The angular acceleration is [tex]6.25 rad/s^2[/tex]
Explanation:
To solve this problem we can use the equivalent of Newton's second law for rotational motions:
[tex]\tau = I \alpha[/tex] (1)
where
[tex]\tau[/tex] is the torque acting on the body
I is the moment of inertia of the body
[tex]\alpha[/tex] is the angular acceleration
In this problem we have:
[tex]\tau = 450 Nm[/tex] is the torque
The moment of inertia of a solid cylinder about its axis is
[tex]I=\frac{1}{2}MR^2[/tex]
where
M = 100 kg is the mass
R = 1.2 m is the radius
Substituting,
[tex]I=\frac{1}{2}(100)(1.2)^2=72 kg m^2[/tex]
And solving eq.(1) for [tex]\alpha[/tex], we find the angular acceleration:
[tex]\alpha = \frac{\tau}{I}=\frac{450}{72}=6.25 rad/s^2[/tex]
Learn more about rotational motions:
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