Rotational Equilibrium and Dynamics Chapter 8 Magnitude of

Rotational Equilibrium and Dynamics Chapter 8

Magnitude of a Torque • What is a torque? – A quantity that measures the ability of a force to rotate an object around some axis. r

• • Example… Imagine opening a door around a hinge. The door is free to rotate about a fixed axis. This axis is the door’s axis of rotation When an individual pulls the door open at the handle with a force that is perpendicular to the door, the door opens. – The measurement of this force is the Torque of the motion.

Torque depends on a force and a lever arm • If someone opens the door from the previous example, but at a point closer to the hinge, the door would be much more difficult to rotate. • How easily an object rotates depends on how much force is applied as well as WHERE the force is applied.

Torque depends on a force and a lever arm • The farther the force is from the axis of rotation, the easier it is to rotate the object and the more torque is produced. • The PERPENDICULAR distance from the axis of rotation to a line drawn along the direction of the force is called the lever arm, or moment arm.

Torque and Lever Arm Angular speed d F axis of rotation Force applied to an extended object can produce a torque. This torque in turn, causes this object to rotate.

TORQUE Equation • Torque also depends of the angle between a force and a lever arm Torque = Force (lever arm)(sin theta) T = Fd(sin 0) **SI unit of torque: Newton meter or (N x m)

Calculating Torque at an angle F d F(sin � ) T = d x (sin � )

Conceptual Question… • 1: A student pushes with minimal force of 50. 0 N on the middle of a door to open it. – What minimum force must be applied at the edge of the door in order for the door to open? – What minimum force must be applied to the dinged side of the door in order for the door to open?

Conceptual Question… • 1: A student pushes with minimal force of 50. 0 N on the middle of a door to open it. – What minimum force must be applied at the edge of the door in order for the door to open? – 25 Newtons – (half a much because the distance in increased by 2) – T=d*F – What minimum force must be applied to the hinged side of the door in order for the door to open? – The door cannot be opened by a force at the hinge location. It can be broken but not opened normally.

Torque can be (+) or (-) • Torque is a vector quantity (i. e. has direction) • To find net torque, simply add up the individual torques. Tnet = Sum of Torques = T 1 + T 2 = (F 1 d 1) + (F 2 d 2) The sign of the net torque value (+ or -) will tell you the direction an object rotates. Clockwise rotation = - torque Counterclockwise rotation = + torque

Example: Torque • A basketball is being pushed by two players during tip-off. One player exerts a downward force of 11 N at a distance of 0. 07 m from the axis of rotation. The second player applies an upward force of 15 N at a perpendicular distance of 0. 014 m from the axis of rotation. • Find the net torque acting on the basketball.

Example: Torque • A basketball is being pushed by two players during tip-off. One player exerts a downward force of 11 N at a distance of 0. 07 m from the axis of rotation. The second player applies an upward force of 15 N at a perpendicular distance of 0. 14 m from the axis of rotation. • Find the net torque acting on the basketball. Givens: F 1 = 15 N d 1 = 0. 14 m F 2 = 11 N d 2 = 0. 07 m Formula: Tnet = T 1 + T 2

Example: Torque • A basketball is being pushed by two players during tip-off. One player exerts a downward force of 11 N at a distance of 0. 07 m from the axis of rotation. The second player applies an upward force of 15 N at a perpendicular distance of 0. 14 m from the axis of rotation. • Find the net torque acting on the basketball. Givens: F 1 = 15 N d 1 = 0. 14 m F 2 = 11 N d 2 = 0. 07 m F 2 = 11 N Formula: Tnet = T 1 + T 2 d 1 = 0. 14 m F 1 = 15 N d 2 = 0. 07 m

• A basketball is being pushed by two players during tip-off. One player exerts a downward force of 11 N at a distance of 0. 07 m from the axis of rotation. The second player applies an upward force of 15 N at a perpendicular distance of 0. 14 m from the axis of rotation. • Find the net torque acting on the basketball. Givens: F 1 = 15 N d 1 = 0. 14 m F 2 = 11 N d 2 = 0. 07 m T 1 = F 1 d 1 = -(15)(0. 14) = -2. 1 Nm T 2 = F 2 d 2 = -(11) (0. 07)= -0. 77 Nm Formula: Tnet = T 1 + T 2 F 2 = 11 N d 1 = 0. 14 m (both forces produce a clockwise rotation Therefore both torques are negative) F 1 = 15 N d 2 = 0. 07 m

• A basketball is being pushed by two players during tip-off. One player exerts a downward force of 11 N at a distance of 0. 07 m from the axis of rotation. The second player applies an upward force of 15 N at a perpendicular distance of 0. 14 m from the axis of rotation. • Find the net torque acting on the basketball. T 1 = F 1 d 1 = -(15)(0. 14) = -2. 1 Nm T 2 = F 2 d 2 = -(11) (0. 07)= -0. 77 Nm Tnet = T 1 + T 2 = -2. 1 – 0. 77 Tnet = -2. 9 N m F 2 = 11 N d 1 = 0. 14 m Negative torque thus Ball rotates in a clockwise direction. F 1 = 15 N d 2 = 0. 07 m

Individual Practice • Take 5 minutes to work the following on your own: *problem 1 (practice 8 A) on page 282

ROTATION and INERTIA • Identify Center of Mass • Distinguish between mass and moment of Inertia • Define the secondition of equilibrium • Solve problems involving the first and seconditions of equilibrium

Center of Mass • Center of Mass – the point at which all the mass of the body can be considered to be concentrated when analyzing translational motion. • Consider a stick thrown in the air while playing fetch with a puppy…

• Since the center of mass is considered to be the point at which all the mass of an object is concentrated… • The stick rotates in the air around its center of mass • The complete motion of the stick in the air is both rotational and translational. The center of mass moves as if the stick were a point mass, with all of its mass concentrated at that point for purposes of analyzing its translational motion.

Moment of Inertia • Moment of inertia: the tendency of a body rotating about a fixed axis to resist a change in rotational motion. • The moment of inertia is a measure of the object’s resistance to a change in its rotational motion about some axis.

Calculating Moment of Inertia • Depends of the shape of the object. • SI units: kg x m 2 See table 8 -1 in textbook for the different formulas.

Rotational Equilibrium • Rotational Equilibrium is different from translational equilibrium – If the net force on an object is zero, then the object is in translational equilibrium – If the net torque on an object is zero, then the object is in rotational equilibrium – THUS, for an object to be completely in equilibrium, the net force and the net torque must be zero – The dependence of equilibrium on the absence of net torque is called the secondition for equilibrium.

ROTATIONAL DYNAMICS • Describe Newton’s 2 nd law for rotation • Calculate the angular momentum for various rotating objects.

Newton’s 2 nd Law for Rotation net = � Net torque = moment of Inertia x angular acceleration Remember…for translational motion, F=ma

Ex: Newton’s 2 nd law for rotation • A student tosses a dart using only the rotation of her forearm to accelerate the dart. The forearm rotates in a vertical plane about an axis at the elbow joint. The forearm and dart have a combined moment of Inertia of 0. 075 kgm 2 about the axis, and the length of the forearm is 0. 26 m. If the dart has a tangential acceleration of 45 m/s 2 just before released, what is the net torque on the arm and dart?

A student tosses a dart using only the rotation of her forearm to accelerate the dart. The forearm rotates in a vertical plane about an axis at the elbow joint. The forearm and dart have a combined moment of Inertia of 0. 075 kgm 2 about the axis, and the length of the forearm is 0. 26 m. If the dart has a tangential acceleration of 45 m/s 2 just before released, what is the net torque on the arm and dart? GIVEN: = 0. 075 kg x m 2 a = 45 m/s 2 d = 0. 26 m =? = �(where � = a/d) = (0. 075)(45/0. 26) = 13 Nm

MOMENTUM • Angular Momentum: the product of a rotating object’s moment of inertia and angular speed about the same axis L = (� ) Angular momentum = moment of Inertia x angular speed

Conservation of Angular Momentum • When the net external torque acting on an object or objects is zero, the angular momentum of the object(s) does not change. This is the: Law of Conservation of Angular Momentum.

• Assuming the friction between skates and the ice is negligible, there is no torque acting on an ice skater. When she brings her hands in and feet closer to her body, more of her mass on average is nearer to her axis of rotation. Thus, the moment of inertia is constant and her angular speed increases to compensate for her smaller moment of inertia.

Rotational Kinetic Energy • Rotational KE = ½ x moment of Inertia x angular speed 2 2 KErot = ½ � Rotational KE = energy of an object due to its rotational motion Rotational KE is conserved in the absence of any external force moving an object.

Be ready for a vocab quiz on Friday • • Torque Work Lever Arm Energy Center of Mass Momentum Moment of Inertia Angular Speed Angular Momentum Angular Acceleration Rotational Kinetic Energy Rotational Motion 6 types of simple machines “SAWLIP”