A 4.00-kg cylindrical reel with a radius of 0.600 m and a frictionless axle, sta

Question

A 4.00-kg cylindrical reel with a radius of 0.600 m and a frictionless axle, starts from rest and speeds up uniformly as a 2.00-kg bucket falls into a well, making a light rope unwind from the reel (Fig. 8.36). The bucket starts from rest and falls for 4.00 s.

Figure P8.36
(a) What is the linear acceleration of the falling bucket?



Your response differs from the correct answer by more than 100%. m/s2 downward
(b) How far does it drop?
m
(c) What is the angular acceleration of the reel?



Your response differs from the correct answer by more than 10%. Double check your calculations. rad/s2

Explanation / Answer

Given, mass of cylinder, M = 4.0kg; Radius,R = 0.60 m;           mass of bucket, m = 2.0 kg Cosider the motion of the bucket alone,                                  mg - T = m.a ---------(1) For motion of the cylinder, Torque applied, =T.R = I. The moment of inertia, I = 0.5M.R2 Since, ang velocity, = v/R Angular acceleration,                           = d/dt =(1/R)dv/dt = a/R Hence, = T.R = I. =I.a/R
                            = 0.5 MR2.a/R = 0.5Ma.R                       T = 0.5M.a --------------(2) Adding (1) and (2)                       mg = (m + 0.5M) a (a) Acceleration,                       a = mg/(m + 0.5M)                          = 2.0x9.8/4.0 = 4.9 m/s2 (b) S = 0 + 0.5a.t2 = 0.5x4.9x16= 39.2 m. (c) Angular Acceleration, = a/R = 4.9/0.6 =8.17 rad/s2

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