Well, consider a case there is a sphere and a ring, of very same mass \$M\$ and radius \$R\$. Lock both beginning rolling down the skinny plane. We understand moments of them as well, \$\$I_ extsphere=frac25MR^2\$\$ and \$\$I_ extring=MR^2\$\$ respectively. So, We recognize that sphere will have much more transitional kinetic energy, so an ext velocity, for this reason it will certainly take much less time to reach at bottom.

The inquiry is while using equation for both, \$\$v^2 - u^2 = 2as, \$\$ initial velocity is \$0\$ because that both, last velocity are different for both, but acceleration and distance travel same. So, whereby is the blunder happening?

And additionally the equation \$\$v=u+at,\$\$ if velocity for round is greater, then what about the time? Why is the time taken less? Where space the equations acquiring wrong or is that me getting it wrong?

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rotational-kinematics moment-of-inertia
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edited jan 7 "19 in ~ 13:48

Kyle Kanos
asked jan 7 "19 in ~ 13:42

Alvin CarterAlvin Carter
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What you have actually left in calculations is acceleration,, a≠g nor a=gsinα. \$\$a= gsinα-F/M\$\$where α is angle of incline, F is pressure of friction and M remains mass.. Since friction acting on both space different, your acceleration are different for very same distance s.

Same goes because that your 2nd equation v=u+at, below a is different. Very same goes for t, not you think one with faster translation kinetic energy reach sooner? hope this helps..

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edited jan 7 "19 at 14:06
answered jan 7 "19 in ~ 14:01

Anubhav GoelAnubhav Goel
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Your kinematic equation the \$v^2-u^2=2as\$ is correct, but similar to your question right here you room neglecting the effects of friction, which offers rise to various accelerations because that each object.

Considering the net pressure acting on every object, us actually have two forces with contents acting under the ramp: gravity (\$mgsin heta\$) and also friction (\$f\$). Without friction, the objects only will have actually the force \$mgsin heta\$ acting under the ramp, and also there would be no network torque acting about the center of every object. Therefore, each object would slide without rolling down the ramp v the very same acceleration and reach the bottom the the incline at the same time!

So, what you need to do is identify the net pressure acting on each object:\$\$F_net=mgsin heta-f\$\$

However, as with the concern of yours i referred to, by imposing the rolling without slipping condition, \$a=alpha R\$, you are constraining friction to it is in a certain value for each object that counts on their minute of inertia \$I\$. This can be seen by considering the net torque on every object:\$\$ au_net=Ialpha=frac aRI=fR\$\$

So we view that in stimulate to have rolling without slipping it have to be the situation that\$\$f=fracaIR^2=gamma ma\$\$for \$I=gamma mR^2\$

So we view that we finish up with various frictional force for every object. Putting it every together:\$\$mgsin heta-gamma ma=ma\$\$\$\$a=fracgsin heta1+gamma\$\$

Showing what you already knew: the larger value of \$gamma\$ reasons a reduced acceleration, and hence a much longer time down the ramp as soon as both objects role without slipping under the ramp.