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Transcript

Moments of INERTIA Review of Inertia • Inertia – Objects with mass always resist a change in there motion (acceleration). From Newton’s second law we see that this resistance depends on the objects’ mass only. • SF = ma or a = SF/m • However this concept of inertia applies to translational does it also apply to rotational motion? The “Roll” of Mass • Let us say there are two wheels, both the same size, one is made out of a light plastic and the other heavy lead. If you to spin them, which would be harder to get to move, and which one would be harder to stop once it started spinning? • The answer many would say the lead one and they would be right, but why? • Obviously the lead wheel has more mass so it will have a greater resistance to change. The point of mass • Looking at the wheel as an object spinning Newton’s 2nd law of motion does not seam to apply because the wheel’s motion is rotational and not translational. • However, if you think of the wheel as a collection of points each with their own mass all moving in circular motion (which is translational) Newton’s 2nd law does seam to apply Rotational motion Rotational motion Translational motion The Shape of things • Who is harder to spin in a circle with a rotational speed of 1 RPM; a 10 g point moving with a radius of 20 cm, or a 10 g point moving with a radius of 1 m? • Since both points complete 1 circle in the same amount of time the outside point must travel 5x faster than the inside point, which means it is harder to get the outside point to move at the same angular rate as an inside point. • So let us say we have to a 1 kg disk, with a radius of 50 cm, whose mass is evenly distributed through from it’s center to edge. And a 1kg hoop, with a radius of 50 cm, that has no mass on the interior and all of it’s mass is on it’s edge. • Which of these two would be harder to get to spin, and which would be harder to stop spinning once they get started? 1 kg disk 1 kg Hoop • There can be no difference due to the quantity of mass since they both have the same mass. • However the hoop has all of its mass moving in the largest possible circle making it a difficult as possible to change the hoop’s rotational motion. • So the difficulty is not just about mass, but how it is distributed in the object that is rotating. • The disk’s mass is spread out meaning it will be easier to change the rotational motion of the inter mass since The sum resistance of each hoop from small to large is Less than the resistance of one large hoop. Moments of Inertia • Moments of Inertia is an object’s resistance to change in it’s rotational motion – It is how well an objects resists angular acceleration – Moments of Inertia play the same roles in rotational motion as mass does in translational motion. Moments of Inertia • The symbol we use for the Moment of Inertia is the letter “I” • The Units for the Moment of inertia are often Kg*m2 – They are always mass*distance2 • Each shape has it’s own unique equation for its moment of inertia – Some examples: • • • • I Thin Hoop = Mass*Radius2 = mr2 I Disk = (1/2) Mass*Radius2 = (1/2)mr2 I solid Sphere = (2/5)Mass*Radius2 = (2/5)mr2 I Hollow Sphere = (2/3)Mass*Radius2 = (2/3)mr2 A “rotational version of Newton’s 2nd Law” • We know that when an object is pulled by an unbalanced force it moves in a translational way with an acceleration that is a = SF/m. But what if the object had an unbalanced torque? • For example let’s say there is a wheel that has a fixed axis in it’s center. And there is a force that pulls an the rim of the wheel. (For simplicity we will ignore gravity) Radius (r) Torque (t) Fixed axis lets the wheel spin, but Not translate (holds the wheel in place) • The net unbalanced torque cause the rotation of the disk to change – The net torque creates and angular acceleration (much like a force creates a translational acceleration) • The relationship between the Net torque and angular acceleration is very logical if you remember the following – Torques are like forces for the “rotational world” – Moments of inertia acts like mass for the “rotational motion” – Accelerations are accelerations • So since for translational motion SF = ma – Or a = SF/m – The rate of an object’s change in motion is a ratio of the Force’s influence to create change in translational motion and the object’s ability to resist change • Then for rotational motion St = Ia – Or a= St /I – The rate of an object’s change in rotational motion is a ratio of the torque’s influence to create change in rotational motion and the object’s ability to resist change A “rotational version of Newton’s 2nd Law” • Now let’s say there is a wheel that has a fixed axis in it’s center. And there is a force that pulls an the rim of the wheel. And we want to find the angular acceleration of the disk. (For simplicity we will ignore gravity) q Force (F) creates the torque Since St =Ia And t = [Fsin(q)]r Radius (r) [Fsin(q)]r = Ia Torque (t) Created by the force a = {[Fsin(q)]r}/I t = Fr = [Fsin(q)]r Fixed axis lets the wheel spin, but Not translate (holds the wheel in place) Remember that for a disk I = (1/2)mr2 Problem Solving • Since rotational dynamics follows its own version of Newton’s 2nd Law (like translational dynamics) we can use pretty much the same strategy • Step 1: Draw free body diagram – Done similarly with forces, but here you need to draw the forces where the act on the object, not just at the object’s center of mass. • Step 2: Write out the net Force equation, AND torque equation – Instead of just SF = ma, we also use St = Ia • Step 3: Find accelerations and Moments of inertia – Since it is not just the object’s mass, but shape determines how an object resists a net torque we will normally have to calculate the object’s moment of inertia. – We normal will have to find the angular acceleration in the system, sometimes we will need to relate the object’s angular acceleration to the tangential acceleration to the object’s surface. • Step 4: Solve the problem Single Object same problem • We have a disk that has a mass of 10 kg and a radius of .5 meters. A string is wrapped around the disk and pulled with a constant force of 50N as shown, for a time of 6 seconds. What is the disk’s final angular velocity, and what is the tangential speed of the wheel’s rim? 50 N 30O Step 1 Free body diagram • Draw all the forces 50 N (Reaction Force from the disk’s holder) Ry 30O (Reaction Force from the disk’s holder) Rx Fg = -98 N Step 1 Free body diagram • Draw Torques 50 N (Reaction Force from the disk’s holder) Ry 30O (Reaction Force from the disk’s holder) Rx Picking the axial as the pivot. Fg = -98 N Notice that all the blue forces go trough the axils so the produce no torque. Step 1 Free body diagram • Draw and find torques 50 N 60O 30O .5m -[(50N Sin(60)).5m)] or -[(50N Cos(30)).5m] Picking the axial as the pivot. Only the 50N force does not go through the axel so it creates a torque around it. Step 2 Write the torque equation 50 N 60O 30O .5m -[(50N Sin(60)).5m)] or -[(50N Cos(30)).5m] Picking the axial as the pivot. St = -[(50N Sin(60)).5m)] = Ia = -[12.5Nm] = Ia Step 3 Find moments of inertia and/or accelerations 50 N 60O 30O .5m -[(50N Sin(60)).5m)] or -[(50N Cos(30)).5m] Picking the axial as the pivot. IDisk = (1/2)Mr2 = (1/2)(10kg)(.5m)2 IDisk = 12.5 kgm2 Step 3 Find moments of inertia and/or accelerations 50 N 60O 30O .5m -[12.5Nm]] Picking the axial as the pivot. St = Ia = -[12.5Nm] = [12.5 kgm2]a a = -1 rad/sec2 Step 3 Find moments of inertia and/or accelerations a = (.5m/s2) .5m a = -1 rad/sec2 a = ar a = (1 rad/sec2)(.5m) Remember we remove the sign for the acceleration because rotational language is different than translational language Step 4: Solve Vf = Vi + at Vf = ( 0m/s) + (.5m/s2)(6s) Vf = 3m/s wf = wi + at wf = ( 0m/s) + (-1 rad/s2)(6s) wf = -6 rad/s wf = 6 rad/s clockwise An Application • Let us look at a disk that is attached and wrap around several times by a rope that is also attached to an object that will fall. The disk has a fixes axel in it’s center. How can we find the falling objects acceleration? (Since the object is attached to the disk it can’t fall freely so the acceleration must be less than 9.8 m/s2.) Radius = r Mass = MD Mass = Mh Step 1: FBD Just like all problems where forces are involved we start with a free body diagram. It is easier to 1st separate the objects and then look at all the forces on the objects. Fn = The reaction force from the axel (since it is fixed) Tension force = FT Radius = r Mass = MD Mass = Mh Tension force = -FT Fg = (MD)(-9.8 m/s2) Fg = (Mh)(-9.8 m/s2) Step 2: Force equations Fn = The reaction force from the axel (since it is fixed) Tension force = FT Radius = r Mass = MD Mass = Mh Tension force = -FT Fg = (MD)(-9.8 m/s2) Fg = (Mh)(-9.8 m/s2) SF: Fn + FTension + Fg = (MD)(0 m/s2) SF: FTension + Fg = (Mh)a SF: Fn + -FT + (MD)(-9.8 m/s2) = 0 N SF: FT + (Mh)(-9.8 m/s2) = (Mh)a Oh no!! Here we would normally find the acceleration by we can’t just yet. We do not have enough information. But when faced to a large problem it is often best to torque you way through it. SF: Fn + -FT + (MD)(-9.8 m/s2) = 0 N FT = (MD)(-9.8 m/s2) + Fn SF: FT + (Mh)(-9.8 m/s2) = (Mh)a FT + (Mh)(-9.8 m/s2) = (Mh)a Step 3a: Use F.B.D to find torques Tension force = FT Mass = Mh Fg = (Mh)(-9.8 m/s2) These forces produce no Torque on the hanging mass So we can ignore this. Step 3a: Use F.B.D to find torques Fn = The reaction force from the axel (since it is fixed) Radius = r Mass = MD Tension force = -FT -FT Fg = (MD)(-9.8 m/s2) Both the gravitational force of the disk and the normal force both act On the disk’s center, so they produce no torque on the disk, This leaves only the tension force to work with. Step 3b: Torque equations Radius = r Mass = MD Tension force = -FT St =(-FT)r = Ia a = (-FT)r /I Remember that for a disk I = (1/2)MDr2 a = (-FT)r /(1/2) MDr2 a = 2(-FT) /MDr Step 4: relating angular to tangential Radius = r Mass = MD Tension force = -FT So the torque equation gives us the angular acceleration: a = 2(-FT) /MDr We can now relate angular acceleration of the disk to the tangential acceleration of it’s rim. That tangential acceleration is equal to the acceleration of the falling object. a = ar so a =a/r a/r = 2(-FT) /MDr a = 2(-FT) /MD Now we can find acceleration a = 2(-FT) /MD FT + (Mh)(-9.8 m/s2) = (Mh)a (-FT) = (1/2)aMD (1/2)aMD = (-FT) (-FT) = (1/2)aMD (Mh)(-9.8 m/s2) = (Mh)a + (1/2)aMD (Mh)(-9.8 m/s2) = a[(Mh) + (1/2)MD] [(Mh)(-9.8 m/s2)]/ [(Mh) + (1/2)MD] = a