(^648) „„„„„ A Textbook of Engineering Mechanics
∴ F=
2
22
Ia Mk a
rr
= ... (Q I = mk^2 ) ...(ii)
and now substituting the value of F in equation (i),
2
sin – 2
Mk a
Mg ma
r
θ=
Mg sin θ=
22
221
Mk a k
Ma Ma
rr
⎛⎞
+=+⎜⎟⎜⎟
⎝⎠
∴ a= 222
22
sin sin
1
gg
krk
rr
θθ

• 
  


• 
  

  The above expression shows that the acceleration of the rolling body is independent of its
  mass. Now substituting the value of (a) in equation (ii),
  F=
  22
  222 22
  2
  Mk gsin Mk gsin
  rrk kr
  r
  θθ
  ×=
  ++
  Now for the body (or wheel) to roll down without slipping, the applied force must be less
  than (or equal to) the available force of friction, in order to fulfil the condition of rolling without
  slipping. Or in other words
  F≤μmg cos θ
  2
  22
  Mk gsin
  kr
  θ

  
  


• 
  

  ≤μMg cos θ
  tan θ ≤
  22
  2
  kr
  k
  ⎛⎞+
  μ⎜⎟⎜⎟
  ⎝⎠
  The above equation may also be written as :
  μ≥
  22
  2
  tan
  kr
  k
  θ
  ⎛⎞+
  ⎜⎟⎜⎟
  ⎝⎠
  Note: The above expression shows that the value of (μ) should be more or equal to the value
  obtained from the right hand side. Thus for the minimum value of (μ), the above expression may be
  written as :
  μ= 22
  2
  tan
  kr
  k
  θ
  ⎛⎞+
  ⎜⎟⎜⎟
  ⎝⎠
  Example 31.16. A sphere rolls down a plane inclined at 30° to the horizontal. Find the
  minimum value of the coefficient of friction between the sphere and the plane, so that the sphere
  may roll without slipping.
  Solution. Given: Inclination of plane θ = 30°
  We know that for a sphere of radius r,
  k^2 = 0.4 r^2
  and minimum value of coefficient of friction,
  μ= 22 22
  22
  tan tan 30 0.5774
  0.165
  0.4 3.5
  0.4
  kr rr
  kr
  θ°

  
  

  ++
  Ans.