Use the worked example above to help you solve this problem. A solid, frictionle

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Use the worked example above to help you solve this problem. A solid, frictionless cylinder reel of mass M= 3.15 kg and radius  R = 0.410 m is used to draw water from a well (see Figure (a)). A bucket of mass m= 1.99 kg is attached to a cord that is wrapped around the cylinder. (Indicate the direction with the sign of your answer.)

Please answer a and b in Practice Section. Also, the exercise section a and b. I really appreciate it. I'll give you a great rating for all correct answers

i safari File Edit View History Bookmarks Window Help Q27% DE Mon 3:45:01 PM webassign.net | 1.5419.09 points IPrevicus Answers SerCP9 8AE.011 My Notes As EXAMPLE 8.11 The Falling Bucket GOAL Combine Newton's second law with Its rotational analog. PROBLEM A solid, frictionless cylindrical reel of mass M 3.00 kg and radius R0.400 m is used to draw water from a well (Figure (a)). A bucket of mass m 2.00 kg is attached to a cord that is wrapped around the cylinder ( Find the tension T in the cord and acceleration of the bucket, (b) If the bucket starts from rest at the top of the well and falls for 3.00 s before hitting the water, how far does it fall? STRATEGY This problem involves three equations and three unknowns. The three equations are Newton's second law applied to the bucket, ma = Fi the rotational version of the second law (a) A water bucket attached to a rope pasma over a applied to the cylinder, 1- ; and the relationship (c) The tension produces a torque on the cylinder about its between linear and angular acceleration, a which connects the dynamics of the bucket and cylinder. The three unknowns are the acceleration aof the bucket, the angular acceleration of the cylinder, and the tension T in the rope. Assemble the terms of the three equations and solve for the three unknowns by substitution. Part (b) is a review of kinematics. frictionless reel. (b) A free-bodY diagram, for the budet. axis of rotation. (d) A falling cyinder SOLUTION (A) Find the tension in the cord and the acceleration of the bucket. (1) ma-mg T Apply Newton's second law to the bucket in Figure (b). There are two forces: the tension T acting upwards and gravity mg acting downwards. = 1 = YZMR2 (solid cylinder) Apply -1 to the cylinder in Figure (c) Notice the angular acceleration is (2) -TR-½MR2

Explanation / Answer

On pulley,

torque = I alpha

R T = (M R^2 / 2)(a / R)

T = 3.15 a / 2 = 1.575 a

on Bucket: 1.99g - T = 1.99 a

1.99g - 1.575a = 1.99 a

a = (1.99 x 9.8) / (1.99 + 1.575)

a = 5.5 m/s^2 ......Ans

T = 1.575a = 8.6 N ...Ans

(B) d = v0 t + a t^2 /2 = 0 + (5.5 x 3.38^2 /2)

d = 31.4 m

EXERCISE:

on clylinder,

Fnet = m g - T = m a

and torque = I alpha

R T = (M R^2)(a/R)

T = m a

putting in equation,

m g - m a = m a

a = g / 2 = 4.9 m/s^2 ......Ans(a)

(v) d = 1 m

vf^2 - vi^2 =2 a d

v^2 - 0 = 2(4.9)(1)

v = 3.13 m/s

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