The figure below (Figure 1) illustrates an Atwood\'s machine. Let the masses of
ID: 1453393 • Letter: T
Question
The figure below (Figure 1) illustrates an Atwood's machine. Let the masses of blocks A and B be 3.50 kgand 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.400 kgm2 and the radius of the wheel be 0.110 m .
A - Find the linear acceleration of block A if there is no slipping between the cord and the surface of the wheel.
B - Find the linear acceleration of block B if there is no slipping between the cord and the surface of the wheel.
C - Find the angular acceleration of the wheel C if there is no slipping between the cord and the surface of the wheel.
D - Find the tension in left side of the cord if there is no slipping between the cord and the surface of the wheel.
E - Find the tension in right side of the cord if there is no slipping between the cord and the surface of the wheel.
Explanation / Answer
given that
mA = 3.5 kg
mB = 2 kg
r = 0.110 m
moment of inertia I = 0.400 kg*m^2
assume that
block A is the left block and block B is the right block.
The weight of block A is greater than the weight of block B. So block A will move downward and block B will move
upward.
The tension in the left side of the cord is greater than the tension in the right side of the cord, because the weight of
block A is greater than the weight of block B.
The tension in the left side of the cord is causing the pulley to rotate counter clockwise.
The tension in the right side of the cord is causing the pulley to rotate clockwise.
we know that
torque = tension * radius
Counter clockwise torque = T1 * 0.110
Clockwise torque = T2 * 0.110
Net counterclockwise torque = (T1 – T2) * 0.110
Net counterclockwise torque = I * alpha
alpha = angular acceleration
Net counterclockwise torque = 0.400 * alpha
(T1 – T2) * 0.110 = 0.4 * alpha
(T1 – T2) = 3.63 * alpha
Angular acceleration alpha = Linear acceleration a / r
alpha = a ÷ 0.110
(T1 – T2) = 3.63 * a / 0.110
(T1 – T2) = 33 * a ......eq1
For block A
(3.5 * 9.8) – T1 = 3.5 * a
T1 = 34.3 – 3.5 * a
For block B
T2 – (2.0 * 9.8) = 2.0 * a
T2 = 19.6 + 2 * a
T1 – T2 = (34.3 – 3.5 * a) – (19.6 + 2 * a)
T1 – T2 = 14.7 – 5.5 * a .......eq2
from equation 1 & 2
33 * a = 14.7 – 5.5 * a
38.5 * a = 14.7
a = 0.38 m / s^2
T1 = 34.3 – 3.5 * 0.38 =32.97
T2 = 19.6 + 2 * 0.38 = 20.36
(T1 – T2) = 33* a
32.97- 20.36 = 33 * a
a = 0.38 m/s^2
angular acceleration = a / r
angular acceleration = 0.38 / 0.110 = 3.47 rad / s^2
answer
(a) linear acceleration of block A = 0.38 m/s^2
(b) linear acceleration of block A = 0.38 m/s^2
(c) angular acceleration of the wheel C = 3.47 rad / s^2
(d) tension in left side of the cord T1 = 32.97 N
(e) tension in right side of the cord T2 = 20.36 N
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