Block slides down hemisphere; time to leave surface?

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A block on a frictionless hemisphere begins to slide down when slightly disturbed, and it will leave the surface at a height of 2r/3. To determine when it leaves the surface, one must integrate the expression for velocity, v = (2gr(1-cos(theta)))^0.5, with respect to time. The relationship between arc length and angular velocity is established as v = ds/dt = r(dθ/dt), allowing for substitution of v into the equation. By separating variables, the integration can be performed to find the time. For those seeking a quicker solution, online tools like Wolfram's integrator can be utilized.
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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