# Linear acceleration and rotational motion

## Homework Statement

You are the technical consultant for an action-adventure film in which a stunt calls for the hero to drop off a 20.0-m-tall building and land on the ground safely at a final vertical speed of 4.00 m/s. At the edge of the building’s roof, there is a 100-kg drum that is wound with a sufficiently long rope (of negligible mass), has a radius of 0.500 m, and is free to rotate about its cylindrical axis with a moment of inertia I. The script calls for the 50.0-kg stuntman to tie the rope around his waist and walk off the roof.

a.Determine an expression for the stuntman’s linear acceleration in terms of his mass m, the drum’s radius r, and moment of inertia I.

## The Attempt at a Solution

So far I'm unable to find an adequate relationship for part a.
So far I have: T-mg =-ma and τ = Tr = Iσ = I$\frac{a}{r}$, where σ is the angular acceleration, T is the tension, and τ is the torque caused by the tension.
Now if I combine these equations for tension I obtain an expression in terms of g (which doesn't belong as indicated by the question).
Any ideas what I'm doing wrong or neglecting?

Last edited:

TSny
Homework Helper
Gold Member
Even though g is not mentioned in the question, your answer will involve g.

Even though g is not mentioned in the question, your answer will involve g.
Then I would obtain a= $\frac{mg}{m+\frac{I}{r^{2}}}$
Could you maybe comment on the validity of this statement..?
(I'm not sure my two initial equations are valid --mainly the signs)

TSny
Homework Helper
Gold Member
And I just had another quick question, another part of the question asks to calculate the moment of inertia of the drum about its axis. Now the question doesn't seem to give us a formula for moment of inertia, so naturally I believe I would calculate it from the above equation after finding the linear acceleration from a = $\frac{v^{2}}{2h}$ where v = 4 m/s and h = 20 m. Would this be the right approach or can one determine the moment of inertia of the drum in another way?