A solid disk and a solid sphere each of mass M and radius R are released at the same time from the top of an inclined plane. which object will reach the bottom first? Justify your answer. Show complete derivation.
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for motion down the inclined plane,
F= mg sina - f ............(1)
since friction prevents slipping and frictional force is causing rotation with angular acceleration @
thus torque, T = f*R = I*@
m*a=mgsina - I*a/R^2
if frictionless [ you did not specify], then same
but if they are rolling, the cylinder will be slower as has a greater I [as it is more "spread out" about the rotating axis]
to prove that, personally i'd use energy conservation for this rather than forces as its a lot less fiddly
a= gsina - I*a/R^2
now if u put the value of moment of inertia for sphere and disc, u get the the linear acc downward...
thank you guys :D
do u know what is moment of inertia for sphere and disc....
greater the linear acc. more faster the object reaches the ground faster.....
yes we've already had an experiment about this one. thank you again :)
good job @lall
their moment of inertia can be calculated mathematically.....