A cylinder is placed on a rough inclined surface of inclination theta. minimum value of coefficient of static friction between cylinder and surface so that cylinder undergoes pure rolling is
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Answer:
Explanation:
Let the frictional force acting on it be f.
Applying force equation we get
mgsinx-f=ma
Again f.r=(mr^2/2)×a/r
Or f=ma/2
Plugging the value of f into the first equation we get a=2gsinx/3
Hence f=mgsinx/3
Or friction coefficient, k=tanx/3
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Krishnam Kumawat
Krishnam Kumawat, Physics student, class 12
Updated May 19, 2017
Hey there! First of all such a good question! Had to use so many concepts to get here…
Here goes,
This is a diagram of the situation. Now, next will be a free body diagram of the Solid Cylinder.
Now, by looking at the situation, we can say that the normal force will be balanced by Mgcosβ, while friction will apply in the opposite direction of the motion. We know that rolling is a motion of the body made possible due to torque, so let's calculate the torque due to all the forces here.
(consider radius of the cylinder to be ‘r’ and point of rotation to be the Centre of the mass of the sphere)
Torq...(more)
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Shaji Kurien
Shaji Kurien, B.Tech. Mechanical Engineering, University of Kerala
Answered Apr 22, 2018
This is a case of rolling friction. The force, F, required for rolling a cylinder of mass, m, and radius, r, on a horizontal plane = m*g*a/r. ——————(1)
where, a = coefficient rolling friction of wheel material and plane material, and
g = acceleration due to gravity.
Since the cylinder is on a plane making an angle , B, with the horizontal, the component of cylinder weight acting parallel to the inclined plane = m*g*sinB—————(2)
By equating equations 1 and 2
Coefficient of rolling friction, a = r*sinB
Answer:tan θ/3
Hope it helps,
Explanation:
