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AP Physics 1: Rotational Motion and Equilibrium. Chapter 8: Rotational Motion and Equilibrium. 8.1 Rigid Bodies, Translation, and Rotation 8.2 Torque, Equilibrium, and Stability 8.3 Rotational Dynamics 8.4 Rotational Work and Energy 8.5 Angular Momentum.
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Chapter 8: Rotational Motion and Equilibrium 8.1 Rigid Bodies, Translation, and Rotation 8.2 Torque, Equilibrium, and Stability 8.3 Rotational Dynamics 8.4 Rotational Work and Energy 8.5 Angular Momentum
8.1 Rigid Bodies, Translation, and Rotation Please Do Now How do you think the tight-rope walkers are able to keep from falling? Write three lines using full sentences.
8.1 Rigid Bodies, Translation, and Rotation rigid body – an object or system of particles in which the distances between particles are fixed and remain constant. Solids can be considered to be rigid bodies for the purpose of analyzing rotational motion. Fluids are not considered to be rigid bodies. translational motion – every particle in the object has the same instantaneous velocity. There is no rotation. Write two examples. rotational motion – motion about a fixed axis. All the particles of an object have the same instantaneous angular velocity and travel in circles about the axis of rotation. Write two examples. rigid body motion – is usually a combination of translational and rotational motion. Write two examples.
8.2 Torque and Equilibrium τ = r F sinϕ
8.2 Torque and Equilibrium The sign convention for torques is as follows: • The torque, which rotates or tends to rotate the body clockwise (CW), is negative. • The torque, which rotates or tends to rotate the body counterclockwise (CCW), is positive. The SI unit of torque is the m·N (meter-newton). •
8.2 Torque and Equilibrium The perpendicular distance r┴ from the axis of rotation to the line of action of a force is call the moment arm (or lever arm) and is equal to r sin θ.
8.2 Torque and Equilibrium The same force in the opposite direction with a smaller moment arm produces a smaller torque in the opposite direction.
8.2 Torque and Equilibrium When a force acts through the axis of rotation, the torque is zero.
8.2 Torque and Equilibrium equilibrium – a state in which things are in balance or are stable. translational equilibrium – the sum of the forces on a body is zero • the body remains at rest or at constant velocity • Fi = F1 + F2 + F3 + …. = 0 rotational equilibrium – the sum of torques on a body is zero • the body is rotationally at rest or rotates with a constant angular velocity • τi = τ1 + τ2 + τ3 + …. = 0 The board is in translational The board is in translational but not and rotational equilibrium. rotational equilibrium. F F F F
8.2 Torque and Equilibrium mechanical equilibrium – when the conditions for translational and rotational equilibrium are satisfied. Fi = 0 (for translational equilibrium) τi = 0 (for rotational equilibrium) static equilibrium – the condition that exists when a rigid body remains at rest What are some examples of static translational equilibrium? What are some examples of static rotational equilibrium? Problem-Solving Hint: Use the convention counter-clockwise (ccw) is positive and clockwise (cw) is negative.
8.2 Torque and Equilibrium Example 8.1: The bolts on a car wheel require tightening to a torque of 90 m·N. If a 20 cm long wrench is used, what is the magnitude of the force required a) when the force is perpendicular to the wrench, b) when the force is 35° to the wrench as shown. c) Why does it need more force in (b) than in (a)?
8.2 Torque and Equilibrium Example 8.2: A uniform board of weight 40 N supports two children weighing 500 N and 350 N, respectively. If the support is at the center of the board and the 500 N child is 1.5 m from the center, what is the position of the 350 N child?
8.2 Torque and Equilibrium Example 8.3: A 10 m long uniform beam weighing 100 N is supported by two ropes at the ends as shown. If a 400 N person sits at 2.0 m from one end of the beam, what are the tensions in the ropes?
8.2 Check for Understanding 3. (Hint: Use worker A as the pivot point.)
8.2 Stability and Center of Gravity center of gravity (CG) – the point located at the object’s average position of weight. ∙ for a symmetrical object, it is at the geometric center (ex: baseball) ∙ for an irregularly shaped object, the CG is at the heavier end (ex: baseball bat) ∙ for objects not made of the same material throughout, the CG is located at the heaviest part (ex: Weebles) center of mass – the average position of all particles of mass that make up an object. ∙ note: on earth, CG = CM
Topple! 8.2 Stability and Center of Gravity Rule: ∙ If the CG of an object is above the area of support, the object will remain upright. ∙ If the CG extends outside the area of support, the object will topple.
8.2 Stability and Center of Gravity unstable equilibrium – any movement lowers its center of gravity. stable equilibrium – any movement raises its center of gravity. neutral equilibrium – movement does not cause a change in CG.
Warmup: Torque Your Way In Physics Warmup # 71 • An object rotates because a torque acts on it. When you exert a force at the end of a wrench in order to rotate it (along with the bolt it is attached to), you may not have thought about the fact that you applied the force perpendicular to the handle. The longer the wrench’s handle, the less force you have to apply, because the amount of torque is equal to the product of the force times the distance to the axis about which it is rotating. • _________________________________________________________________ • Use the relationship of torque to force and distance to explain the following two small mysteries in your kitchen: • Why is it easier to open a cabinet door when the doorknob is at the end of the door than when it is in the middle of the door? • 2. Why is a big doorknob easier to turn than a small one?
8.3 Rotational Dynamics • Example 8.5: A solid cylinder of mass 10 kg is pivoted about a frictionless axis through its center O. A rope wrapped around the outer radius R1 = 1.0 m, exerts a force of F1 = 5.0 N to the right. A second rope wrapped around another section of radius R2 = 0.50 m exerts a force of F2 = 6.0 N downward. • What is the angular acceleration of the disk? • If the disk starts from rest, how many radians does it rotate through in the first 5.0 s? F1 R1 O R2 F2
8.4 Rotational Work an Kinetic Energy Rotational Translational Work: W = W = Fd = Fx Kinetic Energy: K = ½ I 2 K = ½ mv2 Power: P = P = Fv Work-Energy Theorem: W = K = ½ I 2 - ½ I o2 W = ½ mv2 - ½ mvo2 The kinetic energy of a rolling body (without slipping) relative to an axis through the contact point is the sum of the rotational kinetic energy about an axis through the center of mass and the translational kinetic energy of the center of mass. K = ½ ICM2 + ½ mvCM2 total = rotational + translational KE KE + KE
8.4 Rotational Work an Kinetic Energy Example 8.6: A solid sphere of mass 1.0 kg and radius 0.010 m rolls with a speed of 10 m/s. How high up an inclined plane can it roll before it comes to rest?
8.4 Rotational Work an Kinetic Energy • Example 8.7: A cylindrical hoop of mass 10 kg and radius 0.20 m is accelerated by a motor from rest to an angular speed of 20 rad/s during a 0.40 s interval. • How much work is required? • What is the power output of the motor?
8.4 Check for Understanding • From W = , the unit(s) of rotational work is • a. watt • b. Nm • c. kgrad/s2 • d. Nrad Answer: b
8.4 Check for Understanding 2. A bowling ball rolls without slipping on a flat surface. The ball has a. rotational kinetic energy b. translational kinetic energy c. both rotational and translational kinetic energies Answer: c
