Answer: The magnitude of the angular acceleration of each drum is [tex]0.585 rad/s^{2}[/tex].
Explanation:
We know that,
[tex]\sum M_{o} = I_{o} \alpha[/tex]
[tex]T \times r = MK^{2} \times \alpha[/tex]
where, r = hub radius, M = mass of the drum,
K = radius of gyration, [tex]\alpha[/tex] = angular gyration
[tex]T \times r = MK^{2} \times \alpha[/tex]
[tex]T \times \frac{8}{12} = \frac{225}{32.2} \times \frac{23}{12}^{2} \times \alpha[/tex]
[tex]T \times 0.66 = 6.98 \times 3.67 \times \alpha[/tex]
T = [tex]38.81 \alpha[/tex] ........ (1)
Now, [tex]\sum F_{y} = ma[/tex]
23 lb - T = m([tex]r \alpha[/tex])
[tex]23 - 38.81 \alpha = \frac{23}{32.2} \times \frac{8}{12} \times \alpha[/tex]
[tex]23 - 38.81 \alpha = 0.714 \times 0.66 \alpha[/tex]
[tex]23 - 38.81 \alpha = 0.471 \alpha[/tex]
23 = [tex]39.281 \alpha[/tex]
[tex]\alpha = 0.585 rad/s^{2}[/tex]
Therefore, we can conclude that the magnitude of the angular acceleration of each drum is [tex]0.585 rad/s^{2}[/tex].
The magnitude of the angular acceleration of each drum = 0.585 rad/s²
Given data:
hub radius ( r ) = 0.66
mass of drum ( M ) = 6.98
radius of gyration ( K ) = 1.92
angular gyration ( ∝ ) = ?
To determine the magnitude of the angular acceleration we will apply the equation below
T * r = MK² * ∝ ----- ( 1 )
where ; r = 0.66 , M = 6.98, K = 1.92, ∝ = angular
Insert values into equation ( 1 )
T = 38.881 ∝
∑Fy = ma ; -- ( 2 ) Fy = 23 Ib - T, a = r∝
Equation ( 2 ) becomes
23 - T = m ( r∝ )
= 23 - 38.81∝ = [ 23 / 32.2 ] * ( 8/12 * ∝ )
∴ 23 = 39.281 ∝
∝ ( magnitude of angular acceleration ) = ( 23 / 39.281 ) = 0.585 rad / s²
Hence we can conclude that The magnitude of the angular acceleration of each drum = 0.585 rad/s² .
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