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Problem 1: Quick Multiple Choice Questions [10 pts] Answer all five problems. You do not need to show any work. Do not spend a long time on this section. (a) What are the dimensions of the universal gravitational constant G? Recall that [M] = mass, [L] = length and [T] = time [M][L][T]
- 2 [M][L] 2 [T] - 1 [M] - 1 [L] 2 [T] - 1 [M] - 1 [L] 3 [T] - 2 (b) A uniform disk of radius R has a circular hole of diameter R cut out of it. The hole lies between the outer edge of the disk and the center of the disk, as shown to the right. Mark with an “X” where the center of mass of this disk lies. (c) A hollow cylinder rolls up an incline without slipping. In which direction is the friction force between the cylinder and incline surfaces acting? Down the incline Up the incline There is no friction force acting on the cylinder Page 2 of 2 0
(d) Consider the one-dimensional potential function shown below. Assume that an object in this potential starts at x = xo at rest. Indicate on the plot the maximum value of x the object can attain and circle the stable equilibrium points in this range.
max x
(e) A ball comes in from the left with speed 1 (in arbitrary units) and causes a series of collisions. The initial motion is shown below (the number in the circles indicates the objects’ relative masses): All collisions are elastic. Which of the following indicates the motions of the balls after all of the collisions are completed? (circle the correct diagram) Page 3 of 2 0
Solutions
(a) This problem can solved by considering the translation of the center of mass and rotation about the center of mass of the ball separately. Referring to the force diagram to the right, the equations of motion are: Note that torque is working to slow down the rotation of the ball in this case, so there must be a minus sign in the angle equation if we set ω 0 > 0. A ball is a sphere, so While the ball is slipping, the friction force is at its maximum (constant) value of Ff = μN = μMg, so the equations are easily solved. Using the initial conditions: Page 5 of 2 0
we find (note the specific direction of velocity): (b) The condition of rolling without slipping occurs when Note the minus sign – this is because we defined positive rotation in the initial conditions such that the object has backspin, but when it is not slipping it must have forward spin. This condition can be used to find the time tf when the ball starts to roll without slipping: Plugging this back into the expressions from (a) we find: Note that after this point, the friction force ceases to act as the surfaces are no longer slipping relative to each other, and so the ball continues to roll/spin at these rates. Page 6 of 2 0
Problem 3: A Generic Potential [25 pts] Consider the following one-dimensional potential in which an object of mass m moves freely: where is a constant with dimensions of energy and s = x/x 0 is a normalized dimension of length. (a) [5 pts] Sketch this potential as a function of x in the range - 3x 0 < x < 3xo. (b) [5 pts] Derive values for the equilibrium points and determine which of these are stable. (c) [5 pts] Determine the period of small oscillations about the stable equilibrium point. (d) [5 pts] If the mass starts from x = - ∞ and moves toward positive x, what is the minimum total mechanical energy it needs to reach the stable equilibrium point? What is its velocity at the stable equilibrium point in this case? (e) [5 pts] What is the magnitude and direction of the force acting on the mass when it is at x = 2xo? Page 8 of 2 0
Solution
(a) A sketch of the potential is shown below: (b) At the equilibrium points the net force goes to zero, hence Page 9 of 2 0
(d) Moving from x = - ∞, the mass must make it over the unstable equilibrium point at x=0, which has potential energy U(0) = - 4U 0. Hence The total mechanical energy remains constant, so the loss in potential energy, U(x 0 ) – U(0) where U(x 0 ) = - 5U 0 , goes into kinetic energy. If the mass has effectively zero kinetic energy at x = 0, then (e) Using the expression for F from part (a) and setting x = 2x 0 we find: The force must point toward negative x, since the potential barrier is increasing here. Page 1 1 of 2 0
Problem 4: A Ballistic Rotator [20 pts] A thin arrow with length R, mass m and uniform linear mass density λ = M/R is shot with velocity v into a circular target of radius R, mass M, uniform surface mass density σ = M/πR 2 , and negligible thickness (i.e., it is essentially a thin disk). The arrow head sticks into the target just off center by a distance x < R as shown above. The target is mounted onto a central frictionless shaft, about which it can spin freely. The target is at rest before it is struck by the arrow. (a) [10 pts] Find the angular rotation rate of the target (with arrow embedded) after the arrow strikes it. You can express you answer in terms of IT, the moment of inertia of the circular target alone about the rotation axis shown. (b) [5 pts] Show that if m << M then to first order the rotation rate depends linearly on both x and v. (c) [5 pts] Derive an expression for IT in terms of M and R. Warning! This is a difficult problem, so save it for last! Page 1 2 of 2 0
(b) It ≈ MR 2 , Iarrow ≈ mR 2
- mx 2 so if m << M, then It >> Iarrow, and the solution to (a) reduces to: these quanitites. This depends linearly on both v and x, since It depends on neither of (c) There are a number of ways of solving this, including using a fun trick the TA demonstrated in the review session. However, the most direct way is to use the standard integral expression for I: where the factors in the integral, in polar coordinates, are: The volume integral is then: Page 1 4 of 2 0
The integral over θ can be computed by substituting the appropriate half-angle trig identity; an easier trick is to note that over 2π the integrals of cos 2 and sin 2 are the same (in both cases they are over a full period). Hence: so Page 1 5 of 2 0
Solution
(a) Let’s simply the notation by letting 3M 0 be the initial mass of the rocket. As there are no external forces, we can apply conservation of momentum when each small piece of fuel is ejected/part of rocket shell is lost: Because the rocket is losing mass, so We can divide both sides by and taking the limit for small differences, this equation reduces to derive the equation of motion: Where we have used to eliminate the variable M(t). Page 1 7 of 2 0
(b) The solution from (a) can be integrated by multiplying both sides by : Note that this is just faster than final speed of a rocket of mass 3M that burns M of its fuel but keeps its shell: Page 1 8 of 2 0
Moment of Inertia Moment of inertia for a uniform bar (about COM) Moment of inertia for a uniform hoop (about COM) Moment of inertia for a uniform disk (about COM) Moment of inertia for a solid uniform sphere (about COM) Parallel axis theorem Taylor Expansion of f(x) Page 2 0 of 2 0
