A cylinder of radius R = 6.0 cm is on a rough horizontal surface. The coefficien

Question

A cylinder of radius R = 6.0 cm is on a rough horizontal surface. The coefficient of kinetic friction between the cylinder and the surface is 0.30 and the rotational inertia for rotation about the axis is given by MR2/2, where M is its mass. Initially it is not rotating but its center of mass has a speed of 7.0 m/s. After 2.0 s the speed of its center of mass and its angular velocity about its center of mass, respectively, are:

1.1 m/s, 0 1.1 m/s, 19 rad/s 1.1 m/s, 98 rad/s 1.1 m/s, 200 rad/s 5.9 m/s, 98 rad/s

Explanation / Answer

the torque acting on the cylinder ? = I * ? or F * R = I * ? or ?kMg * R = I * ? or ? = (?kMg * R/I) =(?kMg * R/MR^2/2) or ? = (2?kg/R) ---------------(1) ?k= 0.30,g = 9.8 m/s^2 and R = 6.0 cm = 6.0 *10^-2 m the initial angular velocity of the cylinder wo= (v/R) v = 7.0 m/s the angular velocity about its center of mass after t = 2.0 sis w = wo + ?t the value of ? is obtained from equation (1) the velocity of its center of mass V = R * w

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