bend? [What keeps the skater in the arc?]A 1000 kg car goes around a bend that has a radius of 100 m, travelling at 50 km/h. What is the centripetal force? What keeps the car on the bend? [What keeps the skater in the arc?] Friction keeps the car and skater on the bend

Car rounding a bend Frictional force of road on tires supplies centripetal force If  s between road and tires is lowered then frictional force may not be enough to provide centripetal force…car will skid Locking wheels makes things worse as  k <  s Banking of roads at corners reduces the risk of skidding…

Car rounding a bend Horizontal component of the normal force of the road on the car can provide the centripetal force If then no friction is required FgFg N Ncos  Nsin  

An automobile makes a turn whose radius is 150 m. The road is banked at an angle of 18°, and the coefficient of friction between the wheels and road is 0.3. Find the maximum speed for the car to stay on the road without skidding the banked road.

x-component: F N sin  + f s cos  = ma = mv 2 /R; y-component: F N cos  – fs sin  – mg = 0. F N cos  –  sF N  sin  = mg, F N = mg/(cos  –  s sin  ).

An automobile makes a turn whose radius is 150 m. The road is banked at an angle of 18°, and the coefficient of friction between the wheels and road is 0.3. Find the maximum speed for the car to stay on the road without skidding the banked road. From the x-equation we get:

Rotational Dynamics Easier to move door at A than at B using the same force FEasier to move door at A than at B using the same force F More torque is exerted at A than at BMore torque is exerted at A than at B A B hinge

Torque Torque is the rotational analogue of ForceTorque is the rotational analogue of Force Torque, , is defined to beTorque, , is defined to be Where F is the force applied tangent to the rotation and r is the distance from the axis of rotation r F  = Fr

Torque A general definition of torque isA general definition of torque is Units of torque are NmUnits of torque are Nm Sign convention used with torqueSign convention used with torque –Torque is positive if object tends to rotate CCW –Torque is negative if object tends to rotate CW r F   = Fsin  r

Condition for Equilibrium We know that if an object is in (translational) equilibrium then it does not accelerate. We can say that  F = 0We know that if an object is in (translational) equilibrium then it does not accelerate. We can say that  F = 0 An object in rotational equilibrium does not change its rotational speed. In this case we can say that there is no net torque or in other words that:An object in rotational equilibrium does not change its rotational speed. In this case we can say that there is no net torque or in other words that:  = 0

An unbalanced torque (  ) gives rise to an angular acceleration (  )An unbalanced torque (  ) gives rise to an angular acceleration (  ) We can find an expression analogous toWe can find an expression analogous to F = ma that relates  and  We can see thatWe can see that F t = ma t and F t r = ma t r = mr 2 and F t r = ma t r = mr 2  (since a t = r  ThereforeTherefore Torque and angular acceleration m r FtFt  = mr 2 

Torque and Angular Acceleration Angular acceleration is directly proportional to the net torque, but the constant of proportionality has to do with both the mass of the object and the distance of the object from the axis of rotation – in this case the constant is mr 2 This constant is called the moment of inertia. Its symbol is I, and its units are kgm 2 I depends on the arrangement of the rotating system. It might be different when the same mass is rotating about a different axis  = mr 2 

Newton’s Second Law for Rotation We now have thatWe now have that Where I is a constant related to the distribution of mass in the rotating systemWhere I is a constant related to the distribution of mass in the rotating system This is a new version of Newton’s second law that applies to rotationThis is a new version of Newton’s second law that applies to rotation  = I 

A fish takes a line and pulls it with a tension of 15 N for 20 seconds. The spool has a radius of 7.5 cm. If the moment of inertia of the reel is 10 kgm 2, through how many rotations does the reel spin? (Assume there is no friction)

The torque and corresponding angular acceleration are A fish takes a line and pulls it with a tension of 15 N for 20 seconds. The spool has a radius of 7.5 cm. If the moment of inertia of the reel is 10 kgm 2, through how many rotations does the reel spin? (Assume there is no friction) If we assume the spool starts from rest (ω i = 0) then which is 3.5 rotations (divide 22.5 by 2π rotation) Using rotational kinematics, the angle through which the spool spins is

Angular Acceleration and I The angular acceleration reached by a rotating object depends on M, r, (their distribution) and T When objects are rolling under the influence of gravity, only the mass distribution and the radius are important T

Moments of Inertia for Rotating Objects I for a small mass m rotating about a point a distance r away is mr 2 What is the moment of inertia for an object that is rotating –such as a rolling object? Disc?Sphere?Hoop?Cylinder?

Moments of Inertia for Rotating Objects The total torque on a rotating system is the sum of the torques acting on all particles of the system about the axis of rotation – and sinceis the same for all particles: and since  is the same for all particles: I   mr 2 = m 1 r m 2 r m 3 r 3 2 +… Axis of rotation

Continuous Objects To calculate the moment of inertia for continuous objects, we imagine the object to consist of a continuum of very small mass elements dm. Thus the finite sum Σm i r 2 i becomes the integral

Moment of Inertia of a Uniform Rod L Lets find the moment of inertia of a uniform rod of length L and mass M about an axis perpendicular to the rod and through one end. Assume that the rod has negligible thickness.

Moment of Inertia of a Uniform Rod We choose a mass element dm at a distance x from the axes. The mass per unit length (linear mass density) is λ = M / L

Moment of Inertia of a Uniform Rod dm = λ dx

Example: Moment of Inertia of a Dumbbell A dumbbell consist of point masses 2kg and 1kg attached by a rigid massless rod of length 0.6m. Calculate the rotational inertia of the dumbbell (a) about the axis going through the center of the mass and (b) going through the 2kg mass.

Example: Moment of Inertia of a Dumbbell

Moment of Inertia of a Uniform Hoop R dm All mass of the hoop M is at distance r = R from the axis

Moment of Inertia of a Uniform Disc R dr Each mass element is a hoop of radius r and thicknessdr. Mass per unit area Each mass element is a hoop of radius r and thickness dr. Mass per unit area σ = M / A = M /πR 2 r We expect that I will be smaller than MR 2 since the mass is uniformly distributed from r = 0 to r = R rather than being concentrated at r = R as it is in the hoop.

Moment of Inertia of a Uniform Disc R dr r

Moments of inertia I for Different Mass Arrangements

Which one will win? A hoop, disc and sphere are all rolled down an inclined plane. Which one will win?

Which one will win? A hoop, disc and sphere are all rolled down an inclined plane. Which one will win? 1. Hoop I = MR 2 2. Disc I = ½MR 2 3. Sphere I = 2/5MR 2  =  / I   =  / MR 2 2.α 2 = 2(  / MR 2 )   = 2.5(  / MR 2 )

I. Rotational variables Rigid body: body that can rotate with all its parts locked together and without shape changes. every point of a body moves in a circle whose center lies on the rotation axis. Every point moves through the same angle during a particular time interval. Rotation axis: every point of a body moves in a circle whose center lies on the rotation axis. Every point moves through the same angle during a particular time interval. Angular position: the angle of the reference line relative to the positive direction of the x-axis. Units: radians (rad) Reference line: fixed in the body, perpendicular to the rotation axis and rotating with the body.

Note: we do not reset θ to zero with each complete rotation of the reference line about the rotation axis. 2 turns  θ =4π Translation: body’s movement described by x(t). Rotation: body’s movement given by θ(t) = angular position of the body’s reference line as function of time. Angular displacement: body’s rotation about its axis changing the angular position from θ 1 to θ 2. Clockwise rotation  negative Counterclockwise rotation  positive

Angular velocity: Average: Instantaneous: Units: rad/s or rev/s

These equations hold not only for the rotating rigid body as a whole but also for every particle of that body because they are all locked together. Angular speed (ω): magnitude of the angular velocity. Angular acceleration: Average: Instantaneous:

Angular quantities are “normally” vector quantities  right hand rule. Object rotates around the direction of the vector  a vector defines an axis of rotation not the direction in which something is moving. Examples: angular velocity, angular acceleration Exception: angular displacements

II. Rotation with constant angular acceleration acceleration Linear equations Angular equations

III. Relation between linear and angular variables Position: θ always in radians Speed: ω in rad/s Since all points within a rigid body have the same angular speed ω, points with greater radius have greater linear speed, v. v is tangent to the circle in which a point moves If ω=cte  v=cte  each point within the body undergoes uniform circular motion Period of revolution:

Acceleration: Tangential component of linear acceleration Radial component of linear acceleration: Responsible for changes in the direction of the linear velocity vector v Units: m/s 2

Kinetic energy of rotation Reminder: Angular velocity, ω is the same for all particles within the rotating body. Linear velocity, v of a particle within the rigid body depends on the particle’s distance to the rotation axis (r). Moment of Inertia

Kinetic energy of a body in pure rotation Kinetic energy of a body in pure translation

Discrete rigid body  I =∑m i r i 2 Continuous rigid body  I = ∫r 2 dm

Parallel axis theorem Proof: Rotational inertia about a given axis = Rotational Inertia about a parallel axis that extends trough body’s Center of Mass + Mh 2 h = perpendicular distance between the given axis and axis through COM. R

Units: Nm Tangential component, F t : does cause rotation  pulling a door perpendicular to its plane. F t = F sinφ Radial component, F r :does not cause rotation  pulling a door parallel to door’s plane. Radial component, F r : does not cause rotation  pulling a door parallel to door’s plane.TorqueTorque: Twist  “Turning action of force F ”.

r ┴ : Moment arm of F r : Moment arm of F t Sign: Torque >0 if body rotates counterclockwise. Torque <0 if clockwise rotation. Torque <0 if clockwise rotation. Superposition principle: When several torques act on a body, the net torque is the sum of the individual torques Vector quantity

Newton’s second law for rotation Proof: Particle can move only along the circular path  only the tangential component of the force F t (tangent to the circular path) can accelerate the particle along the path.

VII. Work and Rotational kinetic energy Translation Rot Rotation Work-kinetic energy Theorem Work, rotation about fixed axis Work, constant torque Power, rotation about fixed axis Proof:

Rotational Kinetic Energy We must rewrite our statements of conservation of mechanical energy to include KE r Must now allow that (in general): ½ mv 2 +mgh+ ½ I  2 = constant Could also add in e.g. spring PE

Example - Rotational KE What is the linear speed of a ball with radius 1 cm when it reaches the end of a 2.0 m high 30 o incline?What is the linear speed of a ball with radius 1 cm when it reaches the end of a 2.0 m high 30 o incline? mgh+ ½ mv 2 + ½ I  2 = constant Is there enough information?Is there enough information? 2 m

Example - Rotational KE So we have that: The velocity of the centre of mass and the tangential speed of the sphere are the same, so we can say that: Rearranging for v f : mgh+ ½ mv 2 + ½ Iω 2 = constant

Example: Conservation of KE r A boy of mass 30 kg is flung off the edge of a roundabout (m = 400 kg, r = 1 m) that is travelling at 2 rpm. What is the speed of the roundabout after he falls off?

Example: Conservation of KE r A boy of mass 30 kg is flung off the edge of a roundabout (m = 400 kg, r = 1 m) that is travelling at 2 rpm. What is the speed of the roundabout after he falls off? Roundabout is a disk: Boy has

During a certain period of time, the angular position of a swinging door is described by During a certain period of time, the angular position of a swinging door is described by θ= t t 2 where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0 and (b) at t = 3.00 s.

During a certain period of time, the angular position of a swinging door is described by During a certain period of time, the angular position of a swinging door is described by θ= t t 2 where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0 and (b) at t = 3.00 s. Solution: (a)

During a certain period of time, the angular position of a swinging door is described by θ= t t 2 where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0 and (b) at t = 3.00 s. Solution: Solution: (b)

The four particles are connected by rigid rods of negligible mass. The origin is at the center of the rectangle. If the system rotates in the xy plane about the z axis with an angular speed of 6.00 rad/s, calculate (a) the moment of inertia of the system about the z axis and (b) the rotational kinetic energy of the system.

(a)

In this case, (b) The four particles are connected by rigid rods of negligible mass. The origin is at the center of the rectangle. If the system rotates in the xy plane about the z axis with an angular speed of 6.00 rad/s, calculate (a) the moment of inertia of the system about the z axis and (b) the rotational kinetic energy of the system.

Many machines employ cams for various purposes, such as opening and closing valves. In Figure, the cam is a circular disk rotating on a shaft that does not pass through the center of the disk. In the manufacture of the cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. The cam, of mass M, is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the cam when it is rotating with angular speed ω about the axis of the shaft?

We consider the cam as the superposition of the original solid disk and a disk of negative mass cut from it. With half the radius, the cut-away part has one-quarter the face area and one-quarter the volume and one-quarter the mass of the original solid cylinder: By the parallel-axis theorem, the original cylinder had moment of inertia The negative mass portion has : The whole cam has: and

Find the net torque on the wheel in Figure about the axle through O if a = 10.0 cm and b = 25.0 cm.

The thirty-degree angle is unnecessary information.

A block of mass m 1 = 2.00 kg and a block of mass m 2 = 6.00 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250m and mass M = 10.0 kg. These blocks are allowed to move on a fixed block-wedge of angle = 30.0  as in Figure. The coefficient of kinetic friction is for both blocks. Draw free-body diagrams of both blocks and of the pulley. Determine (a) the acceleration of the two blocks, and (b) the tensions in the string on both sides of the pulley.

For m 1 : For pulley,

For m 2 :

(a) Add equations for m 1, m 2, and for the pulley:

(b)

A uniform rod 1.1 m long with mass 0.7 kg is pivoted at one end, as shown in Fig., and released from a horizontal position. Find the torque about the pivot exerted by the force of gravity as a function of the angle that the rod makes with the horizontal direction.

From the diagram we see that the moment arm of the force of gravity is Thus θ mg

A seesaw pivots as shown in Fig. (a) What is the net torque about the pivot point? (b) Give an example for which the application of three different forces and their points of application will balance the seesaw. Two of the forces must point down and the other one up.

(a)We choose the clockwise direction as positive. (b) We see that changing the 7-N force to ~5.4 N will make the torque zero. An upward force of 8.4 N at the pivot will make the resultant force equal to zero.

Four small spheres are fastened to the corners of a frame of negligible mass lying in the xy plane (Fig. 10.7). Two of the spheres have mass m = 3.1kg and are a distance a = 1.7 m from the origin and the other two have mass M = 1.4 kg and are a distance a = 1.5 m from the origin. (a) If the rotation of the system occurs about the y axis, as in Figure a, with an angular speed ω = 5.1rad/s, find the moment of inertia I y about the y axis and the rotational kinetic energy about this axis. Suppose the system rotates in the xy plane about an axis (the z axis) through O (Fig. b). Calculate the moment of inertia about the z axis and the rotational energy about this axis.

Four small spheres are fastened to the corners of a frame of negligible mass lying in the xy plane (Figure). Two of the spheres have mass m = 3.1kg and are a distance a = 1.7 m from the origin and the other two have mass M = 1.4 kg and are a distance a = 1.5 m from the origin. (a) If the rotation of the system occurs about the y axis, as in Figure a, with an angular speed ω = 5.1rad/s, find the moment of inertia I y about the y axis and the rotational kinetic energy about this axis. Suppose the system rotates in the xy plane about an axis (the z axis) through O (Fig. b). Calculate the moment of inertia about the z axis and the rotational energy about this axis. (a)

Four small spheres are fastened to the corners of a frame of negligible mass lying in the xy plane (Fig. 10.7). Two of the spheres have mass m = 3.1kg and are a distance a = 1.7 m from the origin and the other two have mass M = 1.4 kg and are a distance a = 1.5 m from the origin. (a) If the rotation of the system occurs about the y axis, as in Figure a, with an angular speed ω = 5.1rad/s, find the moment of inertia I y about the y axis and the rotational kinetic energy about this axis. Suppose the system rotates in the xy plane about an axis (the z axis) through O (Fig. b). Calculate the moment of inertia about the z axis and the rotational energy about this axis. (b)

The reel shown in Figure has radius R and moment of inertia I. One end of the block of mass m is connected to a spring of force constant k, and the other end is fastened to a cord wrapped around the reel. The reel axle and the incline are frictionless. The reel is wound counterclockwise so that the spring stretches a distance d from its unstretched position and is then released from rest. (a) Find the angular speed of the reel when the spring is again unstretched. (b) Evaluate the angular speed numerically at this point if I = 1.00 kg·m 2, R = m, k = 50.0 N/m, m = kg, d = m, and θ= 37.0°.

(a)

(b)

A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on a horizontal section of a track, as shown in Figure. It rolls around the inside of a vertical circular loop 90.0 cm in diameter, and finally leaves the track at a point 20.0 cm below the horizontal section. (a) Find the speed of the ball at the top of the loop. Demonstrate that it will not fall from the track. (b) Find its speed as it leaves the track. (c) Suppose that static friction between ball and track were negligible, so that the ball slid instead of rolling. Would its speed then be higher, lower, or the same at the top of the loop? Explain.

(a) Energy conservation for the system of the ball and the Earth between the horizontal section and top of loop:

A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on a horizontal section of a track, as shown in Figure. It rolls around the inside of a vertical circular loop 90.0 cm in diameter, and finally leaves the track at a point 20.0 cm below the horizontal section. (a) Find the speed of the ball at the top of the loop. Demonstrate that it will not fall from the track. (b) Find its speed as it leaves the track. (c) Suppose that static friction between ball and track were negligible, so that the ball slid instead of rolling. Would its speed then be higher, lower, or the same at the top of the loop? Explain. (a) Energy conservation for the system of the ball and the Earth between the horizontal section and top of loop: