GOAL Combine Newton\'s second law with its rotational analog. PROBLEM A solid, f
ID: 1494854 • Letter: G
Question
GOAL Combine Newton's second law with its rotational analog. PROBLEM A solid, frictionless cylindrical reel of mass M = 3.00 kg and radius R = 0.400 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. (a) Find the tension T in the cord and acceleration a 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? mg STRATEGY This problem involves three equations and three unknowns. The three equations are Newton's second law applied to the bucket, ma = Fú the rotational version of the second law applied to the cylinder, , and the relationship The tension produces a torque on the cylinder about its axis between linear and angular acceleration, a - ra, which connects the dynamics of the bucket and cylinder. The three unknowns are the acceleration of 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 mg (a) A water bucket attached to a rope passing over a frictionless reel. (b) A free-body diagram for the bucket. (c) of rotation. (d) A falling cylinderExplanation / Answer
M =3 kg , m= 2kg , R = 0.4 m
Net fore along vertical direction on buket
mg -T = ma ...(1)
Net force along vertical direction on reel
T +T = Ma
T = (1/2) Ma
from (1)
mg - (1/2)Ma = ma
mg = a (M/2 +m)
a = mg/(M/2 +m )
a = (2*9.8)/(3/2+2)
a = 5.6 m/s^2
from (1)
(a) 2*9.8 - T = 2*5.6
T = 8.4 N
(b) t = 3 s
u =0
from kinematic equation
s = ut+(1/2)at^2
s = 0 +(0.5*5.6*3*3)
s= 25.2 m
a = mg/(M/2 +m )
if M is increased, then a is decreased.
If M is increased, T is increased
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