ROTATIONAL MOTION

Slide 1

ROTATIONAL MOTIONNCEA Level 3PhysicsROTATIONAL MOTIONThis unit is made up of the following: Rotational Kinematics (page 62 - 65) Force and Torque (page 66). Rotational Inertia (page 67 - 68). Rolling Motion (page 69 - 72). Rotational Kinetic Energy (page 78 - 81). Angular momentum (page 81 - 86).This is important to understand with respect to aspects of circular motion.In orbital motion, the object that is rotating stays a fixed distance from the centre of rotation.In spinning motion the concept of radius has no meaning because each position on a spinning object is rotating with a different radius to an adjacent position:1234ROTATIONAL KINEMATICSPURE TRANSLATION:Movement of the CoM without any rotation about CoM. Force is applied through the CoM.PURE ROTATION:Circular motion around CoM, while CoM remains stationary.CoMPushPushBoth translation and rotation can occur on an object when a resultant force does not act through the CoM. The object will move through a straight line through the CoM while rotating about the CoM.CoMPushTranslational motionRotational motionANGULAR MEASUREMENTSBAB1A1To move from A to A1 requires an object to move through certain angle of the circle, .To move from B to B1 requires the object to move same angle, .However, the distance covered from A to A1 is far greater than the distance from B to B1. = angle of rotationIn order for A & B to leave their start position and finish at their end positions they move through the same angle of rotation over the same time but there is a difference in their motion.A moves faster, over a larger distancerdIN RADIANSlength of the arc [ d ] divided by the radius [ r ] produces the angular displacement ()

1 Radian = 57o 1 revolution = 2 radiansANGULAR DISPLACEMENT

IN A FULL CYCLE

Hence 3600 = 2 Radians1800 = radiansRADIANSExample 1:A bicycle wheel turns through 4 revolutions:What is the angular displacement of the wheel?What distance does a point on the rim travel, if the wheel has a radius of 40cm?SOLUTION:a. = 4.5 revolutions = 4.5 x 2 = 28 radb. Distance d travelled around the rim is calculated by = d/r d = x r = 0.40 x 28.2743 = 11 mvvrdAA1Suppose the distance, d, from A to A1 takes a time t.

v = tangential velocity

= the ANGULAR VELOCITY[DIVIDING EACH SIDE BY t]ANGULAR VELOCITYvvr

v = tangential velocity

= the ANGULAR VELOCITY in RADIANS PER SECONDConsidering a full circle:

T = Periodic time, f = frequencyExample 2:An aircraft propeller rotates at a speed of 4000 revolutions per minute & has a radius of 0.80m. Find: The angular velocity of the propeller. The linear speed of the tip of the propeller bladeSOLUTION:a. Angular speed = 4000rpm = 4000 x 2 rad per minute[1rev = 2 rad] = 4000 x 2 60 rad per second = 420 rads-1b. Linear speed v = r = 0.80 x 418.879 = 340ms-1ANGULAR ACCERLERATIONIf an object is moving around a circle then even if its angular velocity is constant its direction is still changing. Thus it must be accelerating and hence has angular acceleration, . = / tUnits = rads-2Just like angular velocity we can combine angular acceleration with translational acceleration using the formula:a = rr is the distance from the centre of rotation.Example 3:A pulley of radius 0.20m has a length of string wrapped around it, attached to a mass. The mass is released from rest & falls with an acceleration of 3.0ms-2. Find: The angular acceleration of the pulley. The angular velocity of the pulley after 5.0s.0.20ma = 3.0ms-2SOLUTION:Since the rim of the pulley has the same acceleration as the string, the angular acceleration is: = a/r= 3.0 / 0.20 = 15rads-2b. The change in angular velocity in 5.0 seconds is: = t= 15 x 5.0 = 75rads-1[NB pulley started from rest f = ]ROTATIONAL KINEMATIC EQUATIONS:As we saw with translational motion graphs can be drawn which resulted in our equations of motion. The same graphs can be drawn for rotational motion.tttGradient = Area = Constant Gradient = Constant Constant Equations involve 5 variables:Angular acceleration, Initial angular velocity, iFinal angular velocity, fAngular displacement, Time taken, t.EQUATIONS ARE:f = i + t = it + t2f2 = i2 + 2 = (f + i)tExample 4:A centrifuge can accelerate from rest at a constant angular acceleration of 7.0rads-2, taking 3 minutes to reach top speed. What is the final angular velocity. What angle does it turn through during this time?SOLUTION:Info given i = 0; = 7.0rads-2 & t = 3 minutes = 180sFinal angular velocity f : f = i + t = 0 + 7.0 x 180 f = 1300 rads-1b. Angular dispalcement : = it + t2= 0 + x 7.0 1802 = 110000 rad FORCE & TORQUEThe effect of applying a torque causes rotational motion. In pure rotation the torque is caused by a force couple, e.g. on an axle where one force acts as a reaction force through centre of rotation (CoR) and the other acts as a torque force a certain distance away. = FdTorque, is a turning force applied on an object pivoting about the CoM. Force is applied perpendicular to pivot point. Thus:Unit = NmF = torque forceF = reaction forcerExample 5:On the rear wheel of a mountain bike the chain varies position from one gear to another. At first the chain is on the largest gear wheel (radius 7.0cm) and pulls with a force of 200N. The rider then changes gear and the chain is shifted to the smallest gear wheel (radius 3.0cm). Calculate the torque the chain applies to the rear wheel when it is on its largest wheel. Calculate the force needed to apply the same torque when the chain is shifted to the smallest wheel.SOLUTION:a. = Fr = 200 x 0.070m = 14Nmb. F = /r = 14 / 0.030 = 470NTurning effects can be caused by large forces on small wheels or small forces on large wheels.Combining translational and rotational quantities gives us the table below:TRANSLATIONALROTATIONALEQUATIONd = d/rd = rv = d/t = /tv = r a = v/t = /ta = r ROTATIONAL INERTIAUnbalanced forces acting on an object causes the object to accelerate. Newton's second law. What determines how likely it is for an object to move is the mass. Increase the mass the force has less effect on accelerating it. The mass resists the force. We call this linear inertia.Likewise this is true for rotational motion. In this case the force applied is a torque which generates an angular acceleration. Hence: = Where = rotationl inertia, or the resistance an object has to change its angular velocity. Unit = kgm2.Rotational inertia depends upon:a. Its mass - mass, .b. How far from the CoR the mass is further away .A tightrope walker will often carry a long bar for balance. If the tightrope walker falls, it will be because he has rotated too far about his foot (the CoR). Increasing his reluctance to rotate would be a great advantage! Carrying the bar means that the system of man and bar not only has more mass, but also has some of its mass a long way from the CoR, thus increasing his rotational inertiaUse the above information to explain why tightrope walkers carry a long bar?

Often a good way to find the rotational inertia of an object is to suspend a mass by a string around the object. This allows the torque and angular acceleration to be measured. From this using = , can be calculated.Example 6:A mass of 0.18kg is used to accelerate a wheel of radius 0.20m. The mass accelerates downwards at 1.2ms-2. Calculate the tension force in the string. Calculate the torque on the wheel. Calculate the angular acceleration of the wheel.Calculate the rotational inertia of the wheel.

1.2ms-20.18kg0.20mSOLUTION:Consider the forces acting on the mass. The tension force in the string acts upwards & the gravity acts down wards. The mass is accelerating downwards, and so the resultant unbalanced force must be downwards, & is given by: FU = Fg FTFT = Fg FUThe gravity force Fg = mg and unbalanced force is found from the acceleration of the mass FU = maFg = 0.18 x 9.8 = 1.764FU = 0.18 x 1.2 = 0.216FT = 1.764 0.216FT = 1.5N0.18kgFTFgFUb. Consider the forces acting on the wheel. The only force acting to turn the wheel is the tension force in the string, and so the torque on the wheel is given by: = FT x r= 1.548 x 0.20 = 0.31Nm

c. a = r therefore = a/r = 1.2 / 0.20 = 6.0 rads-2

d. = therefore = / = 0.3096 / 6.0 = 0.052 kgm2

COMPLETE EXERCISEPAGE 67 - 68RUTTERROLLING MOTIONAs with any form of motion energy must be involved and obviously rotational motion is no exception. Place a boulder up a hill and let it go. If you are coming up the hill in the path of the boulder you will get squished!!!Meeting the boulder at the bottom compared to halfway up or at the top leads to an increase in squisheyness!!!Why? more energy is involved.Energy when spinning is rotational kinetic energy, Ekr. Measured in joules and transformed from one form into another. Made up of similar components to Ek.Ekr = 2When a rolling object is moving it has two forms of energy:1. Translational Ek = mv2 &2. Rotational Ekr = 2These two combine to give the total Ek of a rolling object.CoM Ek = mv2 Ekr = 2+=Total gravitational potential energyWhen rolling shapes down a slope not all shapes will move the same. What is important is how the mass is distributed. The further away the mass is from the centre the more inertia it has, thus is harder to move. A hollow cylinder is harder to move than a ball bearing of the same mass. This is due to the cylinder having its mass further away from the centre thus increasing its rotational inertia.Check out the practical on rolling objects.

For a body that is both translating and rotating:Calculate the Ek using the velocity of the CoM.Calculate the Ekr using the about the CoM.Add these two to obtain the total kinetic energy.Example 7In an experiment, a heavy roller, M, is accelerated as a mass, m, falls to the ground. A timer is used to record the speed of the roller after it has moved distance h.Sample results are:Mass of roller, M = 0.70kgHanging mass, m =0.10kgHeight h = 0.60mFinal speed of roller v = 1.0ms-1a. Calculate the rotational kinetic energy of the roller when its speed is 1.0ms-1.MmmhtimervThe roller has a radius of 2.0cm. b. Calculate the rotational inertia of the roller.SOLUTION:The gravitational potential energy lost by the hanging mass is:Ep = mgh= 0.10 x 9.8 x 0.60Ep = 0.588JThis gravitational potential energy is changed into linear kinetic energy of the mass and of the roller plus rotational kinetic energy of the roller; Ep = Ek + Ekr.The linear kinetic energy gained by the roller and hanging mass is:Ek = (M + m)v2= x (0.70 + 0.10) x 1.02Ek = 0.40JThe rotational kinetic energy of the roller is:Ekr = Ep Ek= 0.588 0.40Ekr = 0.19Jb. When the roller is moving forward at a speed of 1.0ms-1, it is also rotating with an angular speed, , given by: = v / r= 1.0 / 0.020m = 50rads-1The rotational inertia of the roller can be calculated from:Ekr = 20.188 = x x 502 = 2 x 0.188 / 502 = 1.5 x 10-4kgm2COMPLETE EXERCISESPAGE 69 - 72RUTTERROTATIONAL ENERGY & MOMENTUM PRACTICALAIM: To investigate rolling different shapes down a slope.

If you drop different masses through the same height, they all take the same time to fall. But, when you roll them down a slope they take different times.Obtain the rolling cylinder set. They all have the same radius and the same mass. But will they take the same time to roll down a slope?

Set up a ramp. Time each cylinder a few times down the slope. Record the average time, t, for each one.

Measure and record the mass, m and the radius, r. Measure and record the two distances, h and d.hdhProve using equations of motion that vf = 2d/tUsing equations of motion to derive final velocity:vf2 = vi2 + 2adBut vi = 0vf2 = 2adSubstituting an equation of motion in for ad = vit + at2But vi = 0d = at2a = 2d/t2

Substituting in a into vf2 = 2advf2 = 2 x 2d/t2 x d vf2 = 4d2/t2[ both sides]vf = 2d/tRESULTS:TrialWood cylinderBrass cylinder12345Average Timeh = _______ md = _______ mUse your slowest object for these calculations:

Calculate the cylinders final speed, v, using v = 2d/t, which is the final speed of the cylinder.Calculate its final linear kinetic energy, Ek(linear)Calculate the initial gravitational potential energy, Ep(grav)Hence find the final rotational kinetic energy, Ek(rot)Use v = r to calculate the cylinders final angular speed, Use your value for the rotational kinetic energy to calculate the rotational inertia, I, of the cylinder.Explain why the cylinders take different times. Why would a small ball bearing be the fastest?ROTATIONAL ENERGYAs work is done on an object that is rotating the energy put in is converted into rotational kinetic energy. We know that EK = mv2but this is only translational energy it also has rotational energy EROT = I2Note: these two energies can be added together for an object that is moving translationally and rotationally.ETOT = EK + EROT = mv2 + I2This basically obeys the law of energy of conservationCOMPLETE EXERCISESPAGE 78 - 80RUTTERANGULAR MOMENTUMJust like all linear motion has speed, acceleration and momentum so does all rotational motion. We call rotational momentum angular momentum as it is mass moving in a circular motion. Thus angular momentum (L) is made up of a mass (m) and angular velocity (). However instead of mass we now talk about rotational inertia (). Hence we get the formula:L = Units = kgm2rads-1Just like angular velocity and acceleration, angular momentum can also be linked to linear momentum. This would occur when an object having linear momentum moves into circular motion e.g. like a child running onto a roundabout. Hence:L = mvrThis obeys the rules that all momentum must be conserved.The law of energy of conservation states that:The total angular momentum before the collision equalsThe total angular momentum after the collision.Just like we learnt at level 2.Try the activity on Page 82 RUTTERExample 8:A turntable (1 = 0.090kgm2) spins freely with an angular velocity I = 4.0 rads-1. A disc (2 = 0.030kgm2) is dropped onto the turntable. Calculate the new angular velocity f.SOLUTION:Since the turntable is spinning freely, the total angular momentum is conserved1I + 2 x 0 = (1 + 2)f [1 + 2 is the rotational inertia of turntable + disc]0.090 x 4.0 = (0.090 + 0.030) x ff = 3.0 rads-1Example 9:A spacecraft is stationary in space. In oredr to turn the spacecraft, two small rockets are fired for a fraction of a second. The spacecraft is a cylinder with a radius of 2.0m and a rotational inertia, , of 1600kgm2.The small rockets each fire 0.40Kg of gas at a speed of 100ms-1. Calculate the angular speed of the spacecraft.SOLUTION:The linear momentum, , of the gas from the rocket is given by: = mv= 0.40 x 100 = 40kgms-1The angular momentum, L, of the gas from each rocket is:L = mvr= 0.40 x 100 x 2.0L = 80kgm2s-1The two rockets produce a total angular momentum of 160Kgm2s-1 clockwise about the spacecraft.Since angular momentum is conserved, the spacecraft will rotate anticlockwise qith an equal angular momentum of 160kgm2s-1.The angular speed of the spacecraft is: = L / = 160 / 1600 = 0.10rads-1To stop the spacecraft rotating would require an equal burst of gas from the rockets in the opposite direction to the original.Try to explain the applications of conservation of rotational angular momentum. This may include: Helicopter rotors Ice-skaters spinning Bicycle wheels.COMPLETE EXERCISESPAGE 81 - 86RUTTER