A solid spherical marble shot up a frictionless 15° slope rolls 2...

Question

A solid spherical marble shot up a frictionless 15° slope rolls 2.50 m to its highest point. If the marble is shot with the same speed up a slightly rough 15° slope, it rolls only 2.30 m. What is the coefficient of rolling friction on the second slope?

Accepted Answer

Hey, everyone. Let's go through this practice problem. A cylindrical cell is given an initial speed of V knot at the bottom of a smooth ramp that makes an angle of 20 degrees with the horizontal. The cell rolls up 1.3 m without slipping before stopping briefly and rolling down, the roughness of the ramp is increased and the cell is given the same initial speed of V knot at the bottom of the ramp. The cell rolls up and without slipping a distance of 1. m, calculate the coefficient of rolling friction of the rough ramp. You have four, multiple choice options. Option A 40.66, option B 0.40, option C 0.43 and option D 0.55. Now, this is a pretty tricky and kind of time consuming problem because there are two different steps. We're analyzing the case of the smooth ramp and the case of the rough ramp, it's necessary to analyze both cases in order to find the, the coefficient for the rough ramp. This is because analyzing the ideal case of the smooth ramp makes it easier for us to discover some of the information that we need in order to solve for the variables regarding the rough ramp. So for that reason, the first step I'm going to take is analyzing the smooth ramp. So I'm just gonna kind of draw a very simple diagram here of what the smooth ramp might look like. And here is the angle theta and the cell is going to move some distance up the ramp that I'm going to label as D sub smooth or D sub s. More simply you make that a little neater. And here are the equations I'm about to write for the smooth case. Now, one thing we can be sure of right off the bat is that we can expect the movement of the cell to follow the law of conservation of energy, which states that in an isolated system, such as this one, the change in mechanical energy is going to be zero. So whatever the amount of energy in the system is before the cell slides up, the ramp is going to stay exactly the same when the cell reaches the top of its motion. In other words, the initial kinetic energy plus the initial potential energy is equal to the final kinetic energy plus the final potential energy, some of these energies may be converted between each other, but the total amount will stay the same. And just to make sure I'm being clear here when I'm saying initial, I'm referring to just the moment when the cell hits the bottom of the ramp. And when I say final, I'm referring to when the cell reaches the top of its motion, when it stops briefly and is at rest. In order for this equation to be more useful to us, I'm going to expand out the terms into the different formulas for kinetic and potential energy. Another way to write this is two is the the initial kinetic energy plus the initial potential energy. And recall that potential energy has a formula. Potential energy is written as the mass multiplied by the gravitational acceleration multiplied by the vertical position of the point we're analyzing. So for our purposes for the initial potential energy for convenience, I'm going to define the point at the bottom of the ramp to be at the point of Y equals zero, which means that the initial potential energy for the cell is its mass multiplied by, by the gravitational acceleration multiplied by its height at that point which is just zero m. So this term cancels out because it goes to zero, this is equal to its final kinetic energy. And recall that kinetic energy has a formula of one half multiplied by the mass of the particle multiplied by the square of its speed. But once again, we can actually simplify that term out with this equation because its final speed when it reaches the top of its motion is going to be zero because the problem even mentions that it stops briefly when it reaches that height before it starts accelerating the opposite direction. So one half M zero m per second squared. So this term also cancels out and goes to zero. So the only thing left on the right hand side of the energy equation is the final potential energy U sub F. Or if we want to rewrite that using our more detailed equations as the mass of the cell multiplied by the gravitational acceleration, G multiplied by its final height, which we can see from the geometry of our diagram is going to be D sub S, the distance it travels along the hypotenuse of the triangle multiplied by the sign of the angle theta. And this is just easy to see from the trigonometry of the situation because the sign of, of a theta in a right triangle is equal to the opposite side or the height divided by the hypotenuse of the triangle, which for our purposes is the D sub S. So most simply our equation is that the initial kinetic energy for the smooth case is equal to the mass multiplied by the gravitational acceleration multiplied by the distance traveled along the smooth ramp multiplied by the sign of theta. Now let's discuss the case for the rough ramp after the roughness of the ramp has increased, this part is going to be a little more complicated because now there's a whole additional force that we're considering acting on the cell. So instead of just drawing a simple basic diagram of a ramp like I did for the smooth case, I'm going to take a step further and draw as more of a force diagram. So we can see the different forces that are acting on the cell. When it's reached some position on the ramp, we're going to have the exact same angle theta. And let's say that we're looking at some case, I'm picking out this point to be where the cell is at some moment in time. And we're analyzing the different forces acting on this point. So of course, there's the downward force from gravity from its weight which has a magnitude of M G. There's also going to be a normal force perpendicular to the surface of the ramp, that's the normal force. And, and assuming that right now, I've drawn a point where the cell is moving up the ramp. That means that there's going to be an opposing force in the opposite direction, pointing down the ramp, which is the frictional force. So this is starting to look like the way our our ramp problems tend to look. So let's set up the coordinate system in a similar way where the x axis is parallel to the surface of the ramp itself and the y axis is perpendicular to the ramp. So we've got kind of a diagonal coordinate system going on here like so now kind of like what we did in the smooth case, let's set up an energy equation except now that we have friction to worry about a non conservative force. We have to consider how that non con conservative force of the rolling friction will play into it. So the initial kinetic energy of the cell for the rough case plus the initial potential energy for the rough case, which is actually going to go to zero. For the exact same reason that we discussed in the smooth case where if we want to do things that are convenient and helpful for us, we'd set this, we'd set the base of the ramp to be the vertical position where Y is equal to zero. So the U sub Y so the U of I term, the initial potential energy term just goes to zero again. And then on the right hand side of the equation, whatever the sum of our initial potential energies are, should be equal to the sum of the final energies. So it should be equal to the final kinetic energy, which again is going to go to zero. For the same reason that we discussed in the smooth case where once it reaches its peak, it's going to be briefly at rest for a moment before it switches direction and then just plus the final potential energy. But there's still a missing piece here because we haven't yet accounted for the non conservative rolling friction force. It is because of this friction that uh it's because of this frictional work acting against the cell that the cell isn't going to be able to reach as highly as it was able to reach before. So I'm going to add a work from the rolling friction W sub roll into the equation. And we can kind of logically expect it to be negative because it's going to effectively decrease the value of the energy in order to lower the total value of the final energy that we would have otherwise expected. So I'm gonna more cleanly write this as case of I R plus W sub roll equals U sub F that said instead of writing U sub F, I'm going to expand it out into the, into similar terms as the one we used in the smooth case to expand out the final potential energy in that case. So I'm going to expand it using the standard gravitational potential energy formula. So I'm gonna write M G multiplied by its height which again, for similar reasons is going to be the distance it travels, which in this case is the is I'm gonna call it D sub R the distance along the ramp it travels in the rough case multiplied by the sign of the same angle theta before we move on though it needs to be pointed out that we can't do much more math with this form with these formulas until we have some way to quantify W sub roll. The work from the rolling friction. So I'm going to quantify that now. So this is work. So work is going is equal to the force multiplied by a displacement. So if we say that the displacement is traveling up of the ramp and the friction is acting against it, then it's easy for us to see that these two forces the displacement and the friction are acting in opposite directions. That is part of the definition of friction. After all, it opposes motion perp like Tati at a 1 80 degree, which means that the ma which means that the work is actually going to be negative. So W sub roll is going to be negative of the frictional force multiplied by the displacement D sub R in order to go even further with this, let's recall, but the frictional force is equal to the coefficient of friction. In this case, the coefficient of rolling friction multiplied by the magnitude of the normal force from our diagram, we can see that the normal force is perpendicular to the ramp as it tends to be. So our W sub roll formula can be expanded out as negative. And then instead of F sub R, I'm going to write the coefficient of friction mu sub R multiplied by M G multiplied by the cosine of theta multiplied by the D sub R displacement. And so this is something that we need can be substituted into our energy equation for W sub roll. Now, the final step is to relate all these equations to one another. And we can do this using the fact the problem tells us that in both cases, in both the smooth case and the rough case, the cell is given the same initial speed. And if it's given the same initial speed and its mass and its mass presumably doesn't change, then that means that the initial kinetic energy in both cases are equal. So in other words, K sub I S, the initial kinetic energy for the smooth case is equal to the initial kinetic energy for the rough case a sub I R. So now we can expand this equation out by plugging in substituting in the other expressions we found we discussed much earlier on case of I S is equal to M G D sub S multiplied by the sign of theta. And this is equal to K sub I sub R. From looking at the energy equation we deduced in the rough section, this should be equal two M G D sub R sine theta minus the W sub roll which he wrote on the side is negative mu sub R M G cosine theta D sub R. Now let's do some simplifying all three terms have an M for the mass. So that cancels out. So the math isn't the mass is inconsequential and so is the G the gravitational acceleration? Also worth noting that on the right hand side of the equation where we have the work, we have the work term, we have a minus and then a negative term. So these can cancel out. So there's actually addition, taking place here, the, the new equation now is D sub s multiplied by the sign of theta is equal to D sub R multiplied by the sign of the plus mu sub R multiplied by the cosine of theta multiplied by D sub R. Now remember our ultimate goal here is to solve for mu sub R. So I'm going to algebraically get the one term that contains a mu sub R on its own. So to do that, I'm going to subtract from both sides of the equation. The D sub R sine theta term. And what we're left with on one hand, on one side of the equation is mu sub R multiplied by the cosine of theta. D sub R is equal to D sub s multiplied by the sine theta minus D sub R multiplied by the sine theta. We can simplify this further by factoring out the sine theta term and then multiplying in parentheses. D sub S minus D sub R. Finally, we can solve from UU R itself by dividing both sides of the equation by cosine data D R. So new sub sub R is equal to D sub S minus D sub R divided by D sub R all multiplied by the sign of theta divided by the cosine of theta. And if you want to simplify this even further, the sin of theta divided by the cosine of theta is just the tangent of theta. So the final equation we're left with is D sub S minus D sub R divided by D sub R multiplied by the tangent of the angle theta. And this is the final equation that we'll want to use. So all that's for us to do is plug in the values we were given in the problem. So D sub S is again the distance that the cell rolled up in the smooth case, which the problem tells us is 1.3 m. D sub R is the distance the cell moved in the rough case. So that's 1.1 m divided by D sub R. So once again, 1.1 m multiplied by the tangent of the angle of inclination, which the problem tells us is 20 degrees. When you put this into your calculator, make sure your calculator is in degree degrees mode so that it can compute the tangent properly. When we put this into a calculator, we find a new sub R A rolling frictional coefficient of about 0.66. So that is the coefficient of rolling friction. And if we look at our multiple choice options, we can see that disagrees with option choice number a letter A. So option A is the correct answer to this problem. And that is it for this problem. That's it for this video. I hope it helped you out if you need more practice. Please consider checking out some of our other videos which will give you more experience with these types of problems, but that's all for now. I hope you all have a lovely day. Bye bye.

A solid spherical marble shot up a frictionless 15° slope rolls 2.50 m to its highest point. If the marble is shot with the same speed up a slightly rough 15° slope, it rolls only 2.30 m. What is the coefficient of rolling friction on the second slope?

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