anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
dont go away i am going to solve this problem..
anonymous
  • anonymous
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anonymous
  • anonymous
for normal reaction, N= mgcosa

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More answers

anonymous
  • anonymous
for motion down the inclined plane, F= mg sina - f ............(1)
anonymous
  • anonymous
since friction prevents slipping and frictional force is causing rotation with angular acceleration @ thus torque, T = f*R = I*@
anonymous
  • anonymous
f= I*@/R = I*a/R^2
anonymous
  • anonymous
m*a=mgsina - I*a/R^2
IrishBoy123
  • IrishBoy123
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
anonymous
  • anonymous
a= gsina - I*a/R^2
anonymous
  • anonymous
now if u put the value of moment of inertia for sphere and disc, u get the the linear acc downward...
anonymous
  • anonymous
thank you guys :D
anonymous
  • anonymous
do u know what is moment of inertia for sphere and disc....
anonymous
  • anonymous
greater the linear acc. more faster the object reaches the ground faster.....
anonymous
  • anonymous
got it?
anonymous
  • anonymous
yes :)
anonymous
  • anonymous
yes we've already had an experiment about this one. thank you again :)
IrishBoy123
  • IrishBoy123
good job @lall
anonymous
  • anonymous
thanks! @IrishBoy123
anonymous
  • anonymous
their moment of inertia can be calculated mathematically.....

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