two-objects-a-sphere-and-a-block-of-the-same-mass-are-released-from-rest-at-the-top-of-an-inclined-p

Question

Two objects, a sphere and a block of the same mass, are released from rest at the top of an inclined plane. The sphere rolls down the inclined plane without slipping. The block slides down the plane without friction. Which object reaches the bottom of the ramp first?
(A) The sphere, because it gains rotational kinetic energy, but the block does not
(B) The sphere, because it gains mechanical energy due to the torque exerted on it, but the block does not
(C) The block, because it does not lose mechanical energy due to friction, but the sphere does
(D) The block, because it does not gain rotational kinetic energy, but the sphere does

Vishal Gupta · Accepted Answer

Hi in the given problem theories. Block and us here both are allowed to fall down an inclined plane, a smooth inclined plane at the same angle and same distance and same vertical height I shown here in this figure envelope, inclination, theater height edge for both of them whether it is a block or it is a solid sphere. Now we have to find out of these two which one will reach the ground earlier. S the block will reach the ground without sleeping means it will roll down, it will rotate about its center of mass so while rolling down the incline. Their sphere will have our rotational kinetic energy also and of course, obtained from the initial gravitational potential energy. But as the block does not revolve while sliding down so wall of its gravitational potential energy will be converted into translational kinetic energy only, so it will reach the ground earlier. Hence you can see. Yeah, our option B is correct. Thank you.

Kristela Garcia · Answer

in this problem. You have two spheres, one hollow, one solid, that are being rolled down the incline and were asked to look at a free body diagram and which one reaches the end of the incline first. So looking at our two spears, we can draw everybody diagram. We&#x27;re gonna drop this from a to D perspective. And so we have a normal force coming from the incline. We have a force of gravity coming from the center of mass of the spheres. Whether it be hollow or solid, the center of mass would still be acting on Thesent er and then you have friction. Friction is in between the two surfaces going up the room. So in this case, um, you can notice that normal force goes through the center of Mass. Oh, my normal force will not exert a torque. My force of gravity also goes through the center of Mass. Oh, my force of gravity will also not exert a torque. So the only thing is causing rotation is my force of friction here, Part B ouse us to determine which reaches the bottom of the inclined plane first. Well, in that case, we need you need to look at the moment of inertia is for these two objects. Now, even if you don&#x27;t know the values of the moment of inertia, it&#x27;s important to know that your moment of inertia I has to do with the way the Mass is being distributed to win right mouse distribution. So because the Hollow one has a has more mass towards its center, the hollow will have a greater moment of inertia than the solid. So the solid one has Maur if it&#x27;s masked closer to the center and the hollow one has more of its mess away from the center. So I&#x27;m gonna write Hollow and I&#x27;m sorry this was hollow and then the other one is solid, so the hollow has a greater moment of inertia than the solid one. So if we look at a torques equal moment of inertia times the angular acceleration, this is the sum of the torques. This is Newton&#x27;s second law for rotation, so we have the sum of the forces equal to mass times acceleration. This is Newton&#x27;s second, and then this is Newton&#x27;s second law for rotation. Well, if friction is exerting the same torque on both situations, the one with the greater moment of inertia is going to have a smaller, angular acceleration, which tells us that the solid sphere will reach the bottom faster or sooner. Now if we look at Part C is asking about the kinetic energies Well, because it&#x27;s not being specific about which type of kinetic energy then we can assume it&#x27;s talking about total kinetic energy. So in terms of kinetic energy, they both have the same amount of potential energy at the top, and this is in terms of gravitational potential energy. And at the bottom you&#x27;re gonna have two types. You will have translational, kinetic energy and rotational kinetic energy. But since they both had the same amount of potential at the top, they will have this total amount be the same at the bottom.

Ben Nicholson · Answer

problem. 11.4. We have a solid sphere rolling down it. And if we want it center of mass to have a linear acceleration of 0.1 g, the find the angle that does that. And then we want to answer whether a frictionless block sliding down is going to have a greater acceleration or less or equal to or whatever and wine. So for a, uh, sphere, its moment of inertia is 2/5 and R squared. And so an equation 11.10 from the chapter. The acceleration of center mass is G sign they, uh, over one plus the moment of inertia divided by that My square. Now we want this to be equal to 1/10. Huh? G putting in our 2/5 m r squared for here we get. This is equal to there&#x27;s a negative side. Z equals negative g sign they divided by 7/5. We want this to be equal to negative 0.1 g does. This is negative because of the coordinates that were used in driving equation 11.10 or 11 dash whatever. And so then solving this for sign of data taking the dark side, we get that it needs to be inclined at 8.0 degrees. Be so they&#x27;re a couple ways of thinking about what happens with the friction this block sliding sliding down here. Probably the the cleanest sort of, you know, most easiest conceptual way, in my opinion, is, uh, thinking about the energy involved. So it&#x27;s going to start with a certain amount of, um, of gravitational potential energy. And now, for the and assuming, like the block of this year have have the same amount, then of the same amount of mass that you know it&#x27;s gonna be equal. But that doesn&#x27;t really matter. Uh, now for the sphere that&#x27;s rolling, it&#x27;s going to have both translational kinetic energy and rotational kinetic energy. They go into that, and it&#x27;s going to have an angular acceleration, along with the linear acceleration. And there&#x27;s only a certain amount of energy that you know is available for for that. And so it&#x27;s going to accelerate more slowly down the implying that block that the slides frictionless Lee. You can also think of it in terms of the forces here. There has to be a force of friction because the rolling without slipping and that produces. You know, this adds to the fearsome tracks to the net force from gravity that is acting on on this fear. Where is if there&#x27;s no friction? And it just lying that this force isn&#x27;t here to affect its acceleration, And so the block accelerates faster. Whichever way you look at it, it&#x27;s still, but it&#x27;s still gonna go faster.

Ze-Han Lee · Answer

So in this problem we know that, boys. Ah, rolling down from from hell. So we know that there is no sleeping. There is There is no sliding. All right, so, um Well, that we want to check out which forces before Mito wrote thio make it rotates. So the choice, eh saves the static static friction friction. So it is actually right, because the static friction ah has direction pointing to this direction. Right? And ah ah, it is not on the center. So this is why this force well, given extra talk to this Ah, to this ball so that it could rotates. So this one is actually correct. But why the other threes? They&#x27;re not correct. So Choice B says that to the kind netted a friction kind net IQ friction. So this is this is incorrect because we from the problem we know that the bull has is no sliding. So in that sense, we do know have any kind attic friction in this problem. Yeah, So choice Say says that Ah, the normal force off the hue on the sphere on the ball. So we know that the normal force is always perpendicular to the incline, right? So it is pointing to this direction and because this is a sphere. So in that sense, the normal force were pointing into the center. All this bowl, right? So, in essence, this force will not give extra talk to this. Ah, sphere. So the number fours? No. Okay. And last one, the gravity. So also, clearly the gravity is performing. The gravity is on the senate, acting on this X on the center of this fear. So there is no it does not before meant any, um, pork on vista sphere. So gravity is also incorrect. So in this sense, only chose says is right.

Averell Hause · Answer

So here we&#x27;re simply for party going to apply the conservation of energy. And this would give us that the initial potential energy would be equal to the, uh, kinetic energy of the center of mass plus the kinetic energy rotational. So this would be giving us m g h. This would be equaling mg l the length of the incline time sign of fate of the angle of the incline. And this would be equal to 1/2 and the the mass times the velocity at the bottom squared plus 1/2 times the moment of inertia times the angular velocity at the bottom squared. And so we can then say that here m g l sign of data would be equaling 1/2 um b so bottom square, two plus 1/2 times the moment of inertia. Atmosphere 2/5 m r squared. And then this would be multiplied by the velocity at the bottom, divided by the radius of the sphere squared. And we finally find that the velocity of the bottom is equaling square root of 10 gh over seven. And so this is equaling square root of 10 g l sign of data over seven and we can solve. So this would be 10 over seven, multiplied by 9.8 meters per second squared multiplied by 10.0 meters and then times sign of 30 degrees. And so we can say that. Then the velocity at the bottom is equaling 8.3 seven meters per second. This would be the linear velocity and then the you can say that the angular velocity would be equaling the velocity of the bottom divided by the radius of the sphere. And we can say that this is equaling 8.367 meters per second. Divided by point to 54 meters, This is equaling 32.9 radiance per second. This would be the angular velocity at the bottom four part B. They wanted the ratio of the kinetic energy of the center of mass, divided by the Connecticut by the rotational kinetic energy. This would be equal in 1/2 times M times velocity at the bottom squared, divided by 1/2 times the moment of inertia. High times of angular velocity at the bottom squared. This is equaling 1/2 em, the bottom squared, divided by 1/2 times 2/5 m r. Squared, multiplied by the velocity at the bottom, divided by the rate the radius of the sphere quantity squared. And we know that this is equal in five. Over to this would be our ratio for part B. And for part C, we can say that this translational speed at the bottom. And so he&#x27;s a bottom and the kinetic energy of the center of mass, divided by the kinetic energy rotational. This ratio and the velocity of the bottom are independent of radius and mass. And then we can say that the rotational speed we can say the angular velocity at the bottom is dependent on the radius. That would be your answer for port. See, that is the end of the solution. Thank you.

Coleen Amado · Answer

in this problem we need to solve for the linear and angular velocities off each sphere after the in back. So that&#x27;s where they ate and for leather me the velocity off each sphere after they have started rolling uniformity. So the Outback is assumed to be perfectly elastic. So Sphere a off Mass, AM and Regis are, which is a ruling without slipping with Velocity V one and Angular Velocity Omega one hits on identical Sphere Me, which is initially a crest. So this produces an impulse and process. Sphere A and sphere be the role with angular velocity, Omega and Omega be, respectively, and the movie Velocity V A and B, respectively. So let&#x27;s first look into the in back on spirit eight. So I have year a sketch off the, um, scenario. So for sphere A. So since, according to the problem, Sphere is initially rolling without slipping so we can see that omega one Seiko toe be won over our now applying the principle off, impose and momentum and taking the moments about pointy, which is the center of this fear, with the clockwise direction, and begin as the positive direction. So we&#x27;ll have. So I Omega one plus zero equals. So I Amanda eight. So this, um, from this, we can see that Omega one is actually equal Do omega A. So which is God through. So this means that omega is also equal to be one over r No writing an equation showing for the linear components off momentum over again for Spirit e were still on spirit. So we have. So am me one my nose. So the impulse equals am me eight. So this is a word equation one. Now let&#x27;s look in tow. Impact onto the impact on this Fear me this time. So I have your A sketch for spirit. Be so again, applying the principle off impulse and momentum and writing an equation for the senior components momentum. So when have so zero plus with the MBA&#x27;s B B B equals I am we mean So this is now our equation on do now I think equations one into so the impulse would be cancelled out. So we&#x27;ll have so end. The one equals O M B A plus M V V. So the masses would actually cancel out. So this will become so be one equals B a plus B B we can have. This is our equation now, writing another equation from the equation for the coefficient of restitution or so we&#x27;ll have the equation for the coefficient of restitution is equal to so easy Will do negative. So be a my nose baby over. We want my nose zero. So it&#x27;s gonna be then S o E v one. It waas maybe my nose we ate and since is equal to one because of the impasse Perfectly elastic. So we have so be one equals we be my nose the A So this is our equation for now Taking the moment about point eat again the center off this fear with clockwise direction taken as positive So we have, um, zero zero equals So I will make a b Iomega Be So this gives us Omega B is equal to zero now solving for equation slee and four simultaneously so we&#x27;ll get it&#x27;s right that that&#x27;s all being poor, racial in sri and equation for until, um B a would be cancelled out so will be left with so v b equals B one. So solving for the eight. So you&#x27;ll see that b A is equal to zero. So hands our answer. Where letter is so the velocity of this fear is good. Zero spear a zero and then it&#x27;s angular. Velocity is equal Doe V one over R in the clockwise direction and then for spear me So its velocity. Zico Group B one And then it&#x27;s angular. Velocity is equal to zero. Sorry, so VB is to the right. Then angular velocity is zero. Now let&#x27;s look onto each arm spheres motion after the impact. So for us fear a So for sphere A. Since this fear rolls without slipping, so would have the A. Brian is a photo omega prime times very juice. So next solvent for the moments about the instantaneous point off rotation since its again rolling without sleeping. So we have, um, I Omega E Plus zero equals Iomega a prime plus M v. A prime times are so given the shape off this fear. So its moment of inertia is given by door over five m r squared. So substituting this and equation for me um, A and Omega A from part A. So are a creation would become so too fibs m r squared times the value of Omega averages be won over or two equals 50 m r squared omega A squared loss So M r squared Omega e squared. So simplifying this will have So do it&#x27;s m or anyone equals 7/5 m r squared Omega e Brian. So from here we can get Omega A. So make a Prime is equal to do V one over seven Art. And since we a single omega, a prime times are so be a primacy Go Duke to be won over seven. Next, we&#x27;ll write the same set of equations. Four. Spear being so bored Sphere be first, Fear bees again, since it&#x27;s also rolling without sleeping. So we be prime. Itwas omega be prime times are and then for the the equation for the moments about the incident in this point off sedation. So we&#x27;ll have. So am M v. Me Times or blood zero equals I omega Be prime plus M v b Prime times or so something something decoration for, um vb Brehm and I so will have m v one r equals over five m r squared omega be prime plus m r squared omega me, Brian. So this would be UM m v. One or because having over five m r squared omega be brain. So from here can get Omega Be prime equals So five b one uber seven R and then scenes. The B prime is a gold toe omega Be Brian times ours a baby prime musical toe. Five Be won over seven hands. The answer or let me is so v a Brian is a go to do be won over seven and then b B Prime is equal. Do five B one times be won over seven.

Problem

In the diagram above, a mass $m$ starting at poin…

Problem 22 Easy Difficulty

Answer

Related Courses

Cracking The AP Physics 1 Exam

Related Topics

Discussion

Top Educators

Recommended Videos

Watch More Solved Questions in Chapter 12

Video Transcript

Robert Franek

Cracking The AP Physics 1 Exam

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Cracking The AP Physics 1 Exam

Robert FranekCracking The AP Physics 1 Exam

Robert Franek

Cracking The AP Physics 1 Exam