Respuesta :
Answer:
The mass of the disk is 0.7179 kg.
Explanation:
Given that,
Radius = 0.453 m
Tangential force = 0.221 N
Angular acceleration = 0.111 rad/s²
We need to calculate the moment of inertia of both the disk and ring
Using formula of moment of inertia
For disk,
[tex]I=\dfrac{1}{2}mr^2[/tex]
For ring,
[tex]I=mr^2[/tex]
The net moment of inertia
[tex]I=\dfrac{1}{2}mr^2+mr^2[/tex]
[tex]I=1.5mr^2[/tex]
We need to calculate the torque
Using formula of torque
[tex]\tau=I\alpha[/tex]
[tex]Fd=1.5mr^2\times\alpha[/tex]
Put the value into the formula
[tex]0.221\times0.111=1.5\times m\times(0.453)^2\times0.111[/tex]
[tex]m=\dfrac{0.221\times0.111}{1.5\times(0.453)^2\times0.111}[/tex]
[tex]m=0.7179\ kg[/tex]
Hence, The mass of the disk is 0.7179 kg.
Moment of inertia is a quantitative measure of a body's rotational inertia . Mass is an important component in moment of inertia formulas. The mass of the disk will be 0.7179 Kg.
What is the moment of inertia?
Moment of inertia is a quantitative measure of a body's rotational inertia the body's resistance to having its speed of rotation about an axis is changed by the application of torque.
The given data in the problem is;
r is the radius = 0.453 m
F is the tangential force = 0.221 N
α is the angular acceleration = 0.111 rad/s²
The moment of inertia of the disk
[tex]\rm I = \frac{1}{2}mr^2[/tex]
The moment of inertia of the ring
[tex]\rm I = mr^2[/tex]
The total moment of inertia will be;
[tex]\rm I = \frac{1}{2}mr^2+\rm mr^2 \\\\ \rm I = 1.5 mr^2[/tex]
The torque is found by;
[tex]\rm \tau =i\alpha \\\\ Fd= 1.5 mr^2 \times \alpha \\\\ 0.221 \times 0.111 = 1.5 m(0.453)^2 \times 0.1111 \\\\ \rm m= \frac{0.221 \times 0.111}{1.5 \times (0.453)^2 \times 0.111} \\\\ \rm m= 0.7179\ Kg[/tex]
Hence the mass of the disk will be 0.7179 Kg.
To learn more about the moment of inertia refer to the link;
https://brainly.com/question/6953943