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Physics 1501: Lecture 21 Today ’ s Agenda. Sec. 1-7. Announcements HW#8: due Oct. 28 Honors ’ students see me after class Midterm 1: average ~ 45 % … Topics Torque Rotational energy Rolling motion . 0 10 20 30 40 50 60 70 80 90 100. Sec. 21-28.

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## Physics 1501: Lecture 21 Today ’ s Agenda

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**Physics 1501: Lecture 21Today’s Agenda**Sec. 1-7 • Announcements • HW#8: due Oct. 28 • Honors’ students • see me after class • Midterm 1: average ~ 45 % … • Topics • Torque • Rotational energy • Rolling motion 0 10 20 30 40 50 60 70 80 90 100 Sec. 21-28 0 10 20 30 40 50 60 70 80 90 100**Summary (with comparison to 1-D kinematics)**Angular Linear And for a point at a distance R from the rotation axis: • x = Rv = R a = R**Rotation & Kinetic Energy...**• 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.**y**x z y x z Direction of Rotation: • In general, the rotation variables are vectors (have a direction) • If the plane of rotation is in the x-y plane, then the convention is • CCW rotation is in the + zdirection • CW rotation is in the- zdirection**Rotational Dynamics:What makes it spin?**TOT=I • This is the rotational version of FTOT = ma • Torque is the rotational cousin of force: • The amount of “twist” provided by a force. • Moment of inertia I is the rotational cousin of mass. • If I is big, more torque is required to achieve a given angular acceleration. • Torque has units of kg m2/s2 = (kg m/s2) m = Nm.**Comment on=I**• When we write =I we are really talking about the z component of a more general vector equation. (we normally choose the z-axis to be the the rotation axis.) z=Izz • We usually omit the zsubscript for simplicity. z Iz z z**Example**• To loosen a stuck nut, a man pulls at an angle of 45o on the end of a 50cm wrench with a force of 200 N. • What is the magnitude of the torque on the nut? • If the nut suddenly turns freely, what is the angular acceleration of the wrench? (The wrenchhas a mass of 3 kg, and its shape is that of a thin rod). 45o F=200N L=0.5m**So =/I = (70.7Nm) / (.25kgm2)**= 283 rad/s2 Example: solution • Torque =Lfsin = (0.5m)(200N)(sin(45)) = 70.7 Nm • If the nut turns freely (no more friction), =I • We know and we want , so we need to figure out I. 45o F=200N L=0.5m **Torque and the Right Hand Rule:**• The right hand rule can tell you the direction of torque: • Point your hand along the direction from the axis to the point where the force is applied. • Curl your fingers in the direction of the force. • Your thumb will point in the directionof the torque. F y r x z**B** A C Review: The Cross Product • We can describe the vectorial nature of torque in a compact form by introducing the “cross product”. • The cross product of two vectors is a third vector: AXB=C • The length of C is given by: C = ABsin • The direction of C is perpendicular to the plane defined by A and B, and inthe direction defined by the right-handrule.**The Cross Product**j • The cross product of unit vectors: ixi=0 ixj =k ix k = -j jxi = -k jx j =0 jx k =i kxi =j kx j =-i kx k =0 AXB = (AX i + AYj+ Azk) X (BX i + BYj+ Bzk) = (AX BX i x i +AX BY i xj+AX BZ i xk) + (AY BX jx i +AY BY jxj+ AY BZ jxk) + (AZ BX kx i +AZ BY kxj +AZ BZ kxk) i k**The Cross Product**• Cartesian components of the cross product: C =AXB CX = AY BZ - BY AZ CY = AZ BX - BZ AX CZ = AX BY - BX AY B A C Note: B XA =- A X B**y**x z Torque & the Cross Product: • So we can define torque as: = r x F = r F sin X = y FZ - z FY Y = z FX - x FZ Z = x FY - y FX F f r**Rotational Work**F • Work done by a constant torque in turning an object by an angle is • Consider a wheel of radius R with a rope wrapped around it • You pull the rope with a constant tension T • W = F d cos • Here d and F are parallel • W = F d • Total displacement is d = s =R so • W = F R • But FR = since F and R are perpendicular • W = s F R**Work**• Consider the work done by a force F acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d: • dW = F.dr= F R d cos() = F R d cos(90-) = F R d sin() = F R sin() d dW = d • We can integrate this to find: W = • Analogue of W= F•r • W will be negative if and have opposite sign ! F R dr=Rd d axis**Work & Kinetic Energy:**• Recall the Work Kinetic-Energy Theorem: K = WNET • This is true in general, and hence applies to rotational motion as well as linear motion. • So for an object that rotates about a fixed axis:**M**R F Example: Disk & String • A massless string is wrapped 10 timesaround a disk of mass M=40 g and radius R=10cm. The disk is constrained to rotate without friction about a fixed axis through its center. The string is pulled with a force F=10N until it has unwound. (Assume the string does not slip, and that the disk is initially not spinning). • How fast is the disk spinning after the string has unwound?**Recall thatIfor a disk about**its central axis is given by: So = 792.5 rad/s Disk & String... WNET = W = 62.8 J = K W = t.q = F x r.q = (10 N)(0.10 m)(10*2p) = 62.8 J M R **Lecture 21, ACT 1Work & Energy**• Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same distance. Disk 1 has a bigger radius, but both are made of identical material (i.e. their density r = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest. • Which disk has the biggest angular velocity after the pull ? w2 w1 (a)disk 1 (b)disk 2 (c)same F F**Rolling Motion**• Now consider a cylinder rolling at a constant speed. VCM CM The cylinder is rotating about CM and its CM is moving at constant speed (vCM). Thus its total kinetic energy is given by :**Rolling Motion**• Consider again a cylinder rolling at a constant speed. Q VCM CM P At any instant the cylinder is rotating about point P. Its kinetic energy is given by its rotational energy about that point. KTOT = 1/2 IPw2**Rolling Motion**• We can find IP using the parallel axis theorem V = 2VCM Q V = r w V = VCM CM IP = ICM + MR2 V = 0 P KTOT = 1/2 (ICM + MR2 ) w2 KTOT = 1/2 ICM w2 + 1/2 M (R2w2 ) = 1/2 ICM w2 + 1/2 M VCM2 !**Connection with CM motion**• For a system of particles, the kinetic energy is : = KCM = KREL • For a solid object rotating about it’s 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 of mass and whose CM is moving: VCM **Rolling Motion**• Cylinders of different I rolling down an inclined plane: K = - U = Mgh v = 0 = 0 K = 0 R M h v = R**Rolling...**• If there is no slipping (due to friction): v v 2v v Where v = R In the lab reference frame In the CM reference frame**The rolling speed is always lower than in the case of simple**sliding since the kinetic energy is shared between CM motion and rotation. Rolling... hoop: c=1 disk: c=1/2 sphere: c=2/5 etc... Use v= R and I = cMR2. c c So: c c**Lecture 21, ACT 2Rolling Motion**• A race !! Two cylinders are rolled down a ramp. They have the same radius but different masses, M1 > M2. Which wins the race to the bottom ? A) Cylinder 1 B) Cylinder 2 C) It will be a tie M1 M2 Active Figure h M? q**v ?**M Example :Rolling Motion • A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? Ball has radius R M M M M M h M q**Example :Rolling Motion**• Use conservation of energy. Ei = Ui + 0 = Mgh Ef = 0 + Kf = 1/2 Mv2 + 1/2 I w2 = 1/2 Mv2 + 1/2 (1/2MR2)(v/R)2 Mgh = 1/2 Mv2 + 1/4 Mv2 v2 = 4/3 g h v = ( 4/3 g h ) 1/2**Consider a roller coaster.**We can get the ball to go around the circle without leaving the loop. Note: Radius of loop = R Radius of ball = r**1**2 How high do we have to start the ball ? Use conservation of energy. Also, we must remember that the minimum speed at the top is vtop = (gR)1/2 E1 = mgh + 0 + 0 E2 = mg2R + 1/2 mv2 + 1/2 Iw2 = 2mgR + 1/2 m(gR) + 1/2 (2/5 mr2)(v/r)2 = 2mgR + 1/2 mgR + (2/10)m (gR) = 2.7 mgR**1**2 How high do we have to start the ball ? E1 = mgh + 0 + 0 E2 = 2.7 mgR mgh = 2.7 mgR h = 2.7 R h = 1.35 D (The rolling motion added an extra 2/10 R to the height: without it, h = 2.5 R)

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