A disk of mass M = 2.0 kg and radius R = 0.10 m is supported by a rope of neglig

Question

A disk of mass M = 2.0 kg and radius R = 0.10 m is supported by a rope of negligible mass, as shown above. The rope is attached to the ceiling at one end and passes under the disk. The other end of the rope is pulled upward with a force FA . The rotational inertia of the disk around its center is MR^2/2 (a) Calculate the magnitude of the force FA necessary to hold the disk at rest. At time t = 0, the force FA is increased to 12 N, causing the disk to accelerate upward. The rope does not slip on the disk as the disk rotates. (b) Calculate the linear acceleration of the disk. (c) Calculate the angular speed of the disk at t = 3.0 s. (d) Calculate the increase in total mechanical energy of the disk from t = 0 to t = 3.0 s. (e) The disk is replaced by a hoop of the same mass and radius. Indicate whether the linear acceleration of the hoop is greater than, less than, or the same as the linear acceleration of the disk. _____Greater than _____ Less than _____ The same as Justify your answer.

Explanation / Answer

a) sum(F) = 0 and T = FA

T + FA -mg = 0 =====> FA = 0.5mg = 0.5*2*10 = 10 N

b) Sum(tau) = I*a

(FA - T)*R = 0.5M*R^2(a/R) ======> T = FA - 0.5*Ma

Also sum(F) = ma

T + FA -mg = ma ==========> (FA - 0.5*Ma) + FA -mg = ma =====> a = (2FA -mg)/1.5M

a = 1.33 m/s^2

c) wo = 0 and alpha = a/R

Then w = (a/R)*t = (1.33/0.1)*3 = 40 rad/s

d) v = rw = 4 m/s

Etot = 0.5*I*w^2 + 0.5*Mv^2 + Mgh = 0.5*0.5MR^2*(v/R)^2 + 0.5*Mv^2 + Mgh

Etot = (3/4)*Mv^2 + Mgh = 24 + 120 = 144 J

e) Ihoop = MR^2

  Ihoop > Icylinder (more M.I. means less a for the same tau = F)

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