Good morning. Today we are going to solve free response question #1 from the 2016 AP
Physics 1 exam. Bobby, please read the problem through part a i.
♪ Flipping Physics ♪
A wooden wheel of mass M, consisting of a rim with spokes, rolls down a ramp that makes
an angle theta with the horizontal, as shown. The ramp exerts a force of static friction
on the wheel so that the wheel rolls without slipping. Part (a) i, On the diagram,
draw and label the forces (not components) that act on the wheel as it rolls down the
ramp, which is indicated by the dashed line. To clearly indicate at which point on the
wheel each force is exerted, draw each force as a distinct arrow starting on, and pointing
away from, the point at which the force is exerted. The lengths of the arrows need not
indicate the relative magnitudes of the forces. - Thank you Bobby. Bo, please answer the question.
Sure. The wheel has mass, so it has a force of gravity acting down on it at the wheel's
center of mass. The wheel is touching the ramp, so there is a force normal acting on
the wheel at the point of contact between the wheel and the incline. The direction of
the normal force is perpendicular to the incline and up.
Therefore, it points toward the center of mass of the wheel.
It states in the problem that there
is a force of static friction on the wheel, so we know the wheel is rolling without slipping.
The direction of the force of static friction has to be parallel to the ramp, so it is either
up or down the incline. If there were no friction, the wheel would slide down the incline, so
the static friction opposes that motion, which means the force of static friction is parallel
to and up the incline and acts at the same location as the force normal, the point of
contact between the wheel and the ramp. - Thank you Bo. Now, some notes about how this
problem was graded by The CollegeBoard. This problem, problem #1, is worth 7 points.
2 of those points are just for getting this free body diagram correct. Yes, roughly 30%
of the points from this problem come from this free body diagram or force diagram. Hopefully
this helps you to understand how important free body diagrams are. Also, when drawing
free body diagrams all forces need a clear indication of where they start and their direction.
If you are terrible at drawing, you can always clarify force start locations and directions
using words. Billy please read part (a)
double i and give your answer. Absolutely. Part (a) double i says As the
wheel rolls down the ramp, which force causes a change in the angular velocity of the wheel
with respect to its center of mass? Briefly explain your reasoning. Okay, well, angular
acceleration equals change in angular velocity over change in time, therefore, change in
angular velocity is caused by angular acceleration. And the rotational form of Newton's Second
Law states that net torque equals rotational inertia times angular acceleration where both
torque and angular acceleration are vectors. Therefore, the angular acceleration of the
wheel is caused by a net torque on the wheel. So, the force which causes a net torque about
the center of mass of the wheel is the one which causes the change in angular velocity
of the wheel. Torque equals rFsine Theta where r is the vector which goes from the axis of
rotation to the location of the force causing the torque, F is the force causing the torque,
and theta is the angle between r and F. The r value for the force of gravity is zero,
so the force of gravity does not cause torque on the wheel about its center of mass. Theta
for the force normal is 180 degrees and the sine of 180 degrees equals zero, so the force
normal causes zero torque about the center of mass of the wheel. So, the only force causing
a torque on the wheel about its center of mass is the force of static friction. So,
the force of static friction is the force which causes a change in angular velocity
of the wheel about its center of mass. - Thank you Billy. I'll take part (b). Part
(b) says, For this ramp angle, the force of friction exerted on the wheel is less than
the maximum possible static friction force. Instead, the magnitude of the force of static
friction exerted on the wheel is 40 percent of the magnitude of the force or force component
directed opposite to the force of friction. Derive an expression for the linear acceleration
of the wheel's center of mass in terms of M, theta, and physical constants, as appropriate.
Okay, the force component directed opposite to the force of friction is the component
of the force of gravity which acts parallel to and down the incline, or the force of gravity
parallel. That means the force of static friction equals 0.4 times the force of gravity parallel
or 0.4 times the mass of the wheel times the acceleration due to gravity times the sine
of the incline angle. I do want to take a moment to point out that you should redraw
your free body diagram to include the components of the force of gravity and not the force
of gravity itself. However, a word of caution that your free body diagram is one of your
answers, so you need to not re-draw that free body diagram over your original answer, if
you do so, you will lose points for your free body diagram. So please, be careful of that.
Now we can sum the forces in the parallel direction to solve for the acceleration of
the center of mass of the wheel in the parallel direction. Defining down the incline as positive,
we get, net force parallel to the incline equals force of gravity parallel minus force of static
friction which equals mass times acceleration of the center of the mass of the wheel in
the parallel direction, which is what we are solving for. Substituting in equations gives
us mass times acceleration due to gravity times sine theta minus 0.4 times mass times
acceleration due to gravity times sine theta equals mass times acceleration parallel. And
Everybody brought mass to the party!
Exactly. Everybody brought mass to the party.
That means the acceleration of the center of mass of the wheel down the incline equals
0.6 times acceleration due to gravity times the sine of the incline angle. Another note
about grading, please make sure you understand what this line means from the scoring guideline:
The expression need not be correct or consistent with the force diagram in part (a). This means
you did not have to get part (a) correct to get points for part (b). As long as your solution
to part (b) is correct, you still get the points, even if it based on an incorrect solution
to part (a). So, please, always answer every question, regardless of whether or not you
think the rest of your solution is correct. Also, please make sure your answer is in terms
of the variables provided. In this problem that is the mass of the wheel, the incline
angle, and any physical constants. If your answer has other variables like the coefficient
of static friction in it for example, that answer is incorrect. Oh, and no need to substitute
in a number for the acceleration due to gravity because the acceleration due to gravity is
a fundamental constant. Okay, now Bo, could you please read all of part (c), and Bobby,
please answer part i and Billy, please answer part double i.
Sure. Part (c) In a second experiment on the same ramp, a block of ice, also with mass
M, is released from rest at the same instant the wheel is released from rest, and from
the same height. The block slides down the ramp with negligible friction. Part i. Which
object, if either, reaches the bottom of the ramp with the greatest speed? The wheel, the
block, or neither; both reach the bottom with the same speed. Briefly explain your answer,
reasoning in terms of forces. And part double i says, Briefly explain your answer again,
now reasoning in terms of energy. - Alright, uh. For part i, the block has no
force of friction acting on it, therefore, the net force in the parallel direction on
the block has a larger magnitude and therefore has a larger acceleration down the incline.
A larger acceleration on the block will result in a larger speed on the block at the bottom
of the incline. So, the block has a larger speed at the bottom of the incline.
For part double i, both the block and the wheel start with the same amount of mechanical
energy and it is all gravitational potential energy. Gravitational potential energy is
converted to kinetic energy as the objects go down the ramp. The block is not rotating;
therefore, the block will not have any rotational kinetic energy as it slides down the incline,
therefore the block will have more translational kinetic energy than the wheel, because some
of the kinetic energy of the wheel has to be rotational kinetic energy, therefore, the
block will have a larger speed at the base of the ramp.
Correct. Very nice everybody. One last note about grading. Both (ci) and (cii) are worth
one point and state, No credit for answer without explanation. In other words, anytime a question
asks you to explain your answer you absolutely must do so! Not explaining your answer will
most likely get you zero points. And one quick final note. The College Board expects full
sentences, not sentence fragments like what is currently on the screen. I used sentence
fragments to make it easier to fit everything on the screen. But, maybe it would have been
better to write everything out instead.
Thank you very much for learning with me today, I enjoyed learning with you.