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A solid sphere, disc and solid cylinder all of same mass and made up of same material are allowed to roll down (from rest) on an inclined plane, then
(a) solid sphere reaches the bottom late
(b) solid sphere reaches the bottom first
(c) disc will reach the bottom first
(d) all will reach the bottom at the same time
choose and explain please!!!!!!!!
(b) solid sphere will reach first....as the bodies are rolling on inclined plane therefore there acceleration will be
a=‹g sinθ›/‹‹1› + ‹I/mr2 › where I is moment of inertia of the body...and thus whose acc will more will reach first...
option dcheck for the formla
acc=gsin(o) / 1 + I/MR^2hence acc is prop to 1/Ihence then compare the acc.
Lets assume all the three are rolling from an height H above the ground
From conservation of energy:
mgH=(1/2)mv^2 + (1/2)Iw^2, where v is the velocity at the bottom and w is the angular velocity
put w=(v/R)
we get,
v^2={2mgH}/{m + (I/R^2)}
now in this relation only moment of inertia will be different from shape to shape(if the radius is same)
then substitute the value of moment of inertia and then compare which has the highest velocity.The one with the highest velocity would be the one to reach the bottom the fastest
Hi Pratiksha,
See all the bodies would loose the same amount of PE and hence all will also gain the same amount of KE.
KE = 1/2(mv^2) + 1/2(Iw^2); with v = rw (fpr pure rolling)
So v (or you can say w as well), will be more for the body with less I value and will reach the bottom first. (Remeber KE will be the same for all the bodies, so more I less v and less I more v).
In this case it would be solid sphere. Option (B).
Regards,
Ashwin (IIT Madras).
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