Question: A hanging mass m = 8 2 kg is released from rest and accelerates downward. The winding drum A of mass ma = 6 5

A hanging mass m=82kg
is released from rest and accelerates downward. The winding drum A
of mass ma=65kg
with an effective diameter of \Phi a=440mm
and a radius of inertia ka=350mm
meshes with gear B
.
>>> Please note: Gear B
has a moment of inertia of 0.9 kgm2
.
Assume \omega b=5\omega a
, with subscript a
related to the winding drum and b
to gear B
.
Analyse the frictionless system and calculate:
The moment of Inertia of the drum Ia=mak2a=
Answer for part 1
7.9625
One possible correct answer is: 7.9625
kgm2
The linear acceleration a=
Answer for part 2
3.2632
One possible correct answer is: 1.1307713621037
m/s/s
The angular accelerations for the drum \alpha a=
Answer for part 3
14.833
One possible correct answer is: 5.1398698277439
rad/s/s
as well as the gear \alpha b=
Answer for part 4
74.159
One possible correct answer is: 25.69934913872
rad/s/s
(*)
The tension in the cord F1=
Answer for part 5
267.566
One possible correct answer is: 711.6967483075
N
.
Assume an observation period of 2 s
from the instant the system is released from rest and calculate the change of linear speed \Delta v=
Answer for part 6
6.526
One possible correct answer is: 2.2615427242073
m/s
and linear distance the hung mass moves \Delta h=
Answer for part 7
6.526
One possible correct answer is: 2.2615427242073
m
(*)
During the 2 s
observation, the following changes of energy with the correct sign can be computer:
Ek=
Answer for part 8
1746.136
One possible correct answer is: 2102.23847006
J
(Kinetic energy) and
Ep=
Answer for part 9
5249.645
One possible correct answer is: -1819.2301982069
J
(Potential Energy)
The perfect, frictionless system described above is replaced with a realistic one with a gear efficiency of 92%
.
Analyse the impact of the efficiency of the dynamics of the system and calculate:
The acceleration of the system under new conditions, a2=
Answer for part 10
One possible correct answer is: 1.0699713843583
m/s/s
,
the tension T=
Answer for part 11
One possible correct answer is: 716.68234648262
N
and
the number of revolutions of the drum during the same period of observation of 2 sec
: N=
Answer for part 12
One possible correct answer is: 1.5481021344318
rev

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