# Calculating Velocity and Angle - Normal & Tangential Accel.

1. Oct 18, 2015

### Cal Ripken

1. The problem statement, all variables and given/known data

A small 1 lb block is at rest at the top of a cylindrical surface. The block is given an initial velocity v0 to the right which causes it to slide on the cylindrical surface. Assume smooth surface and no friction.

a. Use any computational software or Excel to calculate the velocity and angle theta at which the block leaves the surface if initial velocity v0 is from 5 ft/s to 15 ft/s (every 1 ft/s).

b. If there is friction between the block and the surface what will be the difference between your equations? Set up the equations without solving. Assume the coefficient of kinetic friction μk between the block and the surface is given.

2. Relevant equations
∑Fn=man
∑Ft=mat

3. The attempt at a solution
For part a, my attempt at a solution involved setting up the equations of motion as a function of theta. Recognizing that the block begins to leave the surface when the Normal force acting on the block is equal to zero, I set up an equation solving for theta when N=0, then plugged that value of theta in to the equation established from the summation of forces in the tangential direction to solve for the final velocity of the block when it leaves the surface.

My equation setup:

Values obtained for values of V0 ranging from 5 ft/s to 15 ft/s:

I would like to know if my equations are set up properly. Further, I would like some help interpreting the data. It makes sense that the velocity of the block would be greater when it leaves the surface because gravity is doing positive work. The angle theta is what's throwing me off, though. The values I obtained for theta when N=0 make sense if measured from the positive x axis, because lower values of V0 should correspond to a greater change in theta. In setting up my second equation, my interval of integration was from 0 to theta. When I tried integrating from 90° to theta, I ended up with VF =√ (- number) . My results make sense to me provided theta is measured from the x axis, but I intended to measure theta from the y axis. What am I missing here and what do I need to do to prevent making this mistake again?

2. Oct 18, 2015

### Scott Redmond

My first thought upon seeing this was that your angle theta is measured from the y-axis. It seems to be measured from your weight vector, as drawn. So for vo=5.00 ft/s, the block leaves the cylindrical surface when theta = 8.93 degrees -- as measured from the y-axis. I'll ask a question: what is the value of theta, according to your diagram, when the block is at the top of the cylinder? Then Δθ=θfi.

3. Oct 19, 2015

### haruspex

Check your expression for the loss in gravitational PE.