Rotation AND/OR Rigid Bodies

  • Thread starter SsUeSbIaEs
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  • #1
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Can somebody help me with this problem, it is probably pretty easy but I can not seem to think of how to solve it.


A object initially located at point A on the rim of a grinding wheel rotating about a horizontal axis. The object is dislodged from point A when the diameter through A is horizontal. It then rises vertically and returns to A the instant the wheel completes one revolution. Find the speed of a point on the rim if the radius is 1.153 m.
 

Answers and Replies

  • #2
We can write an equation for the time of the object to rise and fall as a function of the initial vertical velocity (which is the same as the speed of the rim) and we can write an equation for the time of one rotation of the wheel as a function of the speed of the rim (it actually depends on the speed of the rim and the radius, but the radius is given). If the object is to contact point A when it comes back down, these two times must be equal, so they are the same variable. Now there are 2 equations and 2 variables, so all that's left is to solve the system of equations.

Do you need help writing the equations or is this enough?

cookiemonster
 
  • #3
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I need help writing the equations.
 
  • #4
[tex]
\Delta y = \frac{1}{2} a t^2 + v_0 t
[/tex]
We note that [itex] \Delta y = 0 [/itex] and [itex] a = g = -9.8 m/s^2 [/itex], so the only two variables left are t and [itex] v_0 [/itex]. That's one equation.

[tex]
t = \frac{distance}{velocity} = \frac{circumference}{speed} = \frac{2 \pi r}{v_0}
[/tex]
We note that r = 1.153m, so we again have two variables.

This gives us two equations and two variables.

cookiemonster
 
Last edited:
  • #5
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THANKYOUTHANKYOU THANKYOU

Thank you so much that makes a lot of sense!!!
 
  • #6
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THANKYOUTHANKYOU THANKYOU

Thank you so much that makes a lot of sense!!!
 

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