Question

pls pls.. explain this question.... urgently!
A cyclinder of height h is placed on an inclined
plane, the angle of inclination of which is slowly
increased. It begins to slip when the angle of
inclination is 45º. What is the radius of the cylinder?
1.h
2.3/4h
3.1/2h
4.1/4h​

Answer 1

Simple explanation

Solution:-There are two chances,,,if angle of inclination will be increased then

1)The cylinder can slip

1)The cylinder can slip2)The top of the cylinder can fall downward,,before the slipping

Now,to maintain the equilibrium......the top of the cylinder should not move downward,,,

This can only be possible if the torque acting due to forces will be zero....

at the maximum inclination upto which it will not slip,,the normal 'N' will be at the lower bottom end of the cylinder

Torque force about the centre of cylinder=N×h/2

this torque will be in clockwise direction (see the first figure)

Now,to maintain the equilibrium there must be another torque,which will act opposite(anticlockwise) to the first torque and eqaul to it, so that they cancel out to he zero

torque about h/2,,due to friction=f×r

now these two torques will be eqaul to zero.to maintain the equilibrium

N×h/2-f×r=0

N×h/2=f×r.........I)

as we know in inclination normal is eqaul to mgcos∅

so,N=mg cos∅

also,

friction force depends on how much force is applied on the object horizontally,

as in inclination force applied to slip=mgsin∅

so,

f=mgsin∅

putting this value in first we get

mgcos∅×h/2 =mgsin∅×r

putting it=45°(given)

mgh/2√2=mgr/√2

h/2=r

so, r =h/2