Chapter 5 and 8 PowerPoint, Mr. Mardit
advertisement
Chapter 5: Circular Motion; Gravitation Chapter 8: Rotational Motion 5-1: Kinematics of Circular Motion • Uniform circular motion: movement in a circle at constant speed 1- Is the object accelerating? YES/NO ? WHY? HINT: π = βπ£ βπ‘ 2- Direction of π ? HINT: What is the direction of βπ£? (What is the direction of π£2 and π£1 ?) Remember vectors add tip to tail! π£2 3- πππππ‘πππππ‘ππ = ; proof see text page 107 π 5-1: Kinematics of Circular Motion • πππππ’ππππ¦ = π = # ππ πππ£πππ’π‘ππππ /ππ¦ππππ π πππππ • ππππππ = π = time required for one revolution/cycle •π = 1 π • Examples: Earth, moon, heart rate, waves breaking on the shore, ….Can you name some more….? 5-2: Applying N2L to Uniform Circular Motion • Remember Newton’s 2nd Law: • πΉπππππ = πΉπππ‘ = ππ • Let’s apply N2L to an object moving in a circle at constant speed? • πΉπππππ = πΉπππ‘ = ππππππ‘πππππ‘ππ • 1-What is the direction of πΉπππ‘ ? • 2-Brainstorm some possible sources of πΉπππ‘ : 5-6: Newton’s Law of Universal Gravitation •πΉππππ£ππ‘π¦ = πΊπ1 π2 π2 • The above gives only the magnitude of the force of gravity NOT the direction, gravity is always an attractive force that acts along the line joining π1 and π2 • G=6.67 x 10-11 units??? ο What are the units of G? 5-7: Applying N2L to Gravity; g • Remember Newton’s 2nd Law: • πΉπππππ = πΉπππ‘ = ππ • Let’s apply N2L to two masses π1 πππ π2 • πΉπππππ = πΉπππ‘ = πΉππππ£ππ‘π¦ = πΊπ1 π2 =π2 π π2 • Now let’s say π1 is the Earth and π2 is a person standing on the surface of the Earth: πΊππΈπππ‘β πππππ ππ πππππ ππ π= ; π2 cancels; substitute 2 π πΊππΈπππ‘β = π2 • π= 2 Nm 6.67×10−11 kg2 ×(5.98×1024 ππ) (6.371×106 πππ‘πππ ) ≅ 9.8 π π 2 ≡π Polar Plot: (r,θ) Good news: We will only be studying rotational motion of CONSTANT radius. Why do we care? Imagine you are walking in a circle, riding a bicycle, twisting a wrench, torqueing a screw driver, rolling a ball, or doing a cart wheel. All of these consist of motion in both the x and y– which must be resolved into two—one dimensional problems before they can be analyzed. However, using polar coordinates—when the radius is held constant— only one variable changes: θ YESSSSS! Much Simpler! 2π πππππππ = 360 πππππππ Example: The hockey puck spins 185 times between defending zone and the attacking zone. How many radians? How many degrees? For every linear (straight line) physical quantity, there is a rotational analogue: To review the linear physical quantities to date: x (distance), v, a, m, F, KE, p, Position ο Velocity ο Acceleration ω = angular velocity = θ2 −θ1 π‘2 −π‘1 compare with linear velocity= α = angular acceleration = compare with = βθ βπ‘ π₯2 −π₯1 π‘2 −π‘1 ω2 −ω1 π‘2 −π‘1 ≡ ω (omega) = βπ₯ βπ‘ = ≡π£ βω βπ‘ ≡ α (πππβπ) π₯2 −π₯1 linear acceleration= π‘2 −π‘1 = βπ₯ βπ‘ ≡π£ Examples: θ, ω, α 1- A small spider walks an arc length of constant radius around a sun dial from 1 o’clock to 8 o’clock. What is the βθ of the ant? ( in “hours”, in degrees, in radians) The journey takes 2.6 minutes what is the ω of the spider? (radians/second) 2- Timmy hops on his bicycle and starts pedaling. He completes 4.2 revolutions of the rear tire in the first 3 seconds. What is the ωππππ‘πππ of the tire? One minute later the rear tire completes 15.6 revolutions in 0.6 seconds. What is the ωπππππ of the tire? What is the angular acceleration α of the tire? Remember this equation from middle school: Circumference of a circle = 2PI *r …well that was just a special case of the following: Manipulating l=rθ to connect linear and angular quantities; a little bit of calculus (PRODUCT RULE) π π π‘ ×π π‘ ππ‘ ππ ππ = × π π‘ + π(π‘) × ππ‘ ππ‘ l=r θ π π π = πθ ππ‘ ππ‘ ππ ππ πθ = •θ+π• ππ‘ ππ ππ‘ ππ‘ πθ = π • ππ‘ v=rω ππ‘ v=rω Example: A cyclist with a 24 inch diameter tire glances down at his speedometer and notices his speed is 28 miles/hour . What is the angular velocity of the wheel in radians/second? Applying the PRODUCT RULE a second time π π π‘ ×π π‘ ππ‘ ππ ππ = × π π‘ + π(π‘) × ππ‘ ππ‘ v=r ω π π π£ = rω ππ‘ ππ‘ ππ£ ππ πω = •ω+π• ππ‘ ππ£ ππ‘ ππ‘ = π• a=rα πω ππ‘ ππ‘ atangetial = r α Example: At the Indianapolis 6000 a x-wing fighter banks a turn around the 4th moon of Xarnex at 18000 m/π 2 . The radius of curvature is 200 meters. What is the angular acceleration? • ππππ’πππ π£ππππππ‘π¦ ω = 2π ∗ πππππ’ππππ¦ = 2ππ = • ππππππ = π = time required for one revolution •π = 1 π # ππ πππ£πππ’π‘ππππ 2π π πππππ 1 Dimensional kinematic equation also in cases of constant angular acceleration with appropriate change in variable π£π = π£0 + ππ‘ → ωπ = ω0 + απ‘ Δπ₯ = π£0 π‘ + 1 2 ππ‘ 2 → Δθ = ω0 π‘ + 1 2 απ‘ 2 π£π2 = π£02 + 2πΔπ₯ → ωπ = ω0 + απ‘ π£= π£π + π£0 2 → ω= ωπ + ω0 2 Example: A cyclist with a 24 inch diameter tire glances down at his speedometer and notices his speed is 28 miles/hour . What is the angular velocity of the wheel in radians/second? He comes to a stop in 6 seconds. How many revolutions does his wheel turn during this time? Torque: the rotational analogue of force suppose we have a rigid, massive body: ο if a force F is applied a distance r away from a fixed axis of rotation, then the rigid body will experience a torque τ when a force is directed in a straight line the result is a push or a pull; a linear force; however when that force causes a rotation it is called a torque It is only when the force is applied perpendicular to the lever arm r, that a torque is generated. What if the force is at an angle? Torque Example: Susan is changing her tire and applies a 20 Newton force perpendicular to the end of a wrench that is 20 centimeters long. What is the magnitude of the torque generated? π = π × πΉ = π πΉ sin π Torque examples everyday life: Center of Mass (cm) or Center of Gravity: average of the masses factored by their distances from a reference point If you take out your cell phone and balance it on the tip of your finger why does it balance and not fall over? Example: π₯ππ π1 π₯1 + π2 π₯2 + π3 π₯3 + = = π1 + π2 + π3 + ππ π₯π ππ π₯π = ππ ππππ‘ππ π¦ππ π1 π¦1 + π2 π¦2 + π3 π¦3 + = = π1 + π2 + π3 + ππ π¦π ππ π¦π = ππ ππππ‘ππ Continuous Mass Distribution Moment of Inertia: the rotational analogue of mass When a force is applied to a mass that then moves in a straight line, we consider as if all of the mass were located at the center of mass; the balance point of mass distribution and only consider the total mass. However, when we consider torques we need to know the position of where the mass is located. It takes more force to rotate a mass around an axis of symmetry when the mass is located 100 meters away then when it is 4 meters away. Moment of intertia= 2 ππ ππ Example: Determine the moment of inertia of a system of 4 point masses located at coordinates (1,2), (-2,3), (-3,4), and (-5,-7) each of mass 2 kilograms when the axis of rotation is about the z-axis. Moment of intertia= ππ ππ2 Continuous mass distributions Newtons’ 2nd Law: πΉπππππ = πΉπππ‘ = ππ → πππππ’ππ = τπππ‘ = πΌα Example: If each of the three rotor blades is 8cm long and has a mass of 95 gram, calculate the amount of torque that the motor must supply to bring the blades up to a speed of 5 rev/sec in 8.0 seconds. Linear Kinetic Energy ο Rotational Kinetic Energy 1 2 mπ£ 2 → 1 2 Iω 2 Example: A pitcher throws a fastball with all translational kinetic energy (ZERO spin) at 44 meters/second. If the pitcher throws the same baseball with a curveball pitch with the same total kinetic energy at 33 meters/second determine how many rotations/second the baseball spins at. NOTE: mass_baseball=143 grams; diameter_baseball=75 millimeters; Angular Momentum Linear Momentum: p=mv ο Angular Momentum: L=Iω What are the units of angular momentum? Example: Determine the angular momentum of the baseball in the previous example. Newton’s 2nd Law revisited: We initially saw N2L like this: π π π‘ ×π π‘ ππ‘ ππ ππ = × π π‘ + π(π‘) × ππ‘ ππ‘ πΉ = ππ βπ ππ π π ππ ππ£ ππ£ however, Issac originally wrote it like: πΉ = = = (π)= (ππ£)= π£ + π =π =ππ βπ‘ ππ‘ ππ‘ ππ‘ ππ‘ ππ‘ ππ‘ SO…use this and our new definition of angular momentum to derive N2L for rotational motion: π= βπΏ βπ‘ = ππΏ π π ππΌ = (πΏ)= (πΌπ)= π ππ‘ ππ‘ ππ‘ ππ‘ + ππ πΌ ππ‘ = ππ πΌ =πΌ ππ‘ πΌ Conservation of angular momentum: L=Iπ = ππππ π‘πππ‘ Explain: Comparison chart: Translation Rotation Connection x π x=rθ v π v=rω a πΌ a=rα m πΌ I=Σmr2 F π Τ=rFsinθ KE=½mv2 KE=½Iω2 p=mv L=Iω W=Fd W=τα ΣF=ma Στ=Iα ΣF=βπ/βπ‘ Στ=βπΏ/βπ‘ Add example problems after each slide where a new concept is introduced that might be unclear Comparison chart at the end; with linear and rotational values
