In our lab, we roll a non-uniform but cylindrically symmetric cylinder down a ro

Question

In our lab, we roll a non-uniform but cylindrically symmetric cylinder down a rough 20° incline. The cylinder rolls without slipping and has mass of 6 kg, radius 1.2 m, and moment of inertia about its center of mass 7.3 kg m2.

1-What is the acceleration of the cylinder's center-of-mass?

m/sec 2

2-What minimum coefficient of friction is required so that there is no slipping?

3-What is the ratio of kinetic energy of the center-of-mass to kinetic energy about the center-of-mass? (KEtrans/KErot)

4-Starting from rest, the cylinder rolls without slipping a distance d = 3.5 m down the incline. What is the final translational speed of the center of mass?

Explanation / Answer

1) Let Ff be the friction exerted by the incline on the cylinder and 'a' be the acceleration of the centre of mass of the cylinder.

Then, Net torque on the cylinder, T = FfR = I(a/R)

=> Ff = Ia/R2

Force equation:

Mg(sin 20o) - Ff = Ma

=> Mg(sin 20o) - Ia/R2 = Ma

=> a = Mg(sin 20o) / (M + I/R2) = [6 * 9.81 * (sin 20o)] / (6 + 7.3/1.22) = 1.82 m/s2

2) For minimum coefficient of friction, Ff = maximum friction

So, Ff = µN = µMg(cos 20o) = Ia/R2

=> µ = Ia / [Mg(cos 20o)R2] = (7.3 * 1.82) / [6 * 9.81 * (cos 20o) * 1.22] = 0.167

3) Let v be the speed of centre of mass.

Then, KEtrans = Mv2/2

KErot = I2/2 = Iv2/2R2

So, KEtrans/KErot = MR2/I = (6 * 1.22) / 7.3 = 1.18

4) Final translational speed (v) is given by the following kinematic equation:

v2 = u2 + 2ad

=> v = (u2 + 2ad)1/2 = [0 + (2 * 1.82 * 3.5)]1/2 = 3.57 m/s

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