Block slides down hemisphere; time to leave surface?

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Discussion Overview

The discussion revolves around determining the time at which a block, sliding down a frictionless hemisphere, leaves the surface after being disturbed from rest at the top. The focus is on the mathematical approach to find this time, involving concepts of angular velocity and integration.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant states they know the height at which the block will leave the surface (2r/3) but seeks to find the time of departure.
  • Another participant suggests that an expression for angular velocity could be used to facilitate the integration needed to find the time.
  • A participant provides a velocity function in terms of the angle theta, indicating a relationship between linear velocity and angular displacement.
  • Further clarification is offered on the relationship between arc length and angular displacement, suggesting a method to substitute values into the equations for integration.
  • One participant mentions using an online integrator as an alternative if they are not interested in the integration challenge.

Areas of Agreement / Disagreement

Participants are engaged in a technical discussion with no clear consensus on the method of integration or the specifics of the approach to find the time of departure.

Contextual Notes

There are unresolved aspects regarding the integration process and the assumptions made in deriving the velocity function. The discussion does not clarify the limits of integration or the specific conditions under which the equations apply.

Dan6500
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A block is at rest at the top of a frictionless hemisphere of radius r. It is slightly disturbed at starts sliding down. I already know where it will leave the surface (height = 2r/3). My question is, WHEN will it leave the surface?
 
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You've obviously got an expression for angular velocity if you've determined "the where." Integrate it.
 
I've got v as a function of theta: v = (2gr(1-cos(theta))^0.5. So now I integrate with respect to TIME? How?
 
Expanding what Bystander said (in case you're not familiar with angular velocity)
v=\frac{ds}{dt}=r\frac{d\theta}{dt}
in which ds is an increment of arc length. r is, of course, a constant.
You can substitute this value for v into your equation, then separate variables.
If I'm not interested in the challenge of the integration I sometimes use the Wolfram on-line integrator.
 

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