ROTATIONAL MOTION

1. ROTATIONAL MOTION NCEA Level 3 Physics

2. ROTATIONAL MOTION • This 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).

3. ROTATIONAL KINEMATICS 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: 4 1 2 3

4. PURE 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. Push CoM Push Both 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. CoM Translational motion Push Rotational motion

5. ANGULAR MEASUREMENTS A1 To move from A to A1 requires an object to move through certain angle of the circle, . To move from B to B1 requiresthe object to move same angle, . However, the distance covered from A to A1 is far greater than the distance from B to B1. B1  B A  = angle of rotation In 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 distance

6. d ANGULAR DISPLACEMENT IN RADIANS length of the arc [ d ] divided by the radius [ r ] produces the angular displacement (θ)  r 1 Radian = 57o 1 revolution = 2 radians

7. IN A FULL CYCLE RADIANS Hence 3600 = 2 Radians 1800 =  radians

8. Example 1: • A bicycle wheel turns through 4½ 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 rad b. Distance d travelled around the rim is calculated by  = d/r d =  x r = 0.40 x 28.2743 = 11 m

9. ANGULAR VELOCITY Suppose the distance, d, from A to A1 takes a time t. v A1 [DIVIDING EACH SIDE BY t] v d  A v = tangential velocity r  = the ANGULAR VELOCITY

10. v = tangential velocity v  = the ANGULAR VELOCITY in RADIANS PER SECOND v  Considering a full circle:  r T = Periodic time, f = frequency

11. Example 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 blade SOLUTION: a. Angular speed  = 4000rpm = 4000 x 2 rad per minute [1rev = 2  rad] = 4000 x 2 60 rad per second = 420 rads-1 b. Linear speed v = r  = 0.80 x 418.879 = 340ms-1

12. ANGULAR ACCERLERATION If 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, .  =  / t Units = rads-2 Just like angular velocity we can combine angular acceleration with translational acceleration using the formula: a = r ‘r’ is the distance from the centre of rotation.

13. 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.  0.20m • SOLUTION: • Since the rim of the pulley has the same acceleration as the string, the angular acceleration is: •  = a/r = 3.0 / 0.20 •  = 15rads-2 • b. The change in angular velocity in 5.0 seconds is: •  = t = 15 x 5.0 •  =75rads-1 [NB pulley started from rest f = ] a = 3.0ms-2

15. 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. Gradient =  Gradient =     Area =  t t t Constant  Constant  Constant  Equations involve 5 variables: Angular acceleration,  Initial angular velocity, i Final angular velocity, f Angular displacement,  Time taken, t.

16. EQUATIONS ARE: f = i + t  = it + ½t2 f2 = i2 + 2  = ½ (f + i)t

17. Example 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 = 180s • Final angular velocity f : • f = i + t =0 + 7.0 x 180 • f =1300 rads-1 • b. Angular dispalcement : •  = it + ½t2 = 0 + ½ x 7.0 1802 •  =110000 rad

19. FORCE & TORQUE Torque, ‘’ is a turning force applied on an object pivoting about the CoM. Force is applied perpendicular to pivot point. Thus:  = Fd Unit = Nm The 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. F = reaction force r F = torque force

20. Example 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 = 14Nm Turning effects can be caused by large forces on small wheels or small forces on large wheels. b. F = /r = 14 / 0.030 = 470N

22. Combining translational and rotational quantities gives us the table below:

23. ROTATIONAL INERTIA Unbalanced 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.

24. Rotational inertia depends upon: a. Its mass -  mass,  . b. How far from the CoR the mass is – further away  . Use the above information to explain why tightrope walkers carry a long bar? 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 inertia

28. 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. 0.20m 1.2ms-2 0.18kg

29. SOLUTION: • 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 – FT • FT = Fg – FU • The gravity force Fg = mg and unbalanced force is found from the acceleration of the mass FU = ma • Fg = 0.18 x 9.8 = 1.764 • FU = 0.18 x 1.2 = 0.216 • FT = 1.764 – 0.216 • FT = 1.5N FT 0.18kg FU Fg

30. b. 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

31. COMPLETE EXERCISE PAGE 67 - 68 RUTTER

32. ROLLING MOTION As 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 = ½2

33. When a rolling object is moving it has two forms of energy: 1. Translational Ek = ½mv2 & 2. Rotational Ekr = ½2 These two combine to give the total Ek of a rolling object. Ekr = ½2 Total gravitational potential energy = + CoM Ek = ½mv2

34. When 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.

35. Example 7 In 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 = 0.70kg Hanging mass, m =0.10kg Height h = 0.60m Final speed of roller v = 1.0ms-1 a. Calculate the rotational kinetic energy of the roller when its speed is 1.0ms-1.  v M timer m h m The roller has a radius of 2.0cm. b. Calculate the rotational inertia of the roller.

36. SOLUTION: The gravitational potential energy lost by the hanging mass is: Ep = mgh = 0.10 x 9.8 x 0.60 Ep = 0.588J This 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 = ½(M + m)v2 = ½ x (0.70 + 0.10) x 1.02 Ek = 0.40J The rotational kinetic energy of the roller is: Ekr = Ep – Ek = 0.588 – 0.40 Ekr = 0.19J

37. b. 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-1 The rotational inertia of the roller can be calculated from: Ekr = ½2 0.188 = ½ x  x 502  = 2 x 0.188 / 502  = 1.5 x 10-4kgm2

38. COMPLETE EXERCISES PAGE 69 - 72 RUTTER

39. ROTATIONAL ENERGY & MOMENTUM PRACTICAL AIM: 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. h d h Prove using equations of motion that vf = 2d/t

40. Using equations of motion to derive final velocity: vf2 = vi2 + 2ad But vi = 0 vf2 = 2ad Substituting an equation of motion in for ‘a’ d = vit + ½at2 But vi = 0 d = ½at2 a = 2d/t2 Substituting in ‘a’ into vf2 = 2ad vf2 = 2 x 2d/t2 x d vf2 = 4d2/t2[ both sides] vf = 2d/t

41. RESULTS: h = _______ m d = _______ m

42. Use 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?

43. ROTATIONAL ENERGY As work is done on an object that is rotating the energy put in is converted into rotational kinetic energy. We know that EK= ½mv2 but this is only translational energy it also has rotational energy EROT = ½I2 Note: these two energies can be added together for an object that is moving translationally and rotationally. ETOT = EK + EROT = ½mv2 + ½I2 This basically obeys the law of energy of conservation

44. COMPLETE EXERCISES PAGE 78 - 80 RUTTER

45. ANGULAR MOMENTUM Just 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-1 Just 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 = mvr This obeys the rules that all momentum must be conserved.

46. The law of energy of conservation states that: The total angular momentum before the collision equals The total angular momentum after the collision. Just like we learnt at level 2. Try the activity on Page 82 RUTTER

47. Example 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 conserved 1I + 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 f f = 3.0 rads-1

48. Example 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-1 The angular momentum, L, of the gas from each rocket is: L = mvr = 0.40 x 100 x 2.0 L = 80kgm2s-1

49. The 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-1 To stop the spacecraft rotating would require an equal burst of gas from the rockets in the opposite direction to the original.

50. Try to explain the applications of conservation of rotational angular momentum. This may include: • Helicopter rotors • Ice-skaters spinning • Bicycle wheels.