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PHYSICS
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Expert’s Solution for Question asked by IIT-JEE Aspirants
1. A homogeneous sphere of radius r rolls without
slipping with constant angular speed ω' over bigger
sphere, of radius R, which is in pure rotation with
constant angular speed ω about its centre O. (Fig.)
Find the time taken.
ω
R
O
ω′
r
C
(i) for the centre C of the rolling sphere to return to its
initial position (with respect to O), and
(ii) for the point of contact of the rolling sphere to make
one full revolution over the bigger rotating sphere.
(b)(i) Determine the acceleration of the contact point of the
rolling sphere, and
(ii) the point of greatest acceleration of the rolling sphere
(both w.r.t. the centre O)
Sol. Suppose that at t = 0, the contact points of lie along
the fixed line (reference line) OX. Let at time t the
line OC makes the angle θ with OX (Fig.)
ω
O
^
t
C
P
θ P′
It is better to express the velocity and acceleration of
any point including contact point P (say) at an
arbitrary instant of time t as,
v po = v co and a po = a co = a pc + a pc + a co (1)
We know that in the case of pure rolling the velocity
of contact point of the rolling body has zero velocity
and zero tangential acceleration relative to the contact
point of the surface on which it rolls. So if P′ be the
contact point of rotating sphere at time t, then
v p′o = v po
So, v p'o = v pc + v co (2)
ω′
X
(a) If t is the direction of common tangent, then from
eqn. (2)
v p'o (t) = v pc (t) + v co (t)
⎛ dθ
⎞
ωR = – ω′R + ⎜ ⎟ (R + r)
⎝ dt ⎠
dθ ωR
+ ω'r
or,
=
(3)
dt R + r
R + r
so, dt =
dθ (4)
( ωR
+ ω'r)
(ii) It is simple to observer if rotating sphere were at rest,
the CM of rolling sphere C will turn by the angle 2 π,
⎛ dθ
⎞
with angular speed ⎜ ⎟ – ω to satisfy the condition
⎝ dt ⎠
of the problem.
dθ
dt
ω
Hence the sought time t′ (say) is given by
2π
2π(R
+ r)
t′ = = [using Eq. (4)]
⎛ dθ
⎞ r( ω−ω ' )
⎜ ⎟ − ω
⎝ dt ⎠
(b) From a po = a pc + a co = a pc (tangential) + a pc (normal) + a co
In our case a pc (tangential) = 0, because ω′ = constant
Hence, a po = a pc (normal) + a co (5)
⎡
2
⎤
2 ⎛ dθ
⎞
a po = ⎢− ω'
r + ⎜ ⎟ (R + r)
⎥
⎢⎣
⎝ dt ⎠ ⎥⎦
dθ ωR
+ ω'r
(using = from equation (3) or part (a))
dt R + r
(ii) From eqn. (5) it is obvious that the maximum value
|apo | ma x = |a pc (normal) | + | a co |
2
= – ω′ 2 ⎛ dθ
⎞
r + ⎜ ⎟⎠ (R + r)
⎝ dt
dθ
where will be substituted from Eqn. (3).
dt
XtraEdge for IIT-JEE 22 DECEMBER 2009