Physics question--any ideas?

Question

Cylindrical pulley w/ the following:
mass, M = 6.9 kg
radius, r = 0.811 m
moment of inertia 1/2 Mr^2

is used to lower a bucket (m=2.9 kg) into a well. The bucket starts from rest and falls for 3.3 s, acceleration of graviy is 9.8 m/s^2.

A) What is the linear acceleration of the falling bucket in units of m/s^2?

B) How far does it drop in units of m?

C) What is the angular acceleration of the cylinder in units of rad/s^2?

Answer 1

(note: for this problem, I assume that the rope is unwinding off of the RIGHT side of the cylinder causing the cylinder to rotate CLOCKWISE. I also assume that "positive" angular rotation is COUNTER-CLOCKWISE and positive movement is UPWARD)

First, draw a diagram of the forces acting on the bucket. You'll have a force of the bucket's weight (m*g) pulling down on it and the tension of the rope (T) pulling up on the bucket. Thus, the net force downward on the bucket is T-mg. That net force is equal to -m*a, where a is the resulting acceleration of the bucket downward (that's why I put a negative sign in front of the ma). That is,

T - mg = -ma

The tension in the rope is set by the rotational inertia (think "mass") of the cylinder. As the cylinder's inertia increases, so will the tension.

Draw a force diagram of the cylinder. Since the cylinder isn't falling, its forces are in static equilibrium, so we're not interested in its forces. Instead, we're interested in its torque. The only torque on the cylinder is the tension of the rope at a distance r away from the center of the rotating cylinder. Thus,

-T*r = I * alpha

where T is the tension in the rope, r is the radius of the cylinder, I is the rotational inertia of the cylinder, and alpha is the angular acceleration of the cylinder. I have put a negative sign in front of the torque because it will tend to rotate the cylinder clockwise and positive angular motion is generally considered counter-clockwise. Note that alpha = -r*a, where a is the acceleration of a point on the rope. I've put a negative sign here too because I'm considering positive acceleration to be DOWNWARD and positive ANGULAR acceleration to be counter-clockwise. Thus, we actually have:

-T * r = -I * r * a

so:

T = I*a

We can plug this into our original equation T-mg=ma to get:

I*a - mg = -ma

Now just solve for a, the acceleration of the bucket:

a = mg/(I + m)

Now replace "I" with your expression for the moment of inertia:

a = mg/(0.5*M*r^2 + m)

And there you have it. You can plug in numbers to get:

A) 5.498 m/s/s

To answer (B), keep in mind that the DISPLACEMENT of an object with acceleration a and initial velocity 0 is 0.5*a*t^2. It dropped for 3.3 seconds, so we get 0.5*(5.498 m/s/s)*(3.3 s)^2 = 29.9367 meters. So the answer to B is:

B) 29.9367 meters

Finally, remember that our expression for angular acceleration was:

alpha = -r*a = -(0.811 m)*(5.498 m/s/s) = -3.616 rad/s/s

So your answer to C is :

C) -3.616 rad/s/s where positive rotation is counter-clockwise (in other words, 3.616 rad/s/s clockwise (or 207.191 degrees/s/s clockwise))

I think that should do it.

Answer 2

Parachute. Take 2 foam rings smaller than the egg. Put the egg between them, and use threads to hold it there. Now attach a large plastic sheet, preferably cut out from a large shopping bag (use the low quality ones! those are lighter), using strings, to one of the rings to function as the parachute. Good luck on your project!