Mass in circular motion with friction. Find max height.

[demenius]

Homework Statement

A mass starts from rest at the upper rim of a semi-circle of radius 8m. In the bottom of this there is a region of size 1m which has a coefficient of kinetic friction of 0.5. What is the maximum height that the mass reaches after passing through this region once?

http://imageshack.us/photo/my-images/848/circularmotionfriction.png/"

Homework Equations

Note: arc length
∫cos(ax)dx = 1/a sin(ax)

The Attempt at a Solution

I do not know where to start. :S

 

[PeterO]

Homework Statement

A mass starts from rest at the upper rim of a semi-circle of radius 8m. In the bottom of this there is a region of size 1m which has a coefficient of kinetic friction of 0.5. What is the maximum height that the mass reaches after passing through this region once?

http://imageshack.us/photo/my-images/848/circularmotionfriction.png/"

Homework Equations

Note: arc length
∫cos(ax)dx = 1/a sin(ax)

The Attempt at a Solution

I do not know where to start. :S

You could start by working out how fast the mass would be traveling through the bottom of the semicircle if there was no friction, and then contemplate what the friction might do.

 

[demenius]

I know how to calculate the speed at the bottom but the friction part is before the bottom and I cannot figure out how to calculate the speed just before the friction. I believe the arc length from the top to the friction section is 4∏ - 0.5m. But do not know where to go from there.

 

[PeterO]

I know how to calculate the speed at the bottom but the friction part is before the bottom and I cannot figure out how to calculate the speed just before the friction. I believe the arc length from the top to the friction section is 4∏ - 0.5m. But do not know where to go from there.

OK so you know how to calculate it, but have you actually calculated it?

Suppose that instead of a semi circle, we had a quarter circle down, a 1 metre flat, then a quarter circle up.
Without friction, the mass would get to the top of the second quarter circle, having covered the flat at a constant speed.
How high would it reach if the 1m flat had friction? only half way up? 1/4 the way up? 98% of the way up?

I suspect the underlying assumption here is that the 1m section is only a small fraction of the semi circle, with no significant rise and fall [change in PE]
The key factor to be taken into account is that the reaction force will be greater than the weight of the mass, due to the required centripetal force/acceleration, so the energy lost to friction will be more than you might otherwise expect - certainly more than the example above, with the flat bottom section.

 

[demenius]

The Speed at the bottom would be √(2gR) = 12.53. Assuming the frictional part was flat the velocity after the the frictional part would be 12.12. And the max height would be 7.499m. But if the frictional part was not flat and was part of the semi circle then I am lost on how to do it.

 

[PeterO]

The Speed at the bottom would be √(2gR) = 12.53. Assuming the frictional part was flat the velocity after the the frictional part would be 12.12. And the max height would be 7.499m. But if the frictional part was not flat and was part of the semi circle then I am lost on how to do it.

I don't get such a big reduction in speed. The acceleration is -g/2 since the coefficient of friction is only 0.5.

When centripetal force is taken into account, the acceleration averages -3g/2 so approx height reached would be 7.25 m

Half a metre each side of the bottom of the slope means each extreme is about 1.5 cm above the lowest point - that is 1.5 in 800 or a very small percentage difference, so I think is could be ignored.
Perhaps of more significance is the reduction in centripetal force due to reduced speed, meaning the speed will not be reduced so much → a slightly higher final position.