Answer: The angular acceleration of the wheel is 47.43 rad/s² and the acceleration of the part of the cord that has already been pulled off the wheel is 10.43 m/s².
Explanation:
To find the angular acceleration of the wheel, we can use the torque equation:
τ = Iα, where:
The torque applied can also be calculated as the product of the force applied and the radius of the object. This is given by the equation:
τ = Fr, where:
Setting these equations equal to each other:
Iα = Fr
We can rewrite the moment of inertia (I) as:
I = [tex]\frac{1}{2} mr^{2}[/tex], since it is the moment of inertia for a solid disk rotating about an axis through its center perpendicular to the plane of the disk.
[tex]\frac{1}{2} mr^{2}\alpha = Fr[/tex]
Plugging in our given values, we get:
[tex]\frac{1}{2} (9.2)(0.22)^{2}\alpha = (48)(0.22)[/tex]
[tex](9.2)(0.22)^{2}\alpha = (2)(48)(0.22)[/tex]
[tex]\alpha = \dfrac{(2)(48)(0.22)}{(9.2)(0.22)^{2}}[/tex]
α = 47.43
The angular acceleration of the wheel is 47.43 rad/s².
To calculate the linear acceleration (a) of the part of the cord that has been pulled off the wheel, we can convert the angular acceleration to linear acceleration. This is given by the formula:
a = α·r
a = 47.43·0.22
a = 10.43
The acceleration of the part of the cord that has already been pulled off the wheel is 10.43 m/s².
Learn more about angular acceleration here: https://brainly.com/question/13014974
