The system is released from rest with the cable taut, and the homogeneous cylind

Question

The system is released from rest with the cable taut, and the homogeneous cylinder does not slip on the rough incline. Determine the angular acceleration of the cylinder and the minimum coefficient of friction for which the cylinder will not slip. The angular acceleration is positive if counterclockwise, negative if clockwise.

The system is released from rest with the cable taut, and the homogeneous cylinder does not slip on the rough incline. Determine the angular acceleration of the cylinder and the minimum coefficient of friction for which the cylinder will not slip. The angular acceleration is positive if counterclockwise, negative if clockwise.

Explanation / Answer

For cylinder, T - mg = ma ; T - 9.2*9.8 = 9.2 a.....(1)

For disk, since no slipping occurs,

angular acceleration = a/r = Moment / I

a/r = T*r/(m*r*r/2) -----> a = 2T/m ----> T = ma/2 = 4.25 a

Putting in 1 and solving,

a = 19.183 m/s2

Angular acc. = a/r = 63.9 rad/s2

Friction = T - mg sin angle = 4.25*19.183 - 8.5*9.8*sin 24 = 47.64 N

Us = Friction/mg cos angle = 0.626 .

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