Rolling motion of an unbalanced disk

[NoYou]

Homework Statement

The mass center G of a 5kg wheel of radius R = 300 mm is located at a distance r = 100 mm from its geometric center C. The centroidal radius of gyration is k = 150 mm. As the wheel rolls without sliding, its angular velocity varies and it is observed that it is = 8 rad/s in position shown. Determine the corresponding angular acceleration of the wheel.

Homework Equations

Ac=Ag=Ac + Ag/c = Ac + (Ag/c)tangent + (Ag/c) Normal
I= mk^2
a=r$$\alpha$$

The Attempt at a Solution

 

[Nate810]

I= 5 kg(150mm)^2 = 1125 kg m^2a=r\alpha = 100 mm (8 rad/s) = 800 rad/s^2Ac = Ag = I \alpha = (1125 kg m^2)(8 rad/s) = 9000 rad/s^2Ag = Ac + (Ag/c) tangent + (Ag/c) Normal = 9000 rad/s^2 + (Ag/300mm)Tangent + (Ag/300mm)NormalAg = 9000 rad/s^2 + (Ag/300mm)(cos30) + (Ag/300mm)(sin30) = 9000 rad/s^2 + (Ag/3)Ag = 9000 rad/s^2 + 3000 rad/s^2 = 12000 rad/s^2 Therefore, the angular acceleration of the wheel is 12000 rad/s^2

 

[kerimek]

To determine the angular acceleration, we can use the formula a=r\alpha, where a is the linear acceleration, r is the radius of the wheel, and \alpha is the angular acceleration. We know that the linear acceleration is equal to the product of the angular velocity and the radius of gyration (a=\omega k), so we can rewrite the formula as \alpha=\frac{a}{k}.

Using the given values, we can calculate the linear acceleration as a=\omega^2 r=\frac{(8\text{ rad/s})^2(100\text{ mm})}{150\text{ mm}}=42.67\text{ mm/s}^2.

Substituting this into the formula for angular acceleration, we get \alpha=\frac{42.67\text{ mm/s}^2}{150\text{ mm}}=0.2845\text{ rad/s}^2.

Therefore, the corresponding angular acceleration of the wheel is 0.2845 rad/s^2. This means that the wheel is accelerating at a rate of 0.2845 radians per second squared, which will cause its angular velocity to increase over time.