42. The coefficient of static friction between a certaincylinder and a horizonta

Question

42. The coefficient of static friction between a certaincylinder and a horizontal floor is 0.40. If the rotational inertia of the cylinder about its symmetryaxis is given by I = (1/2)MR2, then the maximum acceleration the cylinder can have withoutslipping is: 42. The coefficient of static friction between a certaincylinder and a horizontal floor is 0.40. If the rotational inertia of the cylinder about its symmetryaxis is given by I = (1/2)MR2, then the maximum acceleration the cylinder can have withoutslipping is:

Explanation / Answer

if velocity increment is v and angular velocity increment is w then v=rw => [a-g]t=rmgtr/I => a= 3g=11.76m/sec*sec

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