Correct Answer - Option 3 :
Both (1) and (2)
CONCEPT:
- The cylinder will roll when there is sufficient friction to do so.
- If we lower the coefficient of friction between the cylinder and the plane until the cylinder just begins to slip (not roll),
EXPLANATION:
- It will begin to slip when-
f = μN
- As there is no motion in a direction normal to the inclined plane:
N = Mgcosθ
- Applying Newton's Second Law to the linear motion of the center of mass, the net force on the cylinder rolling down the inclined plane is
F = Ma = Mg sinθ - f
- It is only the force of friction f wich expert torque τ on the cylinder and makes it rotate with angular acceleration.
- It acts tangentially at the point of contact P and has lever arm equal to R.
τ = Force × force arm = f. R
τ = M.I × angular acceleration = Iα
fR = Iα
f = \(\frac{Iα}{R^2}\)
putting the value of f in equation (i)
Ma = Mg sinθ - \(\frac{Iα}{R^2}\)= a = g sinθ - \(\frac{Iα}{MR^2}\)= a + \(\frac{Iα}{MR^2}\)= g sinθ
Moment of Inertia of the solid cylinder about its axis = MR2/2
By putting value, \(\frac{1}{2}MR^2\)
a =\(\frac{gsinθ}{1+\frac{I}{MR^2}}\) = \(\frac{2}{3}\)g sinθ
- Clearly, the linear acceleration of a solid cylinder rolling down an inclined plane is less than the acceleration due to gravity g (a <g)
- The linear acceleration of the cylinder is constant for a given inclined plane and is independent of its mass M, and Radius R.
However, for a hollow cylinder.
I = MR2, the value of a would decrease to 1/2(gsinθ)
From equation (i), the value of the force of friction is,
f = Mgsinθ - Ma by putting the value of a from equation (ii), we get
f = Mgsinθ - M × 2/3(gsinθ) = 1/3(Mgsinθ)
if μs is the coefficient of friction between the cylinder and the inclined plane, then
μs = f/N = \(\frac{\frac{1}{3}Mgsinθ}{Mgcos θ}\)= (1/3) tanθ
-
To prevent slipping, the coefficient of static friction must be equal to or greater than the above value.
That is, μs > 1/3 (tanθ ) or μs = 1/3 (tanθ)
