A wheel rolls uniformly without slipping on a horizontal ground.A mud particle from its topmost point gets detached from the wheel and the center of the wheel reaches above it after 1 sec.Then the radius of the wheel is?

Question

anonymous · Answer

This is how I proceeded: Let the radius of the disc be R.

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The time of flight of mud particle is 2R = (1/2)*g(t^2)
=> 2sqrt(R/10) = t .
Let w be the angular velocity.
The horizontal range of mud particle = t*2Rw.
This must be equal to the velocity of Center of disc*1sec.
So I get R= 0.62m.
But the options are 2m , 2.5m, 4m, 4.5m.
Can someone suggest me where I have went wrong.

anonymous · Answer

@experimentX

experimentx · Answer

didn't understand the last part of Q
"the center of the wheel reaches above it after 1 sec."

experimentx · Answer

those two points coincide? (mud and the center of wheel)?

anonymous · Answer

the wheel runs over the mud particle after 1 second.

experimentx · Answer

let $ \omega $ be the angular velocity of wheel, then $ \omega r $ is the velocity of the wheel.

the velocity of mud is $ 2 \omega r$, there is no way they make contact on air.

experimentx · Answer

the distance traveled by the mud is twice the distance traveled by wheel ... so mud particle should have fallen before 1s

experimentx · Answer

the flight time of the mud particle is half second.

experimentx · Answer

h=2r
g=9.8
t=0.5
u=0

anonymous · Answer

Then r = 9.8/16 = 0.6125 m No such option is in the answer.I am also getting the same.

experimentx · Answer

what are your options?