Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The angular acceleration for first drum [tex]\alpha = 0.792 rad/s^2[/tex]
The angular acceleration for the second drum is [tex]\alpha =1.262[/tex]
Explanation:
From the question we are told that
Their radius of the drum is [tex]r = 13 in = \frac{13}{12} ft = 1.083ft[/tex] each
The weight is [tex]W = 210 lb[/tex]
The mass is [tex]M = \frac{210 lb}{32.2 ft /s^2} = 6.563\ lb s^2 ft^{-1}[/tex]
Their radius of gyration is [tex]z=30 in= \frac{30 }{12} = 2.5 ft[/tex]
The free body diagram of a drum and its hub and 30lb and in the case the weight is connect to the hub separately is shown on the second uploaded image
The T in the diagram is the tension of the string
Now taking moment about the center of the the drum P we have
[tex]\sum M_p = I_p \alpha[/tex]
=> [tex]T * r = Mz^2 * \alpha[/tex]
Where r is the radius ,z is the radius of gyration about the center O , M is the mass of the drum including the hub, and [tex]\alpha[/tex] is the angular acceleration
Inputting
[tex]T * 1.083 = 6.563 * 2.5^2 \alpha[/tex]
=> [tex]T = 37.87\alpha[/tex]
Considering the force equilibrium in the vertical direction (Looking at the second free body diagram now )
The first on is
[tex]\sum F_y = ma[/tex]
=> [tex]30lb - T = m(r \alpha )[/tex]
Where m is the mass of the hanging block which has a value of
[tex]m = \frac{30lb}{32.2 ft/s^2} = 0.9317 \ lb ft^{-1} s^2[/tex]
a is the acceleration of the hanging block
inputting values we have
[tex]30- 37.87 \alpha = 0.9317* 1.083 \alpha[/tex]
[tex]30 = 37.87\alpha + \alpha[/tex]
[tex]\alpha = \frac{30}{38.87 }[/tex]
[tex]\alpha = 0.792 rad/s^2[/tex]
So the angular acceleration for first drum [tex]\alpha = 0.792 rad/s^2[/tex]
The free body diagram of a drum and its hub when the only on the string is 30lb is shown on the third uploaded image
So here we would take the moment about O
[tex]\sum M_o = I_O \alpha[/tex]
So [tex]\sum M_o = 30* 1.083[/tex]
and [tex]I = M z^2[/tex]
Therefore we will have
[tex]30 * 1.083 = (Mz^2 )\alpha[/tex]
inputting values
[tex]30 * 1.083 = 6.563 * 2.5^2 \alpha[/tex]
[tex]32.49=41.0\alpha[/tex]
[tex]\alpha =\frac{41}{32.49}[/tex]
[tex]\alpha =1.262[/tex]
So the angular acceleration for the second drum is [tex]\alpha =1.262[/tex]
