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Problem 1: Masses and Rod [15 pts]
A block with mass 3M connected to a massless rod of length L lies at rest on a fixed frictionless table. A second block of mass M impinges on the system with
speed v 0 , strikes the opposite end of the rod at a right angle, and sticks. For this
problem, assume that the dimensions of the blocks are much smaller than the length of the rod.
(a) [5 pts] What is the final velocity of the center of mass of the block-rod-block system?
(b) [5 pts] What is the final angular velocity of the block-rod-block system about its center of mass?
(d) [3 pts] What is the initial velocity of the more massive block after the collision? Be sure to indicate its direction and speed.
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SOLUTIONS
(a) Linear momentum of system is conserved (no external forces), so
(b) Angular momentum of system is conserved (no external torques) so
where is the distance from the center of mass (COM) to the end of the bar:
and I is the moment of inertia of two point masses about the COM:
and hence
(c) at the position of 3M, total velocity is combination of linear and rotation motion, so
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SOLUTIONS
(a) First calculate potential energy:
Now calculate the equilibrium points, where :
(ignoring solution)
so are the finite equilibrium points. We can evaluate these points and the limits for x:
(note how highest power term is < 0)
The sketch can then be drawn as below
a 3a x
U 0
U
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By inspection, we can see that x=a is unstable and x=3a is stable. We can also determine this from the second derivative of U:
so x=3a is stable.
(b) The effective spring constant about the stability point is
The period of small oscillations is therefore:
(c) At x = a, the total mechanical energy is
At x = 3a,
As this is a conservative force, total mechanical energy is conserved, so
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SOLUTIONS
(a) The minimum initial (in the LAB frame) occurs when the emergent particles have zero velocity in the COM frame, or when they are moving together in the LAB frame. So there are two ways to calculate this. First, move to COM frame:
and
In both the LAB or COM frame, since there are no external forces:
where the prime variables refer to after the collision. Note that this equation only works when we compare energies in the same frame! If the emergent particles have no velocity in COM frame, then
then the minimum kinetic energy of the first particle in the LAB frame is
We can also solve this entirely in the LAB frame by noting that if the emergent
particles are at rest in the COM frame, they most both have velocity in the LAB frame; hence:
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(b) Assume:
In the COM frame, the emergent particles scatter with velocities and. Since the total linear momentum in the COM frame is 0,
and hence the magnitudes are:
Conservation of energy in the COM frame gives:
in the COM frame, then
Moving back to LAB frame, the velocities of the scattered particles are just the
COM velocities
If the particles scatter with angle
The angle in the LAB frame can be then found as:
where
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A uniform hoop with mass M, outer radius R and inner radius R/2 is initially
spinning with angular velocity 0. The hoop is placed onto a uniform disk, also
of mass M and outer radius R, that is initially at rest on a fixed frictionless table. The hoop and disk are aligned along their centers of mass. There is a coefficient of friction between the hoop and disk where their surfaces make contact. Gravitational acceleration is assumed to be constant and points downward.
(a) [10 pts] What is the initial torque (about its center of mass) acting on the disk? What is the initial torque (about its center of mass) acting on the hoop?
(b) [5 pts] The friction force between the hoop and disk cause them to eventually spin at the same angular velocity. What is this angular velocity?
(c) [5 pts] How much time does it take for the disk and hoop to reach this common angular velocity?
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SOLUTIONS
(a) The torque on the hoop from the disk arises from friction , where is the normal contact force between the disk and hoop. First consider a small section where the hoop and disk make contact.
where areal density.
is a small mass segment on the hoop with uniform The total torque is an integral over the area of the loop:
but the mass of the hoop is:
Hence the torque on the hoop as the result of frictional forces from the disk is
Following Newton’s third law, the frictional force on the disk from the hoop is equal and opposite to that of the hoop from the disk, and the force acts over the same surface area. Hence, the torques are equal and opposite:
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again, substituting the value from part (b) gives:
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Problem 5: Spring-loaded Tube [20 pts]
A block of mass M and velocity v 0 is moving through a fixed tube with frictionless
walls. At the bottom of a loop of radius R, the block collides perfectly inelastically (i.e., it sticks) with a plunger of mass M connected to a long fixed spring with spring constant k and equilibrium length πR. As the plunger and block travel
around the loop, they are also acted on by a drag force ; i.e., a force that acts opposite to the direction of motion. Gravitational acceleration acts downward. Assume that the dimensions of the block and plunger are much smaller than R, so that their rotation about their centers of mass are negligible.
(a) [10 pts] What is the incoming block’s velocity if the plunger and block come to rest at = π/2?
(b) [10 pts] Continuing from (a), after the block/plunger pair stop at = π/2, they swing back around the loop and out into the straight, frictionless linear tube. If there are no adhesive forces that keep the block and plunger bound to each other, at what point does the block move away from the plunger and with what speed?
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of both objects at that point. At the bottom of the tube, U = 0 and the total mechanical energy lost is simply twice the work done by the drag force on the way up:
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