How to find the angular acceleration?

[Elson]

Guys i need some help here. Does anyone know how to find the angular acceleration?

Gear A (radius = 150mm, mass= 4kg and radius of gyration = 100mm) and gear B(radius = 450mm, mass= 10kg and radius of gyration= 300mm) rotate about their respective centrers. A constant clockwise torque of 0.75Nm is applied to the gear A. neglecting bearing friction torque, determine the angular acceleration of each gear and the tangential contact forces between the gears.

For Gear A, i used
1)torque = I(moment of inertia) x a(angular acceleration)
0.75 = (1/2)(4)(0.01) x a, a = 0.75/0.02, a=37.5?

2)torque = force times radius = ma times radius ->mra times radius.
torque = 2I times angular acc? For gear A, 0.75= 2(0.02) times angular acc. Therefore i get angular acc = 0.75/0.04 = 18.75?

The correct answer should be 5.4.

 

[AvroArrow]

The equation for angular acceleration is Torque/Moment of Inertia. Moment of Inertia is mass times radius of gyration squared (I = m * r^2). So your equation should be:a = T / (m * r^2) For Gear A, the equation would be:a = 0.75 / (4 * 0.01^2) = 5.4

 

[m2003]

To find the angular acceleration of a rotating object, you can use the formula:
angular acceleration = torque / moment of inertia
In this case, the torque is given as 0.75Nm and the moment of inertia can be calculated using the formula:
moment of inertia = mass x radius of gyration^2
For Gear A, the moment of inertia would be:
I = 4kg x (0.1m)^2 = 0.04kgm^2
Therefore, the angular acceleration of Gear A would be:
a = 0.75Nm / 0.04kgm^2 = 18.75rad/s^2

For Gear B, the moment of inertia would be:
I = 10kg x (0.3m)^2 = 0.9kgm^2
Therefore, the angular acceleration of Gear B would be:
a = 0.75Nm / 0.9kgm^2 = 0.833rad/s^2

To determine the tangential contact forces between the gears, you can use the formula:
torque = force x radius
For Gear A and Gear B, the tangential contact force would be the same, as they are in contact and have the same radius of rotation (0.15m).
Therefore, the tangential contact force would be:
F = 0.75Nm / 0.15m = 5N

It is important to note that these calculations are based on ideal conditions and neglecting factors such as bearing friction torque. In a real-world scenario, the values may differ.