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PHYSICS 231 Review problems for midterm 2 Topic 5: Energy and Work and Power Topic 6: Momentum and Collisions Topic 7: Oscillations (spring and pendulum)

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Presentation on theme: "PHYSICS 231 Review problems for midterm 2 Topic 5: Energy and Work and Power Topic 6: Momentum and Collisions Topic 7: Oscillations (spring and pendulum)"— Presentation transcript:

1 PHYSICS 231 Review problems for midterm 2 Topic 5: Energy and Work and Power Topic 6: Momentum and Collisions Topic 7: Oscillations (spring and pendulum) Topic 8: Rotational Motion

2 PHY 231 2 Move to the middle when you enter C Abreha Boulos D Bradley Corbitt E Danner Feng F Figarra Hicks G Hoebecke Kato H Kearfott Lawrence I Lee Lyons J Mackie Mlot K Mobley Piper L Pokriefka Rosekrans M Saini Sumbal N Tamm Watt O Weidenaar Zhou 22-16

3 Some Housekeeping The 2nd exam will be Wednesday October 22. The exam will take place right here in BPS 1410. We will have assigned seating, so show up early. You need to show up 10 min early to guarantee your seat. If you are sick, you MUST have a note from a doctor on the doctor’s letterhead or prescription pad. You CANNOT use cell phones during the exam. You can (should!) bring two hand written equation sheets on a 8.5x11 sheet of paper – two one-sided or one two-sided 3

4 (A) (B) (C) (D) (E)

5 Energy Potential energy (PE) Energy associated with position. Gravitational PE: mgh Energy associated with position in grav. field. Kinetic energy KE: ½mv 2 Energy associated with motion Elastic PE: ½kx 2 energy stored in stretched/compressed spring. Mechanical energy ME: ME = KE + PE Conservative forces ME i – ME f = 0 Non-conservative forces ME i - ME f = E nc

6 Work (the amount of energy transfer) W=(Fcos  )  x To measure how fast we transfer the energy we define: Power: The rate of energy transfer Power = P = (Work/time) = (W/  t) (J/s=Watt) P = (Fcos  ) ·  x/  t = (Fcos  ) · v average

7 Equations of motion Linear motion Angular motion x(t) = x o + v o t + ½at 2 v(t) = v o + at  (t) =  o +  o t + ½  t 2  (t) =  o +  t Angular motion = Rotational Motion!

8 Momentum and Impulse F = m aNewton’s 2nd law F = m  v/  t a=  v/  t F = m (v final -v inital )/  t Define p = mv p: momentum (kg m/s) F= (p final -p initial )/  t F=  p/  t Definition:  p = Impulse CONSERVATION OF MOMENTUM (for a closed system) m 1 v 1f + m 2 v 2f = m 1 v 1i + m 2 v 2i or p 1i + p 2i = p 1f + p 2f addition of vectors example p i = p 1f + p 2f

9 Inelastic Solution v f = [m 1 v 1i + m 2 v 2i ] / (m 1 +m 2 ) With m 2 = c m 1 v f = [v 1i + cv 2i ] / (1+c) Special case m 2 = m 1 (c=1) v f = [v 1i + v 2i ] / 2 Conservation of momentum: m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f After the collision m 1 and m 2 form one new object with mass M = m 1 + m 2 and one velocity v f =v 1f = v 2f m 1 v 1i + m 2 v 2i = v f (m 1 + m 2 ) KE i - KE f = E nc

10 Elastic Solution v 1f = [(m 1 –m 2 ) v 1i + 2m 2 v 2i ] / (m 1 +m 2 ) v 2f = [(m 2 –m 1 ) v 2i + 2m 1 v 1i ] / (m 1 +m 2 ) For m 2 = c m 1 v 1f = [(1-c) v 1i + 2cv 2i ] / (1+c) v 2f = [(c-1) v 2i + 2v 1i ] / (1+c) Special case m 2 = m 1 (c=1) v 1f = v 2i v 2f = v 1i KE i - KE f = 0

11 Spring Total ME at any displacement x: ½mv 2 + ½kx 2 Total ME at max. displacement A: ½kA 2 Conservation of ME: ½kA 2 = ½mv 2 + ½kx 2 So: v=±  [(A 2 -x 2 )k/m] position Xvelocity Vacceleration +A0-kA/m 0±A  (k/m)0 -A0kA/m

12 A -A velocity v x x(t)=Acos(  t) v(t)=-  Asin(  t) a(t)=-  2 Acos(  t)  =  (k/m) A  (k/m) -A  (k/m) -kA/m a kA/m time (s)

13 pendulum vs spring parameterspringpendulum restoring force F F=-kxF=-(mg/L)s angular frequency  =  (k/m)  =  (g/L) the pendulum

14 circular motion

15 An Example A B 5m 6m The angular velocity of A is 2 rad/s. 1) What is its tangential velocity? If B is keeping pace with A, 2) What is its angular velocity? 3) What is its tangential velocity? 1) v =  r = 2  6 = 12 m/s 12 m/s 2) Must be the same: 2 rad/s  3) v =  r = 2  5 = 10 m/s 10 m/s

16 Translational equilibrium:  F=ma=0 The center of gravity does not move! Rotational equilibrium:   =0 The object does not rotate Mechanical equilibrium:  F=ma=0 &   =0 No movement! Torque:  =Fd Center of Gravity:

17  =I  Moment of inertia I: I=(  m i r i 2 ) I = mr 2 all mass is at the same distance r compare with: F=ma The moment of inertia in rotations is similar to the mass in Newton’s 2 nd law. m1m1 m2m2 m3m3 r1r1 r2r2 r3r3 Moment of inertia

18 Rotational kinetic energy Consider a object rotating with constant velocity. Each point moves with velocity v i. The total kinetic energy is: KE r =½I  2 Conservation of energy for rotating object must include: rotational and translational kinetic energy and potential energy [PE+KE t +KE r ] initial = [PE+KE t +KE r ] final mimi riri v

19 Which one goes fastest Use conservation of mechanical energy. At top (at rest): [½mv 2 +mgh+½I  2 ] = mgh At bottom: [½mv 2 + ½I  2 ] = [½mv 2 + ½(kmr 2 )v 2 /r 2 ] (used: I=kmr 2 and  =v/r) Simplify: [½mv 2 +½kmv 2 ]=(1+k)½mv 2 Combine: E top =E bottom mgh=(1+k)½mv 2 v=  [2gh/(1+k)] No mass dependence!! objectIk ACylindrical shell Mr 2 1 BSolid cylinder(1/2) mr 2 0.5 CThin spherical shell (2/3) mr 2 0.666 DSolid sphere(2/5) mr 2 0.4

20 Angular momentum Conservation of angular momentum If the net torque equals zero, the angular momentum L does not change I i  i = I f  f

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