Question

A cylindrical 6.00-kg reel with a radius of 0.60 m and a frictionless axle, starts from rest and speeds up uniformly as a 4.00-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 5.00s. (a) What is the linear acceleration of the falling bucket? (b) How far does it drop? (c) What is the angular acceleration of the reel?

Answer 1

we use newton's second law and apply it to the bucket and the rotational form to the reely 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/(0.5x6.0 kg + 4.00kg) a=1.4m/s/s standard kinematics tells you distance fell = 1/2 at^2= 1/2(1.4m/s/s)(5s)^2 =15m angular accel = a/R=1.4m/s/s / 0.6m = 2.333rad/s/s