The acceleration a of the supporting surface required to keep the centre G of the circular cylinder in a fixed position during the motion if there is no slipping between the cylinder and the support will be :________________
Ans : 2g sin@
21net force acting on the cylinder parallel to incline
for motion of cylinder
fr=frictional force between cylinder and support
ma'=mgsin@-fr
COM of cylinder is stationary that means a'=d2xdt2=0(assuming x axis along incline)
so,
fr=mgsin@
torque on cylinder abt point G=frR=(mR2/2)α
fr=mRα/2=ma/2
ma/2=mgsin@
a=2gsin@
1Excellent Eragon, ridden the dragon well.
But i've one little doubt here.
Could we have done it by a procedure where we first take the block's acceleration as a...and a pseudo acceleration a for the Cylinder ??
And why are we considering that the cylinder and the block move together such dat der's a common acceleration a ???? If it's so then there should be no force of friction between the cylinder and the block !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
