A hollow verticle cylinder of radius r and height h has a smooth internal surface. A small particle is placed in contact with the inner side of the upper rim, at point A, and given a horizontal speed u, tangential to the rim?

It leaves the lower rim at point B, vertically below A. If n is an integer then?

A) #u/(2pir)(sqrt((2h)/g)) = n#
B) #h/(2pir) = n#
C) #u/sqrt(2gh)=n#

1 Answer
A08
Jan 7, 2018

I get
(A)

Explanation:

We see that vertical and horizontal speed of the particle are orthogonal and therefore can be treated independently. We also see that in all the three choices time #t# is not present and we are to find #n#

A . Vertical speed.
The particle falls freely under action of gravity #g#
It falls through height #h#. This is related by the kinematic expression

#s=ut+1/2at^2#

Inserting various values we get

#h=0xxt+1/2g t^2#
#=>h=1/2g t^2#
#=>h=1/2g t^2#
#=>t=sqrt((2h)/g) ......(1)#

B. Horizontal speed.
The particle is given horizontal speed #u#, tangential to the rim. As such its movement is around the walls of the cylinder as it rolls down in a spiral. The particle leaves the lower rim at point B, vertically below A. Therefore, at the time of exit it would have covered integral multiple #n# of its circumference. As such horizontal distance covered is #=n(2pir)#.
Time taken for this #=t#
The applicable expression is

#2npir=ut#
#=>t=(2npir)/u# .......... (2)

Equating RHSs of (1) and (2) we get

#sqrt((2h)/g)=(2npir)/u#
#=>n=(usqrt((2h)/g))/(2pir)#

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