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Problem 1: Quick Short Answer Problems [10 pts]
Answer all five problems. You do not need to show any work. Be sure to record your answers in your solution book, NOT on this exam.
(a) [2 pts] A bicycle rider pedals up a hill with constant velocity v. In which direction does friction act on the wheels?
Friction does not act while the
Uphill Downhill
bike moves at constant velocity
(b) [2 pts] An 8.012 student pushes a heavy object up a hill, and is prevented from slipping by friction between her shoes and the surface of the hill. While she is walking up, she picks up one of her feet. Will she be:
Neither, as there is no change
in the friction force acting
More likely to slip Less likely to slip
(c) [2 pts] A ball attached to rope is twirled around a stick as shown in the diagram at right. Ignore gravity and friction. Which of the following quantities is conserved in the motion of the ball? Be sure to write down all of the choices below that apply.
Angular None of these are
Energy Momentum
Momentum conserved
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(d) [2 pts] (Challenging) Consider two uniform disks of mass M, radius R and negligible thickness, connected by a thin, uniform rod of mass M. The centers of the disks are separated by a distance 4R. Reproduce the diagram at right in your solution book and draw the principle axes (^) 4R of this object centered at its center of mass [1 pt], indicating the axis about which torque-free rotations are unstable [1 pt]. Note: you do not need to calculate the moment of inertia tensor to solve this problem.
unstable
axis
⊗
North
(e) [2 pts] A car at latitude λ on a rotating Earth drives straight North with constant velocity v as indicated in the diagram. In (^) West
which direction does friction between the East
car’s tires and the road act on the car to ⊗ counteract the Coriolis force on the car?
R^ R M M
M
λ
East West Coriolis force does not act on the car
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SOLUTION
x
L/6 COM
(a) The motion of the bar involves both translation of its center of mass
(at distance L/6 from the pivot point) and rotation about the center of
mass. This motion is determined by both gravitational force and the
constraint forces at the pivot. Assume that there are both horizontal (Fx)
and vertical (Fz) components of the constraint force acting on the
swinging bar (the forces the bar exerts on the fixed post will be equal
and opposite these forces). The figure above shows the force diagram.
Because the motion is rotational, use the polar coordinates r and θ
centered at the pivot point to describe the translational motion, where at
the instant that the bar is horizontal the positive r and θ directions
correspond to x and – z in the prescribed rectangular frame. θ is also the
rotational coordinate of the bar about its center of mass, and points in the
same direction as the translational motion (i.e. the bar rotates and swings
in the same direction).
The equations of motion are (the coordinate r = L/6 is constant):
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Combining the last two equations yields:
To solve for Fx, we can use conservation of mechanical energy since
only a conservative force (gravity) is doing work on the bar. The kinetic
energy of the bar (translation of center of mass and rotation about center
of mass) is drawn from its gravitational potential energy when it starts at
vertical and its center of mass if L/6 higher:
then using the r equation of motion:
The force that the swinging bar exerts on the fixed post is therefore:
Note that we could have also derived the rotation and energy equations
as pure rotation about the pivot. In this case the rotational equation of
motion, with torque coming from gravity acting at the center of mass,
becomes:
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Problem 3: Cue Ball Spin [15 pts]
M
μ
∆p
α
R
A billiards player strikes a cue ball (a uniform sphere with mass M and radius R) with a cue stick at the middle of the ball (i.e., at a height R above the table) and at an angle α with respect to horizontal. The strike imparts an impulse ∆p on the ball in the direction of the strike, causing it to move toward the right as well as “backspin” – spin in the direction opposite of rolling motion. The coefficient of kinetic friction between the ball and table surface is μ. Assume that the ball does not rebound off of the table after the strike, and that constant gravitational acceleration g acts downwards.
(a) [5 pts] What is the initial speed and angular rotation rate of the cue ball after it is struck?
(b) [5 pts] For what angle α will the ball eventually come to rest?
(c) [5 pts] For the case of part (b), how far does the ball travel before it comes to rest?
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SOLUTION
∆p
α
Mg
N
Ff
(a) There are two solutions for this part based on the ambiguity of
whether ∆p is the impulse imparted onto the ball or delivered by the
stick. Both solutions were accepted. In the former (assumed) case, the
horizontal translational momentum of the ball arises purely from the
horizontal impulse, so that:
The vertical impulse imparts no vertical translation because of the
normal force from the floor; however, it does impart an angular impulse
on the ball:
Alternately, if we consider ∆p to be purely the impulse imparted by the
stick over a short time ∆t, then one needs to also consider the other
forces acting on the ball during that time, in particular friction between
the ball and the floor. The friction force acting on the ball during the
time ∆t is:
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(c) The distance traveled by the ball in this case is given simply by our
usual expression for ballistic motion with constant acceleration:
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Problem 4: The Accelerated Atwood Machine [10 pts]
R
M
M
M
An Atwood machine consists of a massive pulley (a uniform circular disk of mass M and radius R) connecting two blocks of masses M and M/2. Assume that the string connecting the two blocks has negligible mass and does not slip as it rolls with the pulley wheel. The Atwood machine is accelerated upward at an acceleration rate A. Constant gravitational acceleration g acts downward.
(a) [8 pts] Compute the net acceleration of the two blocks in an inertial frame of reference in terms of g and A. Do not assume that tension along the entire string is constant.
(b) [2 pts] For what value of A does the block of mass M remain stationary in an inertial frame?
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To determined the accelerations in the inertial frame, add the system
acceleration A to the accelerations of the blocks in the accelerated
frame:
These are downward accelerations, so the block of mass M/2 actually
accelerates upwards in the inertial frame.
(b) The first mass will remain stationary in the inertial frame if
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Problem 5: What is the Best Way to Move a Heavy Load up a Hill? [15 pts]
2 α
μ 1
2M
M
μ 2
2 α
μ 1
2M
M
μ 2
A B
Two students, each of mass M, are attempting to push a block of mass 2M up a symmetric triangular hill with opening angle 2α. Student A pushes the load straight up; student B pulls the load up by running a massless rope through a massless, frictionless pulley at the top of the hill, and pulling on the rope from the other side. The maximum coefficient of friction (assumed here to be equal to the coefficient of kinetic friction) is μ 1 between the students’ shoes and the hill, and μ 2 between the block and the hill. Assume μ 1 > 2μ 2. Constant gravitational acceleration g acts downwards.
(a) [5 pts] For what minimum angle αmin (i.e., maximum steepness) does neither student need to apply any force to hold the load in place?
(b) [5 pts] Calculate and compare the forces each student must exert on the block to move it up the hill at constant velocity. Does either student have an advantage here?
(c) [5 pts] Calculate and compare the minimum angles α < αmin that each student is able to move the block up the hill at constant velocity without their shoes slipping on the hill surface. Does either student have an advantage here?
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Neither student has an advantage in this case.
(c) The force diagrams on each student are shown in diagrams (c) and
(d) above, for the critical case where they almost slip. The force acting
on the students is equal and opposite to the force the students exert to
pull the blocks because the string is massless. For the first student, the
equations of motion in the z and x directions are:
(Student 1)
For the second student, the equations of motion in the z direction is the
same and in the x direction:
(Student 2)
This angle is smaller than that for the first student, so Student 2 can pull
up a block on a steeper incline than Student 1 can without slipping.
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Problem 6: Ball Rolling in a Bowl [15 pts]
θ
L
M R
μ
A solid uniform ball (a sphere) of mass M and radius R rolls in a bowl that has a radius of curvature L, where L > R. Assume that the ball rolls without slipping, and that constant gravitational acceleration g points downward.
(a) [5 pts] Derive a single equation of motion in terms of the coordinate θ (the position angle of the ball with respect to vertical) that takes into account both translational and rotational motion, for any point along the ball’s trajectory. Be careful with your constraint equation!
(b) [5 pts] Find the position angle of the ball along the bowl’s surface as a function of time in the case that θ is small. Assume that the ball is started from rest at a position angle θ 0.
(c) [5 pts] At what maximum initial position angle θ 0 can the ball be placed and released at rest and still satisfy the rolling without slipping condition throughout its motion? Note that θ 0 does not have to be small in this case.
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Note that the fixed radial coordinate r of the center of mass is (L-R), not
L. The constraint equation is:
We can use these equations to solve for the friction force Ff:
which gives a single equation for θ:
This solution could also be found by appealing to energy conservation,
since in the case of rolling without slipping there are no dissipative
forces acting. Taking the zero point for gravitational potential energy to
be at the bottom of the bowl, the total mechanical energy at any point
along the bowl is (using the constraint equation):
The change in energy with time is 0, hence:
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(b) For small angles θ, the equation of motion reduces to:
which is the equation for simple harmonic motion with frequency
The general solution of this equation is:
with the initial conditions
which gives:
(c) There are a number of ways to derive the solution for this part, but
consider that the condition for rolling without slipping implies that Ff ≤
μN at every point along the motion. The maximum angular
acceleration due to gravity, and hence the maximum slippage between
the ball and bowl surfaces, occurs at the top of the roll at which point
= 0. From the radial equation of motion:
and from the expression for Ff above we get:
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