11. A rigid object is rolling down an
inclined plane. Derive expressions for
the acceleration along the track and
the speed after falling through a certain
vertical distance.
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Answer:
ANSWER
if a is the acceleration of the centre of mass of the rigid body and f the force of friction between sphere and the plane, the equation of translatory and rotatory motion of the rigid body will be.
mg sin θ - f = ma (Translatory motion) fR = I α (Rotatory motion)
f=
R
Iα
I = mk
2
, due to pure rolling a = αR
mgsinθ−
R
Iα
=mαR
Mg sin θ=mαR+
R
Iα
mgsinθR+
R
mk
2
α
Mg sin θ=ma+
R
mk
2
α
mgsinθ=a[
R
2
R
2
+k
2
]
a=
[
R
2
R
2
+k
2
]
gsinθ
a=
(1+
R
2
k
2
)
gsinθ
f=
R
Iα
f=
R
2
mk
2
a
⇒
R
2
+k
2
mgk
2
sinθ
fleq∪N
k
2
mk
2
a≤∪≤mgcosθ
R
2
R
2
k
2
×
(k
2
+R
2
)
gsinθ
=∪gcosθ∪≥
[1+
k
2
R
2
]
tanθ
μ
min
=
[1+
k
2
R
2
]
tanθ
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