Class 11 Physics Laws of Motion Rolling friction

Rolling friction

  • Rolling friction is applicable for bodies whose point of contact keeps changing
  • It is the force that opposes the motion of a body which is rolling over the surface of another
  • Bowling balls, rotating wheels are examples illustrating Rolling friction

                                 

  • Coefficient of Rolling friction is lesser than that of Kinetic friction.

fr < fk < fs

Problem: Calculate the force required for pushing a  30 kg wooden bar  over a wooden floor at a constant speed. Coefficient of friction of wood over wood = 0.25

Solution.

M = 30 kg

μ = 0.25

(Fa – f) = ma

For constant speed, a = 0

So, Fa = f= μN

From free body diagram, N = mg

Therefore, Fa = f= μN = μmg = 0.25 * 30 * 9.8 = 73.5 N

 

Problem 2: A homogenous chain of length L lies on a table. What is the maximum length l of the part of the chain hanging over the table if the coefficient of friction between the chain and the table is u, the chain remaining at rest with the table?

Solution.

Let ρ be the mass per unit length of the chain

Ρ = m/l or, m = ρl

Weight of hanging part = W2 = ρlg

Weight of chain over the table = W1 = ρ(L-l)g

For equilibrium,

R = ρ(L-l)g --------------- (i)

 f= W2 = ρlg

μR = ρlg -------------------(ii)

(i)Divided by (ii) gives

1/μ = (L-l)/l

l = μL – μl

l + μl = μL

Therefore, l = μL/( μ + 1)

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