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Well, Consider a situation there is a sphere and a ring, of same mass $M$ and radius $R$. They both starts rolling down the inclined plane. We know moments of them as well, $$I_\text{sphere}=\frac{2}{5}MR^2$$ and $$I_\text{ring}=MR^2$$ respectively. So, We know that sphere will have more transitional kinetic energy, so more velocity, so it will take less time to reach at bottom.

The question is while using equation for both, $$v^2 - u^2 = 2as, $$ initial velocity is $0$ for both, final velocity are different for both, but acceleration and distance traveled same. So, where is the blunder happening?

And also the equation $$v=u+at,$$ if velocity for sphere is greater, then what about the time? Why is the time taken less? Where are the equations getting wrong or is it me getting it wrong?

Kyle Kanos
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What you have left in calculations is acceleration,, a≠g nor a=gsinα. $$a= gsinα-F/M$$ where α is angle of incline, F is force of friction and M remains mass.. Since friction acting on both are different, their acceleration are different for same distance s.

Same goes for your second equation v=u+at, here a is different. Same goes for t, dont you think one with faster translation kinetic energy reach sooner? Hope this helps..

Anubhav Goel
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  • But, I guess Friction for both are also equal as friction is $$F = u(mgcosα), $$ which is same for both ? – Player Minus One Jan 07 '19 at 14:05
  • Its not so simple dear... https://youtu.be/UDYjAF-GA9g No it isnt same, this video show how to calculate friction for sphere, you can extend that to ring.. – Anubhav Goel Jan 07 '19 at 14:14
  • From non mathematical, theoretical point of view you can understand this as, rolling is being caused by friction, yes or no?, and since both rotate at different speed, they must have different frictional force to cause so. – Anubhav Goel Jan 07 '19 at 14:19
  • @AnubhavGoel it can't be F/M. Friction does not act on the center of mass of the object. – Karthik Jan 07 '19 at 14:28
  • @Kartik V Then what it should be? – Anubhav Goel Jan 07 '19 at 14:34
  • See Aaron Stevens answer above. The friction causes the torque on the object, so it is related to the moment of inertia (and hence the radius of gyration). – Karthik Jan 07 '19 at 14:44
  • Saw, but got confused, if I should use $\gamma$ , or not? – Anubhav Goel Jan 07 '19 at 14:48
  • I believe its correct for now, may be I will change it some day later when I am convinced.. – Anubhav Goel Jan 07 '19 at 14:54
  • @KarthikV In my answer the term of acceleration due to friction is also $f/m$. I don't understand why you say it is not. For using Newton's second law to determine linear acceleration is does not matter where the force is acting – BioPhysicist Jan 07 '19 at 15:26
  • @AaronStevens and AnubhavGoel: OK. I'm sorry for it. Maybe a few things for me to work on as well. – Karthik Jan 07 '19 at 15:58
  • @KarthikV I cant remember where I saw your name before? But it seems like we interacted before as well.. – Anubhav Goel Jan 07 '19 at 19:05
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Your kinematic equation of $v^2-u^2=2as$ is correct, but just like your question here you are neglecting the effects of friction, which gives rise to different accelerations for each object.

Considering the net force acting on each object, we actually have two forces with components acting down the ramp: gravity ($mg\sin\theta$) and friction ($f$). Without friction, the objects only will have the force $mg\sin\theta$ acting down the ramp, and there would be no net torque acting about the center of each object. Therefore, each object would slide without rolling down the ramp with the same acceleration and reach the bottom of the incline at the same time!

So, what you need to do is determine the net force acting on each object: $$F_{net}=mg\sin\theta-f$$

However, just like the question of yours I referred to, by imposing the rolling without slipping condition, $a=\alpha R$, you are constraining friction to be a certain value for each object that depends on their moment of inertia $I$. This can be seen by considering the net torque on each object: $$\tau_{net}=I\alpha=\frac aRI=fR$$

So we see that in order to have rolling without slipping it must be the case that $$f=\frac{aI}{R^2}=\gamma ma$$ for $I=\gamma mR^2$

So we see that we end up with different frictional force for each object. Putting it all together: $$mg\sin\theta-\gamma ma=ma$$ $$a=\frac{g\sin\theta}{1+\gamma}$$

Showing what you already knew: the larger value of $\gamma$ causes a lower acceleration, and hence a longer time down the ramp when both objects roll without slipping down the ramp.

BioPhysicist
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I get this solution:

The equations of motion are:

$I\,\ddot{\vartheta}=F_c\,R\qquad (1)$

$M\,\ddot{s}=-F_c\,R +M\,g\,\sin(\alpha)\qquad (2)$

and rolling without slipping

$\ddot{s}=\ddot{\vartheta}\,R\qquad (3)$

with $F_c$ is constraint force.

We have 3 equations for 3 unknowns $\ddot{s}\,,\ddot{\vartheta}$ and the constraint force $F_c$.

we get for the incline acceleration:

$\ddot{s}=\frac{M\,R^2\,g\,\sin(\alpha)}{M\,R^2+I}$

with $I_s=\frac{2}{5}\,M\,R^2$ and $I_r=M\,R^2$ we get:

$\frac{\ddot{s}_{\text{sphere}}}{\ddot{s}_{\text{ring}}}=\frac{10}{7}$

so the incline acceleration of the sphere is $\frac{10}{7}$ greater then the incline acceleration of the ring.

Eli
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