2005 Free Response

Question 3

Before Collision After Collision TOP VIEWS Mech. 3. A system consists of a ball of mass M2 and a uniform rod of mass Ml and length d. The rod is attached to a horizontal frictionless table by a pivot at point P and initially rotates at an angular speed o , as shown above left. The rotational inertia of the rod about point P is —Mld2. The rod strikes the ball, which is initially at rest. As a result of this collision, the rod is stopped and the ball moves in the direction shown above right. Express all answers in terms of Ml , M2, (D , d, and fundamental constants.

(b) Derive an expression for the speed v of the ball after the collision.

(b) 4 points For any indication of onservation of angular momentum Lb=Lr For the correct expression for Lb For substitution for Lr consistent with part (a) M2Dd = —Mid20 For the correct final expression for v v — {ado 2


(c) 4 points For any indication of conservation of kinetic energy No oints were awarded for conservation of mechanical energy. Kb=Kr For the correct expressions for both kinetic energies M2V2 = —1m M2D2 = 10 For correct substitutions for I and for v consistent with part (b) 1112 = -Mid2 d202 = -Mid202 9 1112 1 M12 1 91112 - 3 For the correct final expression for the ratio Ml /M2 = 3 (3:1 was also acceptable)

Before Collision (d) A new ball with the same mass Ml as the rod is now placed a distance x from the pivot, as shown above. Again assuming the collision is elastic, for what value of x will the rod stop moving after hitting the ball?

For the correct equation for conservation of an MIDX = —Mid o For solving this equation for v MID2 = —10 ar momentum For the correct equation fo conservation of kinetic energy —Mid 2 1 22 For the correct substitution of the above expression for v from momentum conservation into the equation for conservation of kinetic energy 1 22 1 d202 3 d2 For the correct final answer

2006 Free Response

Question 3

Mech 3. A thin hoop of mass M, radius R, and rotational inertia MR2 is released from rest from the top of the ramp of length L above. The ramp makes an angle 9 with respect to a horizontal tabletop to which the ramp is fixed. The table is a height H above the floor. Assume that the hoop rolls without slipping down the ramp and across the table. Express all algebraic answers in terms of given quantities and fundamental constants. (a) Derive an expression for the acceleration of the center of mass of the hoop as it rolls down the ramp.

(a) 5 points For use of Newton's 2nd law in rotational form and the parallel axis theorem ET = lacm and 1 = + Mh2 For a correct rotational inertia about the point of contact using the parallel axis theorem 1 = MR2 + MR2 = 2MR2 For a correct to ue about the point of contact = RMgsin9 For a correct relationship between linear and angular acceleration for rolling without slipping a a Substituting for ET , I, and acm into the rotational equation above RMgsin9 = 2MR2 acm For the correct answer a = —sin9 2

(d) Suppose that the hoop is now replaced by a disk having the same mass M and radius R. How will the distance from the edge of the table to where the disk lands on the floor compare with the distance determined in part (c) for the hoop? Less than The same as Greater than Briefly justify your response.

(d) 3 points For checking the space next to "Greater than" For a sufficiently detailed justification containing no incorrect statements. Such an answer logically concludes, at a minimum, that the linear speed or velocity at the bottom of the ramp is greater for the disk because the rotational inertia of the disk is less. It is not necessary to state that the time of fall is the same. One point was awarded for a minimal or partially correct answer. No justification points were awarded if the space next to "Greater than " was not checked. Examples of 2-point answers: A disk will have smaller rotational inertia and will therefore have velocity. This will lead to a greater translational velocity, and a greater distance x. The rotational inertia is less thanüfiÖfi6ÖTüGfiüFéäiFacceleration and more final speed at the end of the table. The acceleration when I = MR2/2 is (2/3) g sing, so the disk will be moving faster at the bottom of the ramp and will travel farther. Examples of I-point answers: A disk has a larger rotational inertia, so it will have a greater kinetic energy and will therefore land farther from the ramp. The moment of inertia for the disk is smaller, thus its rotational velocity is bigger, causing it to go further. Less energy will be used to spin the disk than the hoop, and I of the disk is less than I of the hoop.

2007 Free Response

Question 2

In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data back to Earth. Assume a circular orbit with aperiod of 1.18 x 102 minutes = 7.08 x 103 s and orbital speed of 3.40 x 103 m/s . The mass of the GS is 930 kg and the radius of Mars is 3.43 x 106 m .

(e) In fact, the orbit the GS entered was slightly elliptical with its closest approach to Mars at 3.71 x 105 m above the surface and its furthest distance at 4.36 x 105 m above the surface. If the speed of the GS at closest approach is 3.40 x 103 m s , calculate the speed at the furthest point of the orbit.

For a correct expression of conservation of angular momentum m st)lh = msV2h ore uivalentsuchas 1101 = 1202 or = rl Rc RM where Rc and RF are the distances of closest and farthest approaches, respectively, and RM is the radius of Mars For explicit substitution of radii (not altitudes) into the equation for the correct numerical answer 3.71 x 105m 34.3 x 105 m = (3.40 x 103 m/s) 4.36 x 105m 34.3 x 105 m = 3.34 x 103 m/s Alternatively, if the longer approach using conservation of energy was taken, 1 point was awarded for a correct statement of conservation of energ' if explicitly written as 2 GmsMM 1 2 Gem M 1 2 — , and 1 point was awarded for the explicit — ms1)2 2 substitution of radii (not altitudes) QC for a correct numerical answer.

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