Physics 151 chapter eight notes
Physics 151 chapter eight notes PHYC 151 001
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PHYC 151 001
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This 5 page Class Notes was uploaded by Breanab on Monday March 21, 2016. The Class Notes belongs to PHYC 151 001 at University of New Mexico taught by Dr. Dave Cardimona in Winter 2016. Since its upload, it has received 19 views. For similar materials see General Physics in Physics 2 at University of New Mexico.
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Date Created: 03/21/16
Chapter 7: Rotational Motion 7.1 Describing Circular and Rotational Motion y s θ (adians ) r Particle = angular displacement (replaces 'x' for rotational motion) radius Arc length Angular r s θ = 2π r = 2π radians = 360 =1 rev position full circlr θ 0 € or 1 radian ≈ 60 x angular displacement Δθ angular velocity =ω = time interval = Δt (replaces 'v') € Note that no matter where the particle is along the radiuθ,is the same, and thus so iω ! € Δθ Δ( )r Δs Δt v ω = = = = ⇒ v = rω Δt Δt r r 7.2 The Rotation of a Rigid Body Rotation will always be circular motion, hence we can bring all our knowledge of ‘uniform’ circular motion to thi€ chapter … Since v = ω r, two points of a rotating object will have different speeds if they have different distances from the axis of rotation, but all points have the same angularωv.locity If the angular velocity is changing (NOT uniform motion anymore), then there is an angular acceleration (similar to linear acceleration), given by: α = Δω (as forω, α does not depend upon ‘ r’) Δt Circular, or rotational, motion is very similar to motion in 1D. Instead of left/right or up/down, we now have clockwise/counterclockwise. Hence for bothω and α , their ‘vector’ nature will merely be a ‘+’ or a ‘–’. € Also, as for linear 1D motion, the acceleration will speed up (slow down) the object if it is in the same (opposite) direction as the velocity, regardless of whether that is a ‘+’ or a ‘–’ direction. Remember, angular acceleration speeds up or slows down the object, but centripetal acceleration points towards the center of the circle, and is due to the object’s rotation. Now that we are allowing the rotating object to speed up or slow down, the acceleration of each point on the object will have both a radial component (the centripetal acceleration) and a tangential component (proportional to the angular acceleration). The tangential component of the acceleration will depend upon the distance from the axis of rotattan α r. If the angular acceleration is constant, then, as for linear motion, we have: ω =ω +0 t Use when you don’t care about angular positioθ, θ =ω t + 12 α t2 Use when you don’t care about final angular veloωity, 0 2 2 € ω −ω = 0 α θ Use when you don’t care about timet € θ = ( +ω ) t/2 Use when you don’t care about angular acceleratioα, 0 2 € θ =ω t − 12α t Use when you don’t care about initial angular velωc0ty, Note that the above equations work for any set of uniθ, ω, andα : € 2 2 rad, rad/s, and rad/s or rev, rev/s, and rev/s € Example 7.6: Spinning up a computer disk The disk in a computer hard drive spins up from rest to a final angular speed of 5400 rpm in 2 s. What is the angular acceleration of the disk? At the end of 2 s, how many revolutions has the disk made? 7.3 Torque A force that produces a rotation. The ability of a force to cause a rotation depends on three factors: 1. The magnitude F of the force 2. The distancerfrom the axis about which the object can rotate to the point at which the force is applied 3. The angle at which the force is applied Torque τ’ (Greek tau) measures the "effectiveness" of a force to cause an object to rotate about a pivot. τ = rF = ( Fsinφ) ⊥ or τ = Fℓ = F rsinφ , where ℓ= “lever arm” ( ) (unit = ‘Nm’) € ℓ Note that to calculate a torque, one needs to know € the point about which the object is being rotated. Also, once again the ‘vector’ nature of the torque is merely a ‘+’ or ‘–’. € (As for up/down, where the book chooses to always take up ‘+’, the book takes counterclockwise to be ‘+’. We will choose our ‘+’ or ‘–’ direction however we wish.) Remember, torque and force are NOT the same. A torque is always in reference to a particular point of rotation. EXAMPLE 7.10: Calculating the torque on a nut Luis uses a 20-cm-long wrench to turn a nut. The wrench handle is tilted 30° above the horizontal, and Luis pulls straight down on the end with a force of 100 N. How much torque does Luis exert on the nut? Net torque: ∑ τi=τ 1τ 2...=τ net EXAMPLE 7.11: Net torque on ‘capston’ pulling up boat anchor € 7.4 Gravitational Torque and the Center of GraviXy 7.5 Rotational Dynamics and Moment of Inertia F = m a = mα r # #→Fr = mα r 2 i i i i i i i i i Fr = mα r = α m r2 ∑ i i ∑ i i ∑ i i 2 € Define "moment of inertia" to be I ∑ m i i Fi€ally, we have∑ Fi iτ net α I € Therefore, Newton’s 2ndLaw for rotation becomes τ = I α(instead ofF = m a) Recall that ‘mass’ €as a measure of inertia, the resistance to changes in velocity. Now, “moment of inertia” or “rotational inertia” is a measure of resistance to changes in angular velocity – how hard is it to get the object rotating. Rotational inertia depends on both mass AND where that mass is relative to the center of rotation! EXAMPLE 7.15: Calculating the moment of inertia Your friend is creating an abstract sculpture that consists of three small, heavy spheres attached by very lightweight 10-cm-long rods as shown in the figure . The spheres have masses m = 1.0 kg, 1 m 2 1.5 kg, and m3= 1.0 kg. What is the object’s moment of inertia if it is rotated about axis a? About axis b? Rotational inertias for some common shapes (see p 210): 1 2 1 2 Ithin rod about center Ithin rod about end 12 3 1 2 2 2 2 2 2 Icylinder or disk Ihollow cylinder or hoop Isolid sphereR Ihollow sphereR 2 5 3 € € € € € € 7.6 Using Newton's Second Law for Rotation Here we consider rotation of rigid bodies about a fixed axis. • Identify the axis about which the object rotates. • Identify the forces and determine their distances from the axis. • Calculate the torques caused by the forces, and find the signs of the torques. Then solve τ = I α net Find the moment of inertia either by direct calculation or from Table 7.1 for common shapes. EXAMPLE 7.18: Starting an airplane engine The engine in a small airplane is specified to have a torque of 500 Nm. This engine drives a 2-m-long, 40 kg single-blade propeller. On start-up, how long does it take the propeller to reach 2000 rpm? The propeller can be modeled as a rod rotating about its center. Ropes and Pulleys: Here’s the key! } EXAMPLE 7.19: Time for a bucket to fall Josh has just raised a 2.5 kg bucket of water using a well 's winch when he accidentally lets go of the handle. The winch consists of a rope wrapped around a 3 kg, 4-cm-diameter cylinder, which rotates on an axle through the center. The bucket is released from rest 4 m above the water level of the well. How long does it take to reach the water? Assume the rope is massless and doesn’t slip. 7.6 Rolling Motion Rolling is when an object rotates about an axis that is moving along a straight-line trajectory.