A uniform solid disk of mass m = 2.0 kg and radius R has a massless string wrapp

Question

A uniform solid disk of mass m = 2.0 kg and
radius R has a massless string wrapped around
it. You pull straight up on the string with a
force F = 9.0 N, which causes the disk to roll
without slipping on flat ground. The moment
of inertia of a solid disk, rotating about the
center, is ½ mR2. Use g = 10.0 m/s2.

(a) Complete the freebody diagram, showing all the forces acting on
the disk. If you draw a force of friction, label it
with either Fs for static friction or Fk for kinetic
friction.

(b) Apply Newton’s second law, and Newton’s second law for rotation, to come up
with some equations you will use in (c) and (d) to find the acceleration and the frictional force.

(c) Determine the acceleration of the disk. State both the magnitude and the direction.

(d) Determine the frictional force applied on the disk by the ground. State both the
magnitude and the direction.

(e) What is the minimum possible value of the coefficient of static friction between the disk and the ground, for the disk to roll without slipping under the conditions described above? You can express your answer as a ratio of integers, if you'd like.

Explanation / Answer

Torque = RF = I*a = I*a/R In this case, the disk turns around the contact point with the ground. And this contact point is in the distance R to the center of mass. The moment of inertia of the disk is then I(center of mass) + mR^2 (theorem of parallel axes) = 1/2*mR^2 + mR^2 = 1.5*mR^2. and a = a/R R*9N = (1.5*2*R^2)*(a/R) 9N = 1.5*2a a = 9/3 = 3 m/s^2 = lateral acceleration. ---------- The friction force points into the same direction as the acceleration. It is the only force acting in horizontal direction. Its magnitude is F = I*a/R = 1/2*mR^2/R*a/R = 1/2*ma = 1/2*2*3 = 3 N ---------- the friction force is F = µ(mg - T) with T = tension in the string = pulling force 3 N = µ(2*10 - 9) µ = 3/11 ---------------

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