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 7.0m/s. After 2.0 s the speed of its center of mass and its angular

velocity about its center of mass, respectively, are:

A. 1.1m/s, 0

B. 1.1m/s, 19 rad/s

C. 1.1m/s, 98 rad/s

D. 1.1m/s, 200 rad/s

E. 5.9m/s, 98 rad/s

Explanation / Answer

frictional force,f = 0.3*m*g

deceleration = 0.3*g = 2.94m/s^2

frictional torque = f*R = I* angular acceleration

angular acceleration = f*R/(M*R^2/2) = 2*0.3*g/R =98rad/s^2

Vcom = u-a*t = 7- 2.94*2 = 1.12m/s

Angular velocity = 98*2 = 196 rad /s

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