A uniform solid sphere of radius r is…

A uniform solid sphere of radius r is released from rest at the top of a rough incline and rolls without slipping down it, descending a vertical…

Single correct option

A uniform solid sphere of radius

r

is released from rest at the top of a rough incline and rolls without slipping down it, descending a vertical height

h=0.70\ \mathrm{m}

. At the foot the incline is smoothly continued by a horizontal table-top whose edge is

H=1.25\ \mathrm{m}

above the floor; the sphere keeps rolling without slipping on the table and leaves its edge horizontally. Take

g=10\ \mathrm{m\,s^{-2}}

and neglect air resistance and energy loss at the junction. The horizontal distance from the table edge to the point where the sphere first strikes the floor is closest to

A. $1.58\ \mathrm{m}$
B. $1.87\ \mathrm{m}$
C. $1.45\ \mathrm{m}$
D. $1.53\ \mathrm{m}$

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Answer: (A) $1.58\ \mathrm{m}$

Rolling without slipping conserves mechanical energy because static friction does no work. With

I=\tfrac25 mr^2

mgh=\tfrac12 mv^2\left(1+\tfrac25\right)

, so

v^2=\tfrac{10}{7}gh=10\ \mathrm{m^2s^{-2}}

and

v=\sqrt{10}\ \mathrm{m\,s^{-1}}

. The speed of the centre of mass does not change along the horizontal table, and at the edge the sphere becomes a projectile launched horizontally with this speed; its spin stays constant in flight. The fall time is

t=\sqrt{2H/g}=0.50\ \mathrm{s}

, hence the range is

vt=\sqrt{10}\times0.5\approx1.58\ \mathrm{m}

. Using

v^2=2gh

(frictionless sliding) gives

1.87\ \mathrm{m}

, a hollow shell (

k=2/3

) gives

1.45\ \mathrm{m}

and a disc (

k=1/2

) gives

1.53\ \mathrm{m}

, which are the distractors. The answer is the first option.

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Original question written in the JEE Advanced 2016 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake