- #1

gus_lyon

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## Homework Statement

A long slender rod has a mass of 0.66 kg and a length of 648.0 mm. The friction torque at the hinge is 0.41 Nm.

What is the angular acceleration of the rod when it is 38 degrees below the horizontal?

## Homework Equations

## The Attempt at a Solution

Torque = moment of inertia x angular acceleration

Moment of inertia for a rod = (1/12)*m*(l^2)

= (1/12)*0.66*(0.648^2) = 0.0231 Kgm^2

Torque = Radius * Force

= (0.648*cos(38))*(9.81*0.66) = 3.306 Nm

Angular acceleration = Torque/Moment of inertia

= 3.306/0.0231 = 143.12

Which is wrong.