Download Final Exam Problems from 8.012 Fall 2006 - Motion in Rotating Frames and Central Forces and more Exams Classical Physics in PDF only on Docsity!
Problem 1: Short Answer Problems [2 pts each]
(a) List Newton’s three laws (order is not important).
(b) What are the units of impulse, using the notation [L] = length, [M] = mass and [T] = time?
(c) An archer in the northern hemisphere on Earth faces North and shoots an arrow at very high velocity. Will the arrow veer off to the left or the right? Justify your answer (note that the angular velocity vector of the Earth points away from the North Pole).
(d) Referring to the figure to the right, how far from the mass on the left is the point at which a test particle feels equal gravitational force from both spheres? Assume L is much larger than the radii of the spheres
(e) A mass is at rest in an inertial frame at a distance R from the origin. An observer sits at the origin of a rotating frame with angular velocity Ω that is initially aligned with the inertial frame. Describe in words the motion of the object from the observer’s perspective, and explain how that motion comes about in terms of fictional forces (there are no real forces acting on the mass).
Page 2 of 2 7
(a) (i) An object at rest or in uniform motion tends to stay at rest or in uniform motion unless acted on by an outside force (e.g., inertial frames exist)
(ii) - acceleration is proportional to force and inversely proportional to mass
(iii) - force exerted by one body onto a second body is equal and directed in the opposite direction to the force exerted by the second body onto the first
(b) Impulse = FxΔt = Δp = [M] [L] [T]-^1
(c) The veering of the arrow results from Coriolis force and is directed toward the
right. which is to the right when facing north.
(d) Total forces cancel at a distance x from the left sphere when:
Page 3 of 2 7
Problem 2: The Double Gyroscope [15 pts]
A gyroscope consists of two identical uniform disks with mass M and radius R mounted on a rigid axle with length 2D. The axle is fixed to the outer disk, while the inner disk is allowed to spin freely but is constrained to remain at a distance D from the pivot (at left) by a collar. The axle spins about the pivot freely on a frictionless mount. The outer disk and axle are initially spun up to an angular
frequency ω 0. Assume that the mass of the axle and pivot mount are negligible.
Gravity points downward. Ignore nutation.
(a) Calculate the precession rate Ω of the gyroscope assuming that the inner disk is
not spinning. Assume that ω 0 is very large and the gyroscopic approximation
applies.
(b) Friction between the axle and the disk closest to the pivot causes the latter to spin up. Assuming that a constant torque τ acts at the interface between the axle and inner hole of this disk, calculate the final spin rate ω of both disks and the final precession rate in terms of the initial precession rate.
(c) How much energy is lost from the gyroscope system during the spinning up of the inner disk?
Page 5 of 2 7
(a) The net torque on the gyroscope with respect to the pivot point is
The gyroscope approximation is:
The angular momentum of the first disk is
Hence
(b) The constant frictional torque is
Hence:
This is the result of the force from the axle acting on the inner wheel; Newton’s third law tells us that an equal and opposite force acts on the axle + outer wheel and slows it down. Hence
The frictional force ceases when the surfaces do not slide relative to one another, i.e., when , so
Page 6 of 2 7
Problem 3: Balanced Poles [10 pts]
Two sticks are attached with frictionless hinges to each other and to a wall, as shown above. The angle between the sticks is θ. Both sticks have the same constant linear mass density λ, and the horizontal stick has length L. Find the force (both horizontal and vertical components) that the lower sticks applies to the upper one where they connect at point A. Assume gravity points downward.
Page 8 of 2 7
The best way to approach this problem is to balance the forces and torques acting on the sticks, since the system is not accelerating and not rotating. A force diagram is shown on the left, indicating forces acting on the horizontal stick as solid arrows and forces acting on the angled stick as dotted arrows. The objective is to determine Fh and Fv.
First measure the torque on the horizontal stick about point B (assuming torque is positive in a clockwise direction on the page):
Now measure the torque on the angled stick about point B:
Page 9 of 2 7
This problem is straightforward if one works in the accelerated reference frame. The force diagrams for the two masses are shown at left. Note the additional fictional forces, and the fact that the friction forces both act toward the right – this is because in the accelerated frame the table is sliding to the right, dragging the blocks with it. The tension force acts via the massless string and is the same for both masses. The equations of motion for the two blocks are:
The positions of the masses are constratined to x 1 + x 3 = constant due to the fixed
length of rope connecting them; hence
Replacing this expression in to the equations of motion yield:
We move to the frame of rest of the table by adding the acceleration A to both equations of motion:
Page 1 1 of 2 7
(b) For the 3M mass to remain stationary in the frame of rest, so
Page 1 2 of 2 7
(a) The torque on the mass about the origin is:
Hence, the angular momentum vector about the origin is a constant of the motion.
(b) The effective potential is
where
We can choose a zeropoint such that U(∞) = 0, which for n > 1 implies C = 0, and hence:
(c) The equilibrium point satisfies
Page 1 4 of 2 7
(d) For a stable orbit, the equilibrium point should be a stable one which requires
Page 1 5 of 2 7
(a) There are two main points to this problem that must be recognized to solve it. The first is that friction must act between the wheels and platform, and between the wheels and incline, and that the former modifies the acceleration of the platform. The second is that the no slipping constraint applies between the wheels and platform and the wheels and incline, so
Where φ is the angular motion of the wheel.
This problem can be solved in two ways: by solving the translational and rotational equations of motion and applying the constraint equations and using energy.
The first solution is built upon the force diagrams shown above. Note the matching of forces between the wheel and platform (ƒ and N). Also note that we have not assumed ƒ = ƒ’, nor have we assumed a relation between ƒ and N based on the (unknown) coefficient of friction. We can restrict analysis to the platform and one wheel, since the other wheels will experience the same motion. The translational equations of motion for the platform are (note the directions of x and y from the figures):
Page 1 7 of 2 7
The translational equations of motion for each wheel are
The wheel also spins, so we need to calculate rotational equations of motion. To do this we choose a fixed point parallel to the incline that is aligned with center of the wheel – we cannot chose the center of the wheel as our reference point because the wheel is accelerating down the incline. Hence,
where the xw terms have dropped out based on the y-axis equation of motion of the wheel above.
Now we apply the constraint equations to the xp, xw and φ equations of motion,
and assume giving:
Using these equations to eliminate ƒ and ƒ’ yields:
The alternate derivation of the solution involves computing the change in energy for the system, which is conserved since the friction forces do no work (the surfaces are not slipping relative to each other). Assuming that the system starts
from rest at xp = xw = 0, then
Page 1 8 of 2 7
Problem 7: Spinning Bouncing Ball [15 pts]
A uniform sphere of mass M and radius R spinning with angular velocity ω is dropped from a height H. It bounces on the floor and recoils with the same vertical velocity. During the bounce, the surface of the ball slips relative to the surface of the floor (i.e., it does not roll) and in the process the ball is acted upon by a friction force with magnitude ƒ = μN, where N is the normal contact force between the ball and the floor and μ is a constant. Hence, the ball experiences impulses in both vertical and horizontal directions. Assume that the duration of contact, Δt, is very short.
(a) What angle α with respect to vertical does the ball recoil?
(b) What is the final rotation velocity of the ball?
(c) What value of H results in the ball bouncing off with no spin?
Page 2 0 of 2 7
(a) The angle α is related to the horizontal and vertical components of the ball’s velocity after the bounce. The vertical component is straightforward, as it is stated in the problem that it rebounds with the same velocity as it struck the ground, which is:
The horizontal component arise from the impulse due to friction acting at the contact surface:
This is related to the vertical impulse by:
Hence:
and:
(b) The frictional force produces a torque that acts in the opposite direction as the spinning of the ball. This torque exerts a rotational impulse (change in angular momentum):
Hence,
Using the momentum of inertia for a sphere:
Page 2 1 of 2 7
