- #1

xaer04

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## Homework Statement

From the text: "A [itex]50 kg[/itex] mass is tied to a massless rope wrapped around a solid cylindrical drum. The drum is mounted on a frictionless horizontal axle. When the mass is released, it falls with acceleration [itex]a = 3.7 \frac{m}{s^2}[/itex]. Find (a) the tension in the rope and (b) the mass of the drum.

Mass of the falling object = [itex]50 kg[/itex]

Acceleration of gravity = [itex]9.8 \frac{m}{s^2}[/itex]

Net acceleration of falling object = [itex]3.7 \frac{m}{s^2}[/itex]

## Homework Equations

Newton's 2nd Law

[tex]\sum \vec{F} = m \vec{a}[/tex]

Definiton of torque in terms of radius and Force applied

[tex]\tau = r \vec{F} \sin (\theta)[/tex]

Definition of torque in terms of rotational intertia and rotational acceleration

[tex]\tau = I \alpha[/tex]

Definition of rotational acceleration in terms of radius and tangential acceleration

[tex]\alpha = \frac{a}{r}[/tex]

Rotational intertia for a cylinder

[tex]I = \frac{1}{2} MR^2[/tex]

## The Attempt at a Solution

I know that [itex]a_\textit{gravity} - a_\textit{rope} = a_\textit{final}[/itex] for the falling block. this gave me an upward acceleration of [itex]6.1 \frac{m}{s^2}[/itex], which gave me my first answer:

[tex] T = -305N[/tex]

For my second answer, i set the definitons of torque in terms of quantities that i knew and the one quantity that i didn't know - mass of the falling object, mass of the drum, acceleration of the falling object, force of the falling object.

My equation looked like this before i started cancelling out things:

[tex] (\frac{1}{2}Mr^2) (\frac {a}{r}) = (rma)[/tex]

The radius values and accelerations canceled out, leaving me with this:

[tex] M = 2m [/tex]

This can't be right because the mass of the larger object can't be entirely dependant on the mass of the smaller object. Could someone help me out?

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