Solve Force & Acceleration Homework

[joemama69]

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

90 degrees = 1.5708 radians
45 degrees = .7854 radians

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


w = 6.3004 rad/s


Should the inital theta be 0.

 

[LowlyPion]

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?

 

[joemama69]

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

 

[joemama69]

am i still inncorect

 

[LowlyPion]

Your drawing shows L = 2

This means your I is

I = 1/12*(.3108)*2² = .3108/3 = .1036

 

[LowlyPion]

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)

 

[joemama69]

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

w = 3.86 rad/s


why did u use 1/2 theta