Difficult: Linear vs. Angular acceleration

[datdo]

Homework Statement

A board of mass m and length L is hinged to the ground. Attached to the board is a cup of negligible mass and is positioned R_c away from the hinge. Also on the edge of the board a ball rests. When the board is dropped from an angle of theta the ball is meant to be caught in the cup.

A) What is the minimum angle required for the ball to lag behind the board? I think cos(theta) > 1/2

B) Where must the cup be placed in order to catch the ball in terms of theta and L? I think the answer to this one was L/2cos(theta)

Homework Equations

F=ma

\tau=Frsin\theta

\tau=I\alpha

\alpha r=a

The Attempt at a Solution

I'm really lost.

I guess for the cup to catch the ball both objects must be in the same spot in time so if I find their positions with respect to time and find their intersection...

board:
\tau=.5mgLsin\theta
\alpha=\frac{3\tau}{mL^2}

cup:
a_c=\alpha R_c

ball:
a_b=g

I don't know where to go from here...

 

[Doc Al]

board:
\tau=.5mgLsin\theta

Why sinθ?

You neglected to point out that the ball starts out balanced on the edge of the board.

Hints:

For A: What's the acceleration of the edge of the board? How must it compare to the acceleration of the falling ball in order for the ball to fall freely?

For B: The ball falls straight down.

 

[datdo]

sorry about that.

its sin because torque is F\times r = Frsin\theta
the angular acceleration increases as the angle decreases
A)

\alpha=\frac{3gsin\theta}{2L}
end of the board:
a_e=\frac{3gsin\theta}{2}

g\leq\frac{3gsin\theta}{2}

\frac{2g}{3g}\leq sin\theta

\frac{\sqrt{5}}{3}\leq cos\theta? but isn't the answer...


B)
R_c=Lcos\theta

 

[Doc Al]

its sin because torque is F\times r = Frsin\theta

It would be sinθ if θ was the angle between r and F, not r and the horizontal.

the angular acceleration increases as the angle decreases
A)

\alpha=\frac{3gsin\theta}{2L}
end of the board:
a_e=\frac{3gsin\theta}{2}

g\leq\frac{3gsin\theta}{2}

\frac{2g}{3g}\leq sin\theta

Redo this with the correct angle.

\frac{\sqrt{5}}{3}\leq cos\theta? but isn't the answer...

Not exactly.

B)
R_c=Lcos\theta

Good.

 

[datdo]

understood.

\alpha=\frac{3gsin(\pi/2 -\theta)}{2L}
end of the board:
a_e=\frac{3gsin(\pi/2 -\theta)}{2}

g\leq\frac{3gsin(\pi/2 -\theta)}{2}

\frac{2g}{3g}\leq sin(\pi/2 -\theta)

\frac{2}{3}\leq cos\theta this more likely...

B still confuses me..

 

[Doc Al]

understood.

\alpha=\frac{3gsin(\pi/2 -\theta)}{2L}
end of the board:
a_e=\frac{3gsin(\pi/2 -\theta)}{2}

g\leq\frac{3gsin(\pi/2 -\theta)}{2}

\frac{2g}{3g}\leq sin(\pi/2 -\theta)

\frac{2}{3}\leq cos\theta this more likely...

Good. So what's the maximum angle?

B still confuses me..

What about it confuses you? The ball, which begins at the edge of the board, falls straight down. Find the horizontal distance between ball and hinge--that will tell you where to put the cup.

 

[datdo]

cos^-^1\frac{2}{3}

I guess it just seems too simple...even for a mechanics problem...

 

[Doc Al]

I guess it just seems too simple...even for a mechanics problem...

Don't complain.

(I see that you tagged the thread with "Cooper Union". Are you a student there?)

 

[datdo]

I could be...