Answer with Explanation:
We are given that
Mass=3.01 kg
Radius,r=0.2 m
Angular frequency,[tex]\omega=5.97 rad/s[/tex]
a.Angular momentum,[tex]L=I\omega=\frac{1}{2}mr^2\omega=\frac{1}{2}(3.01)(0.2)^2(5.97)=0.36 kgm^2/s[/tex]
b.When the axis of rotation passes through a point midway between the ceneter and rim then moment of inertia,I=[tex]\frac{3}{4}mr^2[/tex]
Angular momentum,[tex]L=\frac{3}{4}(3.01)(0.2)^2(5.97)=0.55 kgm^2/s[/tex]
(a) The value of angular momentum when axis of rotation is passing through the center of mass is, [tex]0.36 \;\rm kg.m^{2}/s[/tex].
(b) The angular momentum for the rotation axis passing through a point midway between the center and rim is [tex]0.55 \;\rm kgm^{2}/s[/tex].
Given data:
The mass of solid disk is, m = 3.01 kg.
The radius of solid disk is, r = 0.200 m.
The angular frequency of disk is, [tex]\omega = 5.97 \;\rm rad/s[/tex].
(a)
The given problem is based on the concept of angular momentum. In rotational mechanics, the angular momentum is determined as the product of moment of inertia of the object and its angular frequency/angular velocity.
The mathematical expression is given as,
[tex]L = I \times \omega[/tex]
here, I is the moment of inertia.
For the axis of rotation passing through the center of mass, the moment of inertia is, [tex]I=\dfrac{1}{2}mr^{2}[/tex] .
Then the moment of inertia is,
[tex]L = (1/2 mr^{2}) \times \omega\\\\L = (1/2 \times 3.01 \times (0.200)^{2}) \times 5.97\\\\L =0.36 \;\rm kgm^{2}/s[/tex]
Thus, the required value of angular momentum when axis of rotation is passing through the center of mass is, [tex]0.36 \;\rm kg.m^{2}/s[/tex].
(b)
When the axis of rotation passes through a point midway between the center and the rim, then the moment of inertia is given as,
[tex]I'=(3/4)mr^{2}[/tex]
Now, calculate the angular momentum as,
[tex]L'=I' \times\omega \\\\L' = (3/4 mr^{2}) \times \omega\\\\L = (3/4 \times 3.01 \times (0.200)^{2}) \times 5.97\\\\L =0.55 \;\rm kgm^{2}/s[/tex]
Thus, we can conclude that the angular momentum for the rotation axis passing through a point midway between the center and rim is [tex]0.55 \;\rm kgm^{2}/s[/tex].
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