Find moment of inertia for a door

[XodoX]

1. A uniform, thin solid door has a height 2.20 m, width 0.870m, and mass of 23.0 kg. Find its moment of inertia for rotation on its hinges. Is any piece of data unneccessary?

2. A grinding wheel is in the form of a uniform solid disk of radius 7.00 cm and mass of 2 Kg. It starts from rest and accelerates uniformly under the action of the constant of torque of 0.6000 NxM that the motor exerts on the wheel.
How long does it take for the wheel to reach its final operating speed of 1200 rev/min ? Through how many revolutions does it turn while accelerating?





Would appreciate any help! Seems easy, but I can't figure it out..like the equation I need to use

 

[LowlyPion]

Looks like you need to start with what the definition is for the moment of inertia.

 

[nlightner]

I would approach this problem by first identifying the necessary data and parameters for calculating the moment of inertia for the door and the time and revolutions for the grinding wheel. In the first scenario, all of the given data seems necessary in order to accurately calculate the moment of inertia. The height, width, and mass of the door are all important in determining the distribution of mass and the distance from the axis of rotation, which are crucial factors in calculating moment of inertia.

To calculate the moment of inertia for the door, we can use the formula I = 1/12 * m * (h^2 + w^2), where m is the mass of the door, h is the height, and w is the width. Plugging in the values given, we get I = 1/12 * 23.0 kg * (2.20 m^2 + 0.870 m^2) = 1.89 kg*m^2. This is the moment of inertia for rotation on the door's hinges.

Moving on to the second scenario, we are given the radius and mass of the grinding wheel, as well as the torque applied by the motor and the final operating speed. To calculate the time it takes for the wheel to reach its final speed, we can use the equation ω = ω0 + αt, where ω is the final angular velocity (in radians per second), ω0 is the initial angular velocity (in radians per second), α is the angular acceleration (in radians per second squared), and t is the time (in seconds). We know that the final speed is 1200 rev/min, which is equivalent to 125.7 rad/s. The initial speed is 0 since the wheel starts from rest. And the angular acceleration can be calculated using the formula τ = Iα, where τ is the torque and I is the moment of inertia. Rearranging the equation, we get α = τ/I. Substituting the given values, we get α = (0.6000 N*m)(0.07 m)/(2 kg*m^2) = 0.021 rad/s^2. Plugging all of these values into the first equation, we get 125.7 rad/s = 0 + (0.021 rad/s^2)t. Solving for t, we get t = 5986 seconds.

To calculate the number of revolutions the wheel makes while