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Force & Acceleration

  • Thread starter joemama69
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
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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?
 
  • #3
399
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Sorry i forgot the picture. My text says r, which i assumed was radius.
 

Attachments

  • #4
399
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am i still inncorect
 
  • #5
LowlyPion
Homework Helper
3,090
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Your drawing shows L = 2

This means your I is

I = 1/12*(.3108)*22 = .3108/3 = .1036
 
  • #6
LowlyPion
Homework Helper
3,090
4
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)
 
  • #7
399
0
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
 

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