A sanding disk with rotational inertia 1.2 x 10⁻³ kg. m² is attached to an electric drill whose motor delivers a torque of 16 N. m. Find (a) The angular momentum and (b) the angular speed of the disk 33 ms after the motor.

The angular momentum of the sanding disk is 0.528 kg.m^2.s^-1 and its angular speed is 440 s^-1

Explanation:

Given:

Torque ( ) delivers by the electric motor =

Time taken = 33 ms (milliseconds) =

Moment of Inertia,MOI ( ) =

We have to find the angular momentum () and angular speed ().

Notes:

Angular momentum in terms of torque and time ,

And

Angular momentum in terms of MOI and speed,

So,

a) Angular momentum.

⇒

⇒

⇒

The angular momentum of the sanding disk is 0.528 kg.m^2.s^-1

b)Angular speed.

⇒  

⇒

Plugging the values of and MOI

⇒

⇒

The angular speed of the disk is 440 s^-1

So the angular momentum of the disk is and the angular speed is .

beta is electron from nucleus

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we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .