# 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!!!!!!!!

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!!!!!!!!

## 12 Answers

Moment of Inertia (I) (2/5)MR2 MR2 /2 MR2 /2

Radious of Gyration (K) Underoot(2/5) R/root2 R/root2

Any of them on rolling on inclined plane will have an acceleration,

a = gsin8/R2 + K2

It means maximum radius of gyration having object will have minimum acceleration.

Decrease in acceleration .i.e decrease in velocity and thus kinetic energy.

Now, Solid sphere has lowest value of ''K'' thus it rolls down with maximum speed.

Hence it will reach the bottom first.

And disc and solid sphere will reach at the same time with same speed.

(b) is correct!

Approve! plz!

(b) solid sphere will reach first....as the bodies are rolling on inclined plane therefore there acceleration will be

a=‹g sinθ›/‹‹1› + ‹I/mr^{2} › where I is moment of inertia of the body...and thus whose acc will more will reach first...

acc=gsin(o) / 1 + I/MR^2

hence acc is prop to 1/I

hence 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).

olid sphere disc solid cylinder

Moment of Inertia (I) (2/5)MR2 MR2 /2 MR2 /2

Radious of Gyration (K) Underoot(2/5) R/root2 R/root2

Any of them on rolling on inclined plane will have an acceleration,

a = gsin8/R2 + K2

It means maximum radius of gyration having object will have minimum acceleration.

Decrease in acceleration .i.e decrease in velocity and thus kinetic energy.

Now, Solid sphere has lowest value of ''''K'''' thus it rolls down with maximum speed.

Hence it will reach the bottom first.

And disc and solid sphere will reach at the same time with same speed.

(b) is correct!

Approve! plz!

__(2)__

__here,__

__V is inversly propirtional to value of k and solid sphere have lowest value of k__

__hence it will reach bootom first.__Please find below the solution to your problem.

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).

Thanks and Regards