Suppose we wrap a string around the surface of a uniform cylinder of radius 38.0 cm that is supported by an axle passing through its center. This cylinder rotates without friction and we tie a mass to the end of the string and let it fall under the influence of gravity. As the weight falls, we measure the constant acceleration of the falling mass to be 9.0 m/s2. Find the angular acceleration that the cylinder will experience under these conditions. (Assume the positive direction is the initial direction of rotation of the cylinder. Indicate the direction with the sign of your answer.)

Angular acceleration = 23.68 rad / s²

Explanation:

Given that,

acceleration = 9m/s²

Therefore acceleration of string is 9m/s²

since string is constant in length

cylinder of radius 38.0 cm = 0.38m

Angular acceleration = a / r

Angular acceleration = 9 / 0.38

                                   = 23.68 rad / s²

Angular acceleration = 23.68 rad / s²

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 .