# Force & Acceleration

## Homework Statement

A 10lb bar is pinned at its center O and connected to a torsional spring. The spring has a stiffness k = 5 lb ft/rad so that the torque develped is M = 5(theta) lb ft where theta is in radians. If the bar is released from rest when it is vertical at theta = 90 degrees, determine its angular velocity at the instand theta = 45 degrees

## Homework Equations

I(o) = .5mr^2
Moment = I(o)a where a is angular acceleration
w^2 = w(o)^2 + 2a(theta - theta(initial))

## The Attempt at a Solution

10lb = .3108 slug

I = .5(.3108)(1^2) = 1.554 sluf ft^2

Moment = Ia 5theta = 1.554a
a = 32.1750theta

w^2 = 0 + 2(32.1750)(.7854)(.7854 - 1.5708)

Should the inital theta be 0.

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LowlyPion
Homework Helper
First of all your I for the bar is incorrect.

I = 1/12*M*L2

You have used 1 for the length of the bar. Is that part of the given?

Sorry i forgot the picture. My text says r, which i assumed was radius.

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am i still inncorect

LowlyPion
Homework Helper
Your drawing shows L = 2

I = 1/12*(.3108)*22 = .3108/3 = .1036

LowlyPion
Homework Helper
Consider using the potential energy in the spring, and determining the rotational kinetic energy imparted by the time it reaches 1/2 θ

PE = 1/2*k*θ2 = 1/2*k*(θ/2)2 + 1/2*I*ω2

Simplifying I get:

ω2 = k*θ2/ (2*I)

w^2 = (k*theta^2)/(2I) where k = 5 lb ft, theta = .7854 radians, I = .1036