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Problem 1: Quick Multiple Choice Questions [10 pts]
For each of the following questions circle the correct answer. You do not need to show any work.
(a) A block with mass M and contact area A slides down an inclined plane with friction, covering a distance L in time T. How much time does it take another block with the same mass and composition, but twice the surface area, to slide down the same length?
0.5T T 2T T^2
(b) A particle moves with constant speed from point A to point E on the path shown to the right. Among the points B, C and D, circle the point where the magnitude of acceleration is greatest.
(c) A tire rolls on a flat surface with constant angular velocity Ω and velocity as shown in the diagram to the right. If V = ΩR, then in which direction does friction from the road act on the tire?
Toward the left Toward the right Friction does not act on the tire
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Less Greater The same Zero
(d) You pull a brick of mass M sitting on a flat table using a thick rope of mass m (see picture to right). Friction between the surface of the table and the brick is sufficient to prevent the brick from moving. Compared to the force you pull one end of the rope, the force that the other end of the rope exerts on the brick is:
(e) As a swinging pendulum passes through its equilibrium point, in which direction does the total net force act?
Radially outward Radially inward There is zero net force
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SOLUTION
Fig 5a
(a) The geometry of the problem and force diagrams on for the blocks are shown above. Note that because the pulley is accelerating, we must keep track of its position in the inertial, one-dimensional coordinate system indicated above. Friction acts in the negative x direction and is simply equal to the normal force of the blocks on the surface of the table (their weight) time μ. The tension force from the string is the same for both blocks since the string is massless. Hence, the equations of motion for the two blocks are:
The constraint equation is the length of the string, L, which must be constant. The string length can be computed as (R is the radius of the pulley wheel):
Taking the equations of motion, dividing by the lead mass terms and adding the equations together gives us:
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And hence:
These are constant acceleration terms, so the positions of the two masses is readily determined (assuming x 1 (0) = x 2 (0) = 0 and that the blocks are initially at rest):
This is only part of the solution, since if A is insufficient to accelerate the larger block (see next part), then x 2 (t) = 0
(b) Based on the equations above, the larger mass will not move if:
The second mass cannot move backward – the friction forces exactly matches the tension from the string for A < μg/2.
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SOLUTION
Fig 2a
(a) We break this problem up by considering the gravitational force acting on the block from one part of the disk dM, and then integrating over the entire disk. From the symmetry of the problem, we need only worry about the vertical component of the gravitational force. The geometry of the problem is illustrated above. The vertical component of force acting from dM is:
Now make the substitution integral can then be solved as:
and hence , then
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The problem asks to solve this problem to first order in ∆h so we can Taylor expand the solution to the integral as:
Hence, the z-component of gravitation force acting on the mass is, to first order in ∆h,
Which is the equation of simple harmonic motion with period
(b) The force exerted on the block by the disk must be equal and opposite to the force exerted on the disk by the block; hence:
The disk should have the same oscillation period as the block as it is always pulled back toward the block when the block is on either side. There is no way for the oscillations to be “out of phase” since there is no third body to induce a new frequency. The equation above does not quite look like it is going to give the same frequency. However, the variables zdisk and ∆h are not the same, describing the offset of the disk with respect to some fixed midplane and the offset of the block
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Problem 4: Rotating Spring [20 pts]
A cylindrical mass M is placed on a post connected to a rotating shaft. The post forces the mass to rotate with the shaft at constant angular velocity Ω. The mass is connected to the shaft by a spring with spring constant k and rest length L. Ignore friction and gravity.
(a) [5 pts] What is the equilibrium distance of the mass from the central shaft?
(b) [10 pts] At time t=0, the mass is struck, giving it an inward radial velocity VR. Derive the period of oscillation of the cylindrical mass and an expression for its distance from the central shaft as a function of time t.
(c) [5 pts] What happens when Ω^2 > k/M?
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SOLUTION
Fig 3a
(a) Because of the constraint of the post, we are only interested in radial motion. The only identifiable force acting on the system is the restoring spring force. Hence, writing the equations down in radial coordinates, and assuming r is fixed at req,
Make the substitution
(b) As r is no longer fixed, the equations of motion become:
and hence
This is the general expression for simple harmonic motion where
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Problem 5: The Flyball Governor [25 pts]
A flyball governor is a device commonly used in steam engines to control the flow of steam. In the simplified version shown above, a rotating shaft is connected to two hinges of mass M through rigid, massless rods of length L. The rods are also attached at the bottom of the device to a larger block of mass 3M which can slide freely up and down the shaft (in a practical flyball governor, this mass would be attached to a valve closer to reduce fuel flow). The shaft rotates at constant angular rate ω. Assume constant gravitational acceleration g acting downward and ignore friction and viscosity effects.
(a) [20 pts] Derive, as a function of ω, the height of the large block above its non- rotating position.
(b) [5 pts] Derive an expression relating the vertical speed of the block to the time rate change in angular rotation rate of the shaft, and confirm that the block moves upward as the shaft spins faster.
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SOLUTION
Fig 5a
(a) Force diagrams for each of the hinges are shown above. Assume that the angle between the rods and the vertical is α, so that the height of the block is:
and the radial distance of the hinges from the central shaft is
We adopt a cylindrical coordinate system, ignoring the direction. For each hinge the equations of motion are identical:
The unknown variable T’ can be found by dividing both equations though by the appropriate trignometric function of α and adding them together, yielding:
We are only concerned about the z component of motion for the block, and this equation of motion is:
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USEFUL EQUATIONS
Trajectory for constant acceleration
Velocity in polar coordinates
Acceleration in polar coordinates
Taylor expansion of function f(x):
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