Physics 1710 —Warm-up Quiz

1. Physics 1710—Warm-up Quiz Answer Now ! 0 39% 55 of 140 0 Approximately how much torque must a major league baseball player exert in order to accelerate a baseball bat in a quarter of a circle swing to a speed at its tip of 48.0 m/s if the bat is 0.80 m long and has a mass of 4.1 kg? • ~1 Nm • ~10 Nm • ~100 Nm • ~1000 Nm

2. Physics 1710—Chapters 6-10 v ω = v/L L 0 τ = I α I = 1/3 mL2 α = ω2/(2Δθ) ω = v/L τ =(1/3 mL2 )(v/L)2/(2Δθ) = 1/6 m v2 /Δθ τ = 1/6 (4.1 kg)(48 m/s)2 /(3.14/2)= 1002 Nm Solution:

3. Physics 1710—Chapters 6-10 0 Translational →: Rotary↻: x v a vfinal=vinitial +at xfinal=xinitial +vinitialt +½ at2 vfinal 2=vinitial 2 +2aΔx m F F = m a • θ = x/R • ω = v/R • α = a/R • ωfinal= ωinitial + α t • θfinal= θinitial + ωinitialt +½ αt2 • ωfinal 2= ωinitial 2 +2 αΔ θ • I = ∫R2dm • τ = R x F • τ = Iα Translational vs Rotational Kinematics

4. Physics 1710—Chapters 6-10 0 Translational →: Rotary↻: p= mv F = dp/dt Ktrans= ½ mv2 Utrans= - ∫F• dr Etrans = Ktrans + Utrans • L = R x p • τ= dL/dt • Krot= ½ I ω2 • Urot= - ∫τ• dθ • Erot = Krot + Urot Linear vs Rotational Kinematics ⇜⇝Center of Mass: RCM = (Σmiri)/ Σmi Parallel Axis Theorem: Iparallel = ICM + MRCM2 Total E = Etrans + Erot

5. Physics 1710—Chapters 6-10 No Talking! Think! Confer! 0 Which has the greater moment of inertia: a ring (of mass M, radius R) rotating about its center or a similar disk rotating about a point on its circumference? Peer Instruction Time

6. Physics 1710—Chapters 6-10 0 42% 59 of 140 Answer Now ! 0 Which has the greater moment of inertia: a ring (of mass M, radius R) rotating about its center or a similar disk rotating about a point on its circumference? • I bigger for ring. • I bigger for disk. • I the same for each.

7. Physics 1710—Chapters 6-10 0 IRing = mR2 IDisk = IDisk-CM + mR2 IDisk = ½ mR2 + mR2 = 3/2 mR2 >Iring Solution:

8. Physics 1710—Chapters 6-10 0 1′ Lecture (Review) Rotary (circular) motion obeys laws that are analogous to those of translational motion. Linear Momentum is conserved in absence of external forces. F = d p/dt Energy is related to the work done or stored Work is the cumulative force times distance moved. Power is the rate of expenditure of work or energy.

9. Physics 1710—Chapters 6-10 ω r v = ω x r 0 Circular Motion: Acceleration: Centripetal:aC= - ω2r; a = v 2/r, toward center Tangential: aT= rdω/dt

10. Physics 1710—Chapters 6-10 0 Work and Power: W = ∫ F•d r +∫ τ•d θ P = dW/dt = F•v + τ•ω Units: Work—Joule Power—Watt = Joule/sec

11. Physics 1710—Chapters 6-10 0 Work and Kinetic Energy: If only change is speed vor rotation rate ω ΣW = Kfinal – K initial Kinetic Energy K: Ktrans = ½ mv2 Krot = ½ I ω2

12. Physics 1710—Chapters 6-10 0 Work and Potential Energy and Force/Torque: ΔU = - ΣW Force and Torque from Potential Energy Fx = - dU/dx Fy = - dU/dy τ = - dU/dθ Gravitational Potential Energy: Ug = mgy Elastic Potential Energy: Ue = ½ kx2 Total Mechanical Energy E = K + U

13. Physics 1710—Chapters 6-10 0 Momentum and ImpulseI p= m v In the absence of external forces momentum is conserved: pinitial,1+ pinitial,2 = pfinal,1+ pfinal,2 pinitial,x,1+ pinitial,x,2 = pfinal,x,1+ pfinal,x,2 pinitial,y,1+ pinitial,y,2 = pfinal,y,1+ pfinal,y,2 I = Δp =∫Fdt

14. Physics 1710—Chapters 6-10 0 Momentum and Force (2nd Law od Motion) F = d p/dt τ = d L/dt

15. Physics 1710—Chapters 6-10 0 Summary (Review) Rotary (circular) motion obeys laws that are analogous to those of translational motion. Linear Momentum is conserved in absence of external forces. F = d p/dt Energy is related to the work done or stored Work is the cumulative force times distance moved. Power is the rate of expenditure of work or energy. Force is the negative of the gradient of the potential.