Physics 1501: Lecture 21, Pg 1 Physics 1501: Lecture 21 Today’s Agenda l Announcements çHW#8: due Oct. 28 l Honors’ students çsee me after class l Midterm 1: average ~ 45 % … l Topics çTorque çRotational energy çRolling motion Sec Sec. 1-7

Physics 1501: Lecture 21, Pg 2 Summary (with comparison to 1-D kinematics) AngularLinear And for a point at a distance R from the rotation axis: x = R  v =  R  a =  R

Physics 1501: Lecture 21, Pg 3 Rotation & Kinetic Energy... Point Particle Rotating System l 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.

Physics 1501: Lecture 21, Pg 4 Direction of Rotation: l In general, the rotation variables are vectors (have a direction) l If the plane of rotation is in the x-y plane, then the convention is çCCW rotation is in the + z direction çCW rotation is in the - z direction x y z x y z

Physics 1501: Lecture 21, Pg 5 Rotational Dynamics: What makes it spin?   TOT = I  l This is the rotational version of F TOT = ma l Torque is the rotational cousin of force: ç The amount of “twist” provided by a force. Moment of inertia I is the rotational cousin of mass. Moment of inertia I is the rotational cousin of mass.  If I is big, more torque is required to achieve a given angular acceleration. l Torque has units of kg m 2 /s 2 = (kg m/s 2 ) m = Nm.

Physics 1501: Lecture 21, Pg 6 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 = I z  z l We usually omit the z subscript for simplicity. z zz zz IzIz

Physics 1501: Lecture 21, Pg 7 Example l To loosen a stuck nut, a man pulls at an angle of 45 o on the end of a 50cm wrench with a force of 200 N. çWhat is the magnitude of the torque on the nut? çIf the nut suddenly turns freely, what is the angular acceleration of the wrench? (The wrench has a mass of 3 kg, and its shape is that of a thin rod). L=0.5m F=200N 45 o

Physics 1501: Lecture 21, Pg 8 Example: solution L=0.5m F=200N 45 o  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.  = 283 rad/s 2 So  =  / I = (70.7Nm) / (.25kgm 2 )

Physics 1501: Lecture 21, Pg 9 Torque and the Right Hand Rule: l The right hand rule can tell you the direction of torque: çPoint your hand along the direction from the axis to the point where the force is applied. çCurl your fingers in the direction of the force. çYour thumb will point in the direction of the torque. r F x y z 

Physics 1501: Lecture 21, Pg 10 Review: The Cross Product l We can describe the vectorial nature of torque in a compact form by introducing the “cross product”. çThe cross product of two vectors is a third vector: A B C A X B = C C l The length of C is given by: C = ABsin  C AB l The direction of C is perpendicular to the plane defined by A and B, and in the direction defined by the right-hand rule. ABC 

Physics 1501: Lecture 21, Pg 11 The Cross Product l The cross product of unit vectors: i i = 0 i j = k i k = -j i x i = 0 i x j = k i x k = -j j i = -k j j = 0 j k = i j x i = -k j x j = 0 j x k = i k i = j k j = -i k k = 0 k x i = j k x j = -i k x k = 0 i + jk) i + jk) A X B = (A X i + A Y j + A z k) X (B X i + B Y j + B z k) i x i + i x j + i x k) = (A X B X i x i + A X B Y i x j + A X B Z i x k) j x i + j x j + j x k) + (A Y B X j x i + A Y B Y j x j + A Y B Z j x k) k x i + k x j + k x k) + (A Z B X k x i + A Z B Y k x j + A Z B Z k x k) i j k

Physics 1501: Lecture 21, Pg 12 The Cross Product l Cartesian components of the cross product: C A B C = A X B C X = A Y B Z - B Y A Z C Y = A Z B X - B Z A X C Z = A X B Y - B X A Y A B C Note: B X A = - A X B

Physics 1501: Lecture 21, Pg 13 Torque & the Cross Product: r F x y z  l So we can define torque as:  r F  = r x F = r F sin   X = y F Z - z F Y  Y = z F X - x F Z  Z = x F Y - y F X 

Physics 1501: Lecture 21, Pg 14 Rotational Work l Work done by a constant torque in turning an object by an angle  is çConsider a wheel of radius R with a rope wrapped around it çYou pull the rope with a constant tension T çW = F d cos  l Here d and F are parallel çW = F d l Total displacement is d = s =R  so çW = F R  l But FR =  since F and R are perpendicular çW =   F s F  R

Physics 1501: Lecture 21, Pg 15 Work F 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  : Fdr  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 =  l Analogue of W = F  r W will be negative if  and  have opposite sign ! R F dr=Rd  dd axis  

Physics 1501: Lecture 21, Pg 16 Work & Kinetic Energy: Recall the Work Kinetic-Energy Theorem:  K = W NET l This is true in general, and hence applies to rotational motion as well as linear motion. l So for an object that rotates about a fixed axis:

Physics 1501: Lecture 21, Pg 17 Example: Disk & String l A massless string is wrapped 10 times around 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). çHow fast is the disk spinning after the string has unwound? F R M

Physics 1501: Lecture 21, Pg 18 Disk & String... W NET = W = 62.8 J =  K Recall that  I  for a disk about its central axis is given by: So  = rad/s  R M W =    = F x r.  = (10 N)(0.10 m)(10*2  ) = 62.8 J

Physics 1501: Lecture 21, Pg 19 Lecture 21, ACT 1 Work & 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  = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest. çWhich disk has the biggest angular velocity after the pull ? (a) (a) disk 1 (b) (b) disk 2 (c) (c) same FF 11 22

Physics 1501: Lecture 21, Pg 20 Rolling Motion l Now consider a cylinder rolling at a constant speed. V CM CM The cylinder is rotating about CM and its CM is moving at constant speed (v CM ). Thus its total kinetic energy is given by :

Physics 1501: Lecture 21, Pg 21 Rolling Motion l Consider again a cylinder rolling at a constant speed. V CM P Q CM  At any instant the cylinder is rotating about point P. Its kinetic energy is given by its rotational energy about that point. K TOT = 1/2 I P  2

Physics 1501: Lecture 21, Pg 22 Rolling Motion We can find I P using the parallel axis theorem V = V CM P Q CM  I P = I CM + MR 2 K TOT = 1/2 (I CM + MR 2 )  2 K TOT = 1/2 I CM  2 + 1/2 M (R 2  2 ) = 1/2 I CM  2 + 1/2 M V CM 2 ! V = 2V CM V = 0 V = r 

Physics 1501: Lecture 21, Pg 23 Connection with CM motion l For a system of particles, the kinetic energy is : = K REL = K CM l For a solid object rotating about it’s center of mass, we now see that the first term becomes: Substituting but

Physics 1501: Lecture 21, Pg 24 Connection with CM motion... l So for a solid object which rotates about its center of mass and whose CM is moving:  V CM

Physics 1501: Lecture 21, Pg 25 Rolling Motion l Cylinders of different I rolling down an inclined plane: h v = 0  = 0 K = 0 R  K = -  U = Mgh v =  R M

Physics 1501: Lecture 21, Pg 26 Rolling... l If there is no slipping (due to friction): v 2v In the lab reference frame v In the CM reference frame v Where v =  R 

Physics 1501: Lecture 21, Pg 27 Rolling... Use v=  R and I = cMR 2. So: The rolling speed is always lower than in the case of simple sliding since the kinetic energy is shared between CM motion and rotation. hoop: c=1 disk: c=1/2 sphere: c=2/5 etc... c c c c

Physics 1501: Lecture 21, Pg 28 Lecture 21, ACT 2 Rolling Motion l A race !! Two cylinders are rolled down a ramp. They have the same radius but different masses, M 1 > M 2. Which wins the race to the bottom ? A) Cylinder 1 B) Cylinder 2 C) It will be a tie M1  h M? M2 Active Figure

Physics 1501: Lecture 21, Pg 29 Example : Rolling Motion l A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? M  h M v ? Ball has radius R M M M M M

Physics 1501: Lecture 21, Pg 30 Example : Rolling Motion l Use conservation of energy. E i = U i + 0 = Mgh E f = 0 + K f = 1/2 Mv 2 + 1/2 I  2 = 1/2 Mv 2 + 1/2 (1/2MR 2 )(v/R) 2 Mgh = 1/2 Mv 2 + 1/4 Mv 2 v 2 = 4/3 g h v = ( 4/3 g h ) 1/2

Physics 1501: Lecture 21, Pg 31 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

Physics 1501: Lecture 21, Pg 32 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 v top = (gR) 1/2 E 1 = mgh E 2 = mg2R + 1/2 mv 2 + 1/2 I  2 = 2mgR + 1/2 m(gR) + 1/2 (2/5 mr 2 )(v/r) 2 = 2mgR + 1/2 mgR + (2/10)m (gR) = 2.7 mgR 1 2

Physics 1501: Lecture 21, Pg 33 How high do we have to start the ball ? E 1 = mgh E 2 = 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) 1 2