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:

A) 1.1 m/s, 0

B) 1.1 m/s, 19 rad/s

C) 1.1 m/s, 98 rad/s

D) 4.7 m/s, 78 rad/s

E) 5.9 m/s, 98 rad/s

The answer is D, please explain. Thank you.

Explanation / Answer

f = - uk m g

a = f /m = - uk g = -0.30 x9.8 = - 2.94 m/s^2

v = v0 + a t = 7 - 2.94 t

and torque = I alpha

r uk m g = (m r^2 / 2) alpha

alpha = 98 rad/s^2

w = alpha t

kinetic friction will act until, v = w r

7 - 2.94 t = 98 t x 0.06

t =0.794 sec

after which v and w will remain constant.

v = 7 - (0.794 x 2.94) = 4.7 m/s

w = 98 x 0.794 = 78 rad/s

Ans(D)

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