Block slides down hemisphere; time to leave surface?

[Dan6500]

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?

 

[Bystander]

You've obviously got an expression for angular velocity if you've determined "the where." Integrate it.

 

[Dan6500]

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?

 

[Philip Wood]

Expanding what Bystander said (in case you're not familiar with angular velocity)
[tex]v=\frac{ds}{dt}=r\frac{d\theta}{dt}[/tex]
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.

 

[PJC01]

The time it takes for the block to leave the surface of the hemisphere can be determined by using the principles of motion and energy conservation. The block's initial potential energy at the top of the hemisphere will be converted into kinetic energy as it slides down. At the point where it leaves the surface, the block's kinetic energy will be equal to its potential energy at that height. This can be calculated using the law of conservation of energy.

Using the equation for potential energy (PE = mgh) and kinetic energy (KE = 1/2mv^2), we can set them equal to each other and solve for the velocity of the block at the point where it leaves the surface. Once the velocity is known, we can use the equation for average velocity (v = d/t) to calculate the time it takes for the block to travel the distance of 2r/3.

It is important to note that this calculation assumes a frictionless surface and neglects any air resistance. In reality, these factors may affect the time it takes for the block to leave the surface. Additionally, the shape and mass of the block may also impact the results. Therefore, it is important to consider these factors when making predictions or conducting experiments related to this scenario.