A solid cylinder of mass m and radius R has a string wound around it. A person holding the string pulls it vertically upward such that the cylinder is suspended in midair for a brief time interval (change in)t and its center of mass does not move. The tension in the string is T, and the rotational inertia of the cylinder about its axis is (1/2)mR2. What is the linear acceleration of the person's hand during the time interval change in t?

α = 2T/mR

Explanation:

We know torque τ = Iα = TR where I = rotational inertia and α = linear (tangential) acceleration, T = tension in string and R = radius of cylinder.

Now T = tension in string and I = (1/2)mR²

So  τ = Iα = TR

α = TR/I

substituting the values of the variables, we have

α = TR/(1/2)mR²

α = 2TR/mR²

α = 2T/mR

So, the linear acceleration of the person's hand on time, t is α = 2T/mR

1/0.14nm = 10^10/1.4 a lot

d. air pressure decreases as altitude increases.