A force F is applied at the centre of disc of mass M .the minimum value of coefficient of friction of the surface for rolling is
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Torque applied due to external force
т₁ = FR
F is the force applied and R is the radius of the disk
Normal force on the disc
N = Mg
Frictional force on the disc
f = μ Mg
Torque due to friction
т₂ = μMgR
Condition for rolling
Torque on the ball due to friction should not be less than torque due to external force.
т₂ ≥ т₁
μMgR ≥ FR
μMg ≥ F
μ ≥ F ÷ Mg
Therefore the minimum value of coefficient of friction of rolling is F ÷ Mg.
т₁ = FR
F is the force applied and R is the radius of the disk
Normal force on the disc
N = Mg
Frictional force on the disc
f = μ Mg
Torque due to friction
т₂ = μMgR
Condition for rolling
Torque on the ball due to friction should not be less than torque due to external force.
т₂ ≥ т₁
μMgR ≥ FR
μMg ≥ F
μ ≥ F ÷ Mg
Therefore the minimum value of coefficient of friction of rolling is F ÷ Mg.
Answered by
124
See the attached file the answer is right
That girl who answered first had done wrong ans is F/3Mg
And not F/Mg
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