A cylindrical 4.00-kg reel with a radius of 0.60 m and a frictionless axle, starts from rest and speeds up uniformly as a 3.00-kg bucket falls into a well, making a light rope unwind from the reel (Fig. P8.36). The bucket starts from rest and falls for 6.00s. (a) What is the linear acceleration of the falling bucket?

Answer: a=2.45m/s/s

Step-by-step explanation:

By newton's second law

The forces on the bucket are the tension in the string and the weight of the bucket; the bucket accelerates down, so we have

T - mg = - ma (eq. 1) (m is mass of bucket)

the tension exerts a torque on the reel of magnitude TR where R is the radius of the reel; this torque causes an angular acceleration A, such that

torque = I A

the moment of inertia of a cylindrical disk is 1/2 MR^2 where M is the mass of the reel, so we have

torque = T R = I A = 1/2 MR^2 A

or T=1/2 MR A

the angular acceleration is related to linear acceleration via

a = RA, so we get

T=1/2 MR (a/R) = 1/2Ma

use this in equation (1) to get

T-mg=-ma

1/2Ma-mg=-ma

a (1/2M + m) = g

a=g / (1/2 M + m)

using relevant values, we get:

a=9.8m/s/s / (0.5x4 kg + 2.00kg)

a=2.45m/s/s