Solving for Force & Fly-Off Time on Rotating Mass in Cup

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SUMMARY

The discussion focuses on calculating the acceleration and force acting on a stone of mass 100 kg placed at the end of a rotating arm of length 3 m, with an angular acceleration of θ (double dot) = 5 rad/s² cos(t). The key equations used include a(p) = (r(double dot) - r*θ(dot)²)e(r) + (r*θ(double dot) + 2r(dot)θ(dot))e(θ). The user initially calculated the acceleration at θ = π/4 rad incorrectly as -14.8e(r) + 2.34e(θ), while the correct answer is -21.71e(r) + 12.64e(θ). The error was identified as a mistake in integration, specifically neglecting the cosine factor.

PREREQUISITES
  • Understanding of angular motion and acceleration
  • Familiarity with vector notation and unit vectors e(r) and e(θ)
  • Basic knowledge of calculus, particularly integration
  • Ability to apply Newton's laws to rotational dynamics
NEXT STEPS
  • Study the principles of rotational dynamics in physics
  • Learn about vector calculus and its application in physics
  • Explore the integration techniques relevant to angular motion
  • Investigate the effects of angular acceleration on objects in circular motion
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Physics students, mechanical engineers, and anyone involved in dynamics and rotational motion analysis will benefit from this discussion.

thoughtclaw
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1. The stone of mass m=100 kg is placed inside the cup at the end of a rotating arm of length l=3 m. If arm is released from rest at theta=0 and its angular acceleration is given as theta (double dot) = 5 rad/s2 cos t (a) find the acceleration of the stone with theta=pi/4 rad; (b) what is the force acting on the stone from the cup at this instant (include gravity); (c) at what time does the stone fly off the arm?



2. I'm thinking the equation to start with is a(p)=(r(double dot)-r*theta(dot)2)e(r) + (r*theta(double dot)+2r(dot)theta(dot))e(theta)



3. Using the above equation, with the assumption that r(dot) and r(double-dot) are zero, I found that the acceleration at pi/4 was -14.8e(r) + 2.34 e(theta). But the answer given is -21.71e(r) + 12.64e(theta). I cannot figure out what I did wrong. Can anyone help me?
 
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thoughtclaw said:
1. The stone of mass m=100 kg is placed inside the cup at the end of a rotating arm of length l=3 m. If arm is released from rest at theta=0 and its angular acceleration is given as theta (double dot) = 5 rad/s2 cos t (a) find the acceleration of the stone with theta=pi/4 rad; (b) what is the force acting on the stone from the cup at this instant (include gravity); (c) at what time does the stone fly off the arm?
2. I'm thinking the equation to start with is a(p)=(r(double dot)-r*theta(dot)2)e(r) + (r*theta(double dot)+2r(dot)theta(dot))e(theta)
3. Using the above equation, with the assumption that r(dot) and r(double-dot) are zero, I found that the acceleration at pi/4 was -14.8e(r) + 2.34 e(theta). But the answer given is -21.71e(r) + 12.64e(theta). I cannot figure out what I did wrong. Can anyone help me?
I am having difficulty following our notation. What does e(r) mean? or e(theta)? What is t? Is that the angle or time?

To do this problem, it is easier to first determine \dot{\theta}, \text{ and } \ddot{\theta} as functions t, whatever t is.

Then work out the acceleration using: a = \ddot{\theta}r

AM
 
Last edited:


Thanks, Andrew. With your help, I've figured out that my problem was in the integration, forgetting to factor in the cosine of zero being 1.

The e(r) and e(theta) are the unit vectors, as my professor uses them in this class. I don't know how to do the fancy equation editor stuff that you just did, but perhaps I should learn. :)
 

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