PHYS 1443 Section 001 Lecture #14 Monday, June 25, 2007 Dr. Jaehoon Yu Torque Vector Product Moment of Inertia Monday, June 25, 2007 Parallel Axis Theorem Torque and Angular Acceleration Rotational Kinetic Energy PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu 1 Announcements Quiz Results Average: 66/100 Top score: 90 Last quiz this Thursday, June 28 Early in the class Covers up to what we learn Wednesday, June 27 Final exam Date and time: 8 10am, Monday, July 2 Location: SH103 Covers: Ch 8.4 what we cover Thursday, June 28 Dont forget the end-of-semester Planetarium show 10:10am, Thursday, June 28 Evaluation today Monday, June 25, 2007 PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu 2 Torque Torque is the tendency of a force to rotate an object about an axis. Torque, , is a vector quantity. F r P

The line of Action d2 d Moment arm F2 Consider an object pivoting about the point P by the force F being exerted at a distance r from P. The line that extends out of the tail of the force vector is called the line of action. The perpendicular distance from the pivoting point P to the line of action is called the moment arm. Magnitude of torque is defined as the product of the force exerted on the object to rotate it and the moment arm. When there are more than one force being exerted on certain points of the object, one can sum up the torque generated by each force vectorially. The convention for sign of the torque is positive if rotation is in counter-clockwise and negative if clockwise. Monday, June 25, 2007 PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu rF sin Fd 1 2 F1d1 F2 d 2 3 Example for Torque A one piece cylinder is shaped as in the figure with core section protruding from the larger drum. The cylinder is free to rotate around the central axis shown in the picture. A rope wrapped around the drum whose radius is R1 exerts force F1 to the right on the cylinder, and another force exerts F2 on the core whose radius is R2 downward on the cylinder. A) What is the net torque acting on the cylinder about the rotation axis? R1 F 1 R2 The torque due to F1 1 R1 F1 and due to F2 2 R2 F2 So the total torque acting on the system by the forces is

1 2 R1 F1 R2 F2 F 2 Suppose F1=5.0 N, R1=1.0 m, F2= 15.0 N, and R2=0.50 m. What is the net torque about the rotation axis and which way does the cylinder rotate from the rest? Using the above result R F R F Monday, June 25, 2007 1 1 2 2 5.0 1.0 15.0 0.50 2.5 N m PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu The cylinder rotates in counter-clockwise. 4 Torque and Vector Product z O r rxF Lets consider a disk fixed onto the origin O and the force F exerts on the point p. What happens? p The disk will start rotating counter clockwise about the Z axis y The magnitude of torque given to the disk by the force F is Fr sin F x But torque is a vector quantity, what is the direction? How is torque expressed mathematically? What is the direction? The direction of the torque follows the right-hand rule!!

The above operation is called the Vector product or Cross product What is the result of a vector product? Another vector Monday, June 25, 2007 r F C A B C A B A B sin What is another vector operation weve learned? Scalar product C A B A B cos PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu Result? A scalar 5 Properties of Vector Product Vector Product is Non-commutative What does this mean? If the order of operation changes the result changes Following the right-hand rule, the direction changes Vector Product of two parallel vectors is 0. A B B A A B B A C A B A B sin A B sin 0 0 Thus, A A 0 If two vectors are perpendicular to each other AB A B sin A B sin 90 A B AB Vector product follows distribution law A B C A B A C The derivative of a Vector product with respect to a scalar variable is

d A B dA dB B A dt dt dt Monday, June 25, 2007 PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu 6 More Properties of Vector Product The relationship between unit vectors, i, j and k i i j j k k 0 i j j i k j k k j i k i i k j Vector product of two vectors can be expressed in the following determinant form i A B Ax Bx j k Ay Az i By Bz Ay Az By Bz j Ax Az Bx

Bz k Ax Ay Bx By Ay Bz Az B y i Ax Bz Az Bx j Ax B y Ay Bx k Monday, June 25, 2007 PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu 7 Moment of Inertia Measure of resistance of an object to changes in its rotational motion. Equivalent to mass in linear motion. Rotational Inertia: A similar quantity? For a group of particles I mi ri 2 i What are the dimension and unit of Moment of Inertia? For a rigid body I r 2 dm ML 2 kg m 2 Determining Moment of Inertia is extremely important for computing equilibrium of a rigid body, such as a building. Monday, June 25, 2007 PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu 8

Example for Moment of Inertia In a system of four small spheres as shown in the figure, assuming the radii are negligible and the rods connecting the particles are massless, compute the moment of inertia and the rotational kinetic energy when the system rotates about the y-axis at angular speed . y m Since the rotation is about y axis, the moment of inertia about y axis, Iy, is b l M O l M x b m I mi ri2 Ml2 Ml 2m 02 m 02 2Ml 2 i This is because the rotation is done about y axis, and the radii of the spheres are negligible. 1 2 1 K R I 2 Ml 2 2 Ml 2 2 2 2 Why are some 0s? Thus, the rotational kinetic energy is Find the moment of inertia and rotational kinetic energy when the system rotates on the x-y plane about the z-axis that goes through the origin O. 2 2 2 I mi ri 2 Ml Ml 2mb2 mb2 2 Ml mb i Monday, June 25, 2007

1 1 K R I 2 2 Ml 2 2mb 2 2 Ml 2 mb 2 2 2 2 PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu 9 Calculation of Moments of Inertia Moments of inertia for large objects can be computed, if we assume that the object consists of small volume elements with mass, mi. I lim ri2 mi r 2 dm The moment of inertia for the large rigid object is m 0 i It is sometimes easier to compute moments of inertia in terms of volume of the elements rather than their mass dm Using the volume density, , replace dm in the above equation with dV. dV i How do we do this? dm dV The moments of inertia becomes I r 2 dV Example: Find the moment of inertia of a uniform hoop of mass M and radius R about an axis perpendicular to the plane of the hoop and passing through its center. y O The moment of inertia is dm R x

Monday, June 25, 2007 What do you notice from this result? I r 2 dm R 2 dm MR 2 The moment of inertia for this object is the same as that of a point of mass M at the distance R. PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu 10 Example for Rigid Body Moment of Inertia Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an axis perpendicular to the rod and passing through its center of mass. M The line density of the rod is y L so the masslet is dm dx M dx L dx L x x The moment of inertia is Monday, June 25, 2007 L/2 1 3 3x L/2 3 3 M L L M L3 ML2 3L 2 2 3L 4 12 What is the moment of inertia when the rotational axis is at one end of the rod. Will this be the same as the above. Why or why not? M

x2M r dm dx I L/2 L L L/2 2 I L M 1 r 2 dm 0 x M dx x 3 L 3 0 L M M 3 ML2 3 L 0 L 3L 3L 3 2 L Since the moment of inertia is resistance to motion, it makes perfect sense for it to be harder to move when it is rotating about the axis at one end. PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu 11 Parallel Axis Theorem Moments of inertia for highly symmetric object is easy to compute if the rotational axis is the same as the axis of symmetry. However if the axis of rotation does not coincide with axis of symmetry, the calculation can still be 2 done in simple manner using parallel-axis theorem. I I CM MD y

2 2 2 x y I r dm dm (1) Moment of inertia is defined yCM y y (x,y) r CM (xCM,yCM) D xCM x x What does this Monday, June 25, 2007 theorem tell you? Since x and y are x xCM x' y yCM y ' One can substitute x and y in Eq. 1 to obtain 2 2 I xCM x' yCM y ' dm 2 2 xCM yCM dm 2 xCM x' dm 2 yCM y' dm x'2 y'2 dm

Since the x and y are the x' dm 0 y ' dm 0 x distances from CM, by definition Therefore, the parallel-axis theorem 2 2 I xCM yCM dm x'2 y'2 dm MD 2 I CM Moment of inertia of any object about any arbitrary axis are the same as the sum of moment of inertiaSummer for a rotation about the CM and that12of PHYS 1443-001, 2007 Dr. Jaehoon Yu the CM about the rotation axis. Example for Parallel Axis Theorem Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an axis that goes through one end of the rod, using parallel-axis theorem. The line density of the rod is y so the masslet is CM L dm dx dx x x The moment of inertia about the CM I CM M L M dx L M x2M r dm dx L/2 L L

2 L/2 L/2 1 3 3 x L/2 3 3 M L L M L3 ML2 3L 2 2 3L 4 12 2 Using the parallel axis theorem I I CM ML2 L ML2 ML2 ML2 2 D M M 12 2 12 4 3 The result is the same as using the definition of moment of inertia. Parallel-axis theorem is useful to compute moment of inertia of a rotation of a rigid object with complicated shape about an arbitrary axis Monday, June 25, 2007 PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu 13 Torque & Angular Acceleration Ft r m Fr Lets consider a point object with mass m rotating on a circle. What forces do you see in this motion? The tangential force Ft and the radial force Fr The tangential force Ft is The torque due to tangential force Ft

is you see from the above relationship? What do What does this mean? Ft mat mr Ft r mat r mr 2 I I Torque acting on a particle is proportional to the angular acceleration. What law do you see from this relationship? Analogs to Newtons 2nd law of motion in rotation. How about a rigid object? The external tangential force dFt is dFt dmat dmr dFt d dFt r r 2 dm The torque due to tangential force Ft dm is The total torque is r 2 dm I r Contribution from radial force is 0, because its What is the contribution due line of action passes through the pivoting O to radial force and why? point, Monday, June 25, 2007 PHYS 1443-001, Summer 2007making the moment arm 0. 14 Dr. Jaehoon Yu Example for Torque and Angular Acceleration A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane. The rod is released from rest in the horizontal position. What are the initial angular acceleration of the rod and the initial linear acceleration of its right end? The only force generating torque is the gravitational force Mg L/2 Mg Fd F L L Mg I 2 2 L

3 M x ML2 Since the moment of inertia of the rod I r 2 dm x 2dx 0 0 3 when it rotates about one end L 3 0 MgL 3g MgL We obtain 2 2ML 2L 2I 3 Using the relationship L L between tangential and angular acceleration Monday, June 25, 2007 3g at L 2 What does this mean? The tip of the rod falls faster than an object undergoing a free fall. PHYS 1443-001, Summer 2007 Dr. Jaehoon Yu 15