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Figure shows a loop track whose lower pa...

Figure shows a loop track whose lower part ends into a circular track of radius R and centre O.
A small solid sphere of mass M rolls without slipping along the loop track from the end A at a height 6R from the bottom of the track. What is the horizontal force acting on the sphere, when it rises up to the point P in level with the centre O of the circular part ?

A

`(30)/(7)Mg`

B

`(40)/(7)Mg`

C

`(50)/(7)Mg`

D

`(60)/(7)Mg`

Text Solution

Verified by Experts

The correct Answer is:
C
The ball descends through a vertical distance `6R-R=5R`
`therefore` Loss in P.E. of the ball = mg (5R) ….(1)
Kinetic energy of the ball at P
`=K_("Translational")+K_("Rotational")`
`=(1)/(2)mv^(2)+(1)/(2)I omega^(2)`
`= (1)/(2)mv^(2)((1)/(2)xx(2)/(5)MR^(2)xx(v^(2))/(R^(2)))=(7)/(10)mv^(2) " "` .....(2)
By the principle of conservation of energy.
`mg(5R)=(7)/(10)mv^(2) " " therefore v^(2)=(50gR)/(7) " "` ....(3)
At point P, the velocity of the ball is tangentially upwards.
`therefore` A centripetal force acts on the ball along PO, i.e, it is along the horizontal.
`therefore` Horizontal force acting on the ball
`=(Mv^(2))/(R )=(M)/(R )((50gR)/(7))=(50)/(7)Mg`
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