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Physics 1A, Physics 1A, Section 2 Section 2 November 11, 2010

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Physics 1A, Section 2. November 11, 2010. Translation / Rotation. Quiz Problem 25. Quiz Problem 25. Answer: a) T = a(mR 2 + I )/[R(R-r)] f = a( I + mrR)/[R(R-r)] b) m min = a(kR + r)/[g(R-r)] c) rolls to the right. Moment of Inertia. Moments: 0 th moment: ∫dm = mass - PowerPoint PPT Presentation

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Page 1: Physics 1A, Section 2

Physics 1A,Physics 1A,Section 2Section 2

November 11, 2010

Page 2: Physics 1A, Section 2

Translation / Rotationtranslational motion rotational motion

position x angular position

velocity v = dx/dt angular velocity = d/dt

acceleration a = dv/dt = d2x/dt2 angular acceleration = d/dt = d2/dt2

mass m moment of inertia I

momentum p = mv angular momentum L = I

force F = ma torque = I

kinetic energy ½ mv2 kinetic energy ½ I2

Page 3: Physics 1A, Section 2

Quiz Problem

25

Page 4: Physics 1A, Section 2

Quiz Problem

25• Answer:a) T = a(mR2 + I)/[R(R-r)] f = a(I + mrR)/[R(R-r)]b) min = a(kR + r)/[g(R-r)]c) rolls to the right

Page 5: Physics 1A, Section 2

Moment of Inertia• Moments:

0th moment: ∫dm = massuseful 1st moment: ∫ r dm

1/M ∫ r dm is the center of massuseful 2nd moment: ∫R2 dm = moment of inertia

R is the distance from the axis of rotationOther ways to write the integral:

∫dm = ∫ dV = ∫∫∫ dx dy dz

• Moment of inertia examples:• Frautschi et al. Table 14.1, page 379

• Parallel axis theorem: I = ICM + Md2

Page 6: Physics 1A, Section 2

Quiz Problem

41

Page 7: Physics 1A, Section 2

Quiz Problem

41• Answer:a) 1 = 8da4

2 = ½ da4

1/2 = 16b) f = 16/17c) No, some lost to friction. Kfinal/Kinitial = 16/17

Page 8: Physics 1A, Section 2

• Quiz Problem 32 (collision with rotation)

• Optional, but helpful, to try these in advance.

Monday, November 15: