Block slides down hemisphere; time to leave surface?

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SUMMARY

A block resting at the top of a frictionless hemisphere of radius r will leave the surface at a height of 2r/3. To determine the time it takes to leave the surface, one must integrate the expression for angular velocity derived from the velocity function v = (2gr(1-cos(theta)))^0.5. The relationship v = ds/dt = r(dθ/dt) allows for substitution and separation of variables to solve for time. For those seeking a simpler solution, utilizing the Wolfram online integrator is recommended.

PREREQUISITES
  • Understanding of angular velocity and its relationship to linear velocity
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of the physics of motion on curved surfaces
  • Experience with online mathematical tools like Wolfram Alpha
NEXT STEPS
  • Study the principles of motion on curved paths in classical mechanics
  • Learn advanced integration techniques for solving differential equations
  • Explore the use of computational tools for solving physics problems
  • Investigate the effects of friction on motion in similar scenarios
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Physics students, educators, and anyone interested in the dynamics of motion on curved surfaces will benefit from this discussion.

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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