Physics 8 Friday, October 25, 2013

Transcription

1 Physics 8 Friday, October 25, 2013 Hold on to your HW8 paper. Don t turn it in yet! What we covered in class this week went much more slowly than I had expected. We spent a lot of time discussing the ideas that we did manage to cover that s good. But we never got through some of the harder ideas. So everyone found the HW8 problems much too difficult, given what we ve had time in class so far to discuss. Let s spend today reviewing ideas & eqns needed for HW8. You can hand it in Monday with a better grasp of the material. I m also available to help by (or phone, if you re working in a group) over the weekend. But I go to sleep early! No reading for Monday (10/28). I ll push Mon reading back to Weds. Reload online-response page in your web browser. We have the next 2 3 weeks to understand & use torque. HW9 is 7 conceptual + 8 calculation problems. Of the 8 calculation problems, 4 or 5 are from Chapter 11.

2 Since the string s length L stays constant, what shape does the ball s path trace out as it moves? Does the ball s acceleration have a component that points along the axis of the string? If so, does its magnitude depend on the ball s speed? What two forces are acting on the ball? Assuming that no energy is dissipated, how can we relate the ball s speed v to its height y? Can you write m a = F for the component of a and F that points along the string? (note: small ball neglect the ball s rotational K.E.)

3 How do I relate angle θ to speed v? E i = mgl E = 1 2 mv 2 + mgy (A) 1 2 mv 2 = mg(l y) = mgl(1 cos θ) (B) 1 2 mv 2 = mg(l y) = mgl(1 sin θ) (C) 1 2 mv 2 = mg(l y) = mgl cos θ (D) 1 2 mv 2 = mg(l y) = mgl sin θ Hint: draw on the figure a vertical line of length Next: write radial component of m a = F to find T L y = y i y

4 Are the angles of the two strings w.r.t. horizontal equal? Are the tensions in the two strings equal? What three forces act on the ball? Is the ball accelerating vertically? Horizontally? Draw a FBD for the ball, showing both horizontal (radial) and vertical component of each force. Notice that the ball s speed v increases with time, until finally one string breaks. Which one?

5 How would you approach this problem? Discuss with your neighbor while I set up a demonstration along the same lines... (A) initial angular momentum of bucket equals final angular momentum of cylinder + bucket (B) initial G.P.E. equals final K.E. (translational for bucket + rotational for cylinder) (C) initial G.P.E. equals final K.E. of bucket (D) initial G.P.E. equals final K.E. of cylinder (E) initial K.E. of bucket equals final G.P.E. (F) use torque = mgr to find constant angular acceleration

6 What is the rotational inertia for a solid cylinder? How do you relate v of the bucket with ω of the cylinder? Why is this true? What is the expression for the total kinetic energy? Why is angular momentum not the same for the initial and final states? (Why is this an energy problem and not an angular momentum problem? )

7 Textbook & equation sheet contain table of rotational inertias.

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9 If an object revolves about an axis that does not pass through the object s center of mass (suppose axis has distance l from CoM), the rotational inertia is larger, because the object s CoM revolves around a circle of radius l and in addition the object rotates about its own CoM. This larger rotational inertia is given by the parallel axis theorem: I = I cm + Ml 2 where I cm is the object s rotational inertia about an axis (which must be parallel to the new axis of rotation) that passes through the object s CoM.

10 The rotational inertia for a solid sphere of radius R about an axis through its center is I = 2 5 mr2 What is its rotational inertia about an axis tangent to the sphere? (A) 1 5 mr2 (B) 2 5 mr2 (C) 4 5 mr2 (D) 7 5 mr2 (E) 12 5 mr2

11 The rotational inertia for a long, thin rod of length l about a perpendicular axis through its center is I = 1 12 ml2 What is its rotational inertia about one end? (A) 1 12 ml2 (B) 1 24 ml2 (C) 1 2 ml2 (D) 1 3 ml2 (E) 1 4 ml2 (F) 1 6 ml2

12 How would you approach this problem? Discuss with neighbors! Which (if any) of these statements is false? (A) I know the change in G.P.E from the initial to the desired final states. So the initial K.E. (translational + rotational) needs to be at least this large. (B) The book (or equation sheet) gives rotational inertia I for a long, thin rod about its center. So I can use the parallel-axis theorem to get I for the rod about one end. (C) The angular momentum, L = I ω, is the same for the initial and final states. (D) Because the rod pivots about one end, the speed of the other end is v = ωl (where l is length of rod) (E) None. (All of the above statements are true.)

13 Is K.E. the same before & after they join hands? Is momentum p = m v the same before & after they join hands? If so, does this tell you what the center-of-mass does (translationally), or does it tell you about the couple s rotation about the CoM? Is angular momentum different before & after they join hands? (Are there external forces tending to rotate the pair?) What are Chapter 10 s two expressions for angular momentum? answer: L = I ω L = r mv How far from each skater is the (joined) couple s CoM? What is the rotational inertia of each skater (and the pair together) for rotation about the CoM?

14 How would you approach this problem? Discuss with neighbors! (A) The final K.E. (rotational+translational) equals the initial K.E. of the ball. (B) The initial momentum m v of the ball equals the final momentum (m + M) v of the door+ball. (C) The initial angular momentum L = r mv of the ball w.r.t. the hinge axis equals the final angular momentum L = I ω of the door+ball.

15 I know that the rotational inertia of a thin rod of length L about a perpendicular axis through its center is I = 1 12 ml2. The rotational inertia I to use for the final state here is (A) I = ML 2 + ml 2 (B) I = 1 12 ML2 + M( L 2 )2 + m( 2 3 L)2 (C) I = 1 12 ML ml2 (D) I = 1 12 ML2 + m( 2 3 L)2 (E) I = 1 12 ML2 + ml 2 Also think how the answer would change if the radius of the putty ball were non-negligible. What if the thickness of the door were non-negligible? Does the height of the door matter?

16 When does the block lose contact with the sphere? (Assume there is no friction between the block and the sphere.)

17 Physics 8 Friday, October 25, 2013 Hold on to your HW8 paper. Don t turn it in yet! Hand in H8 Monday with a better grasp of the material. I m also available to help by (or phone, if you re working in a group) over the weekend. But I go to sleep early! No reading for Monday (10/28). I ll push Mon reading back to Weds. Reload online-response page in your web browser. I ll update the Canvas & online-response pages tomorrow. We have the next 2 3 weeks to understand & use torque. HW9 is 7 conceptual + 8 calculation problems. Of the 8 calculation problems, 4 or 5 are from Chapter 11. Only a few use new material from Chapter 12. What I handed back to you today includes (a) graded quiz6, (b) graded HW7, and (c) new HW9 handout.