1)a hoop of radius r,mass mkg it rolls along a horizontal floor so dat its centre of mass has a speed vm/s how much work has to b done to stop it?

2)d centre of a weel rolling on a plane surface moves wid a speed v₀.a particle on d rim of d weel at d same level as the centre will be moving at speed . 

3) a sphere is rotating about a diametre

4) a cubical block of mass M and edge a  slides down a rough inclined plane of inclination  with a uniform velocity .  the torque of the normal force on the block about its centre has a magnitude.

5) a rod of length L is hinged from one end.  it is brought to a horizontal position and than released . The angular velosity of the rod when it is in vertical position is

6) a uniform rod A B of mass M and length L at rest on a smooth horizontal surface and impulse P is applied to the end B .  The time taken by the rod to turn through a right angle is

7) A hollow straight tube of length  2L and mass M can turn freely about its centre on a smooth horizontal table . another smooth uniform rod of same length and mass is fitted into the tube so that their centre co-inside.  the system is set in motion with an angular velosity ₀ .  find the angular velosity of the tube at the instant when the rod slips out of the tube. 

8) a fly wheel rotates about an axis . due to friction at the axis , it experiences angular retardation proportional to its angular velocity.  if its angular velocity falls to half the value while its makes n revolution , how many more revolutions will it make before coming to rest. 

hey guys lampard , gaurav , computer001 , & other guys toooooooooo plz answer d questions

let P be the point. then vel of P= OP X w = (r i+r j)X (w k) =rw( i - j) | vel | = rw/root2 where w=v₀/r so the ans is v₀/root2

3) do u want the moment of inertia.....it is  for a solid sphere and  for a hollow sphere... 

4) the one importamt thing that should be noted is that N does not act along the center of mass.there is a trrque about the point A due to mg to balance this torque N will shift towards A. so that it can balance the torque due to mg the max point up to which N can move is A. if the torque is balanced within that then the object does not tople. and torque due to that will be same as that due to mg but in opp. dir. in this case torque due to mg about A =  k so torque due to N=  (-k)