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# A bullet of mass M moving with velocity U just grazes the top of a solid cylinder of mass M and radius R resting on a rough horizontal surface .If the cylinder rolls without slipping then find the angular velocity of cylinder and the final velocity of the bullet

Asked by mvenkataram 22nd August 2019, 8:01 AM
when the bullet which is moving initially with velocity v, just grazes the cylinder ,
contact point on the cylinder starts moving with velocity v.

We know that  if cylinder is rolling on rough horizontal surface without slipping and
the top most point moves with linear speed v , then centre of mass moves with speed v/2
and point of contact of cylinder with horizontal surface has zero velocity.

angular speed of cylinder = ω = v/(2R)  , where R is radius of cylinder

Initial kinetic energy Ei of bullet is given by,   Ei  = (1/2)Mv2 ..................................(1)

where M is mass of bullet

Kinetic energy EC of rolling cylinder is given by,   EC = (1/2)M(v/2)2 + (1/2)Iω2 .......................(2)

where M is mass of cylinder, I is moment of inertia

with substitution of I = MR2/2,  eqn.(2) is simplified  as,  EC = (3/16)Mv2 ....................(3)

Hence kinetic energy of bullet after grazing the cylinder,

Ei - EC = (1/2)Mvf2 = (1/2)Mv2 - (3/16)Mv2 = (5/16)Mv2...................(4)

where vf is the bullet velocity after grazing the cylinder and it is obtained from eqn.(4) as,

vf = 0.56 v
Answered by Expert 23rd August 2019, 10:17 AM
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