Work-Energy(Polar Coordinates)

[Dynamic24]

Homework Statement

A 5kg box is started at the top of a spiral slide at t=0 with a speed of 15m/s. The spiral has a radius, r=2m, and it's height is given by z=-.4Θ where z is in meters and Θ is in radians. The slide has a wall and a floor with a coe. of friction of .3

a) With no friction, what is the speed of mass when Θ=2pi?
b)With no friction, what is the speed of the mass when t=.5s?

c)With friction only between the floor and mass where does the mass stop and at what time?

e)With friction only between the wall and the mass, what is the final speed of the mass?

Homework Equations

z=-.4Θ
W=ſFds
W=1/2mv^2-1/2mv^2

The Attempt at a Solution

Basically the polar coordinates throw me off. With out friction the only force is gravity. So can I just integrate to get my function which defines the position?

 

[Chrisas]

With out friction the only force is gravity. So can I just integrate to get my function which defines the position?

There is another force from the outside wall of the slide. Otherwise nothing would be causing the box to rotate as it goes down the slide.

 

[Chrisas]

With no friction in part a, you can just use conservation of energy to get the speed.

 

[Dynamic24]

Yes, I got that now do I need to put z=-.4Θ with a radius 2m into vectors to find the velocity at t=.5sec

 

[Chrisas]

I'm not sure this is the best way to do it, but I made a position vector in cylindrical coordinates using the givens of the problem. Next I differentiated the position vector twice to get acceleration. Multiplied that by mass and set equal to the sum of forces. Equating components gives you three equations of which you only need to use two to find the angular acceleration which is constant. Integrate that to find angular velocity as a function of time. Use the velocity vector along with initial speed to find the initial angular velocity needed to find the constant of integration. Now you can find the angular velocity at any time t and thus the speed.

However, that seems a lot of work for a problem that is labeled as "Work-Energy". I'm not sure how to use work-energy considerations to simplify the above.