During a TV report about ground water contamination, a reporter stands next to a

Question

During a TV report about ground water contamination, a reporter stands next to an old style well which works by lowering a bucket at the end of a rope into a deep hole inthe ground to get water. At the top of the well, a single vertical pulley, which is essentially a heavy stell ring supported by light spokes, is mounted to help raise and lower the bucket. To demonstate the depth of the well, the reporter completely wraps the thin rope around the pulley, then releases the bucket, which starts at rest near the pulley then falls to the bottom of the well, unwinding the rope from the pulley as it falls. The fall takes t=2.0 s . Although the depth of the well is never mentioned, you decide to calculate it. You estimate that the pulley has the same mass of the bucket and is well-oiled, since you didn't hear any squeaking.

Explanation / Answer

Ans:-

As a quick estimate, let’s first assume a freely falling bucket. How deep would a well be if it takes the bucket 2.5 sec to reach the bottom?

y = 1 /2 gt^ 2 = 1/ 2 *9.8 (2.0s) ^2 = 19.6m

This gives us the upper limit o the well depth. Our well will be less deep as the bucket’s acceleration will be less than 9.8 m/s 2 . How do we know? Well, this bucket is connected to a rope, which is wrapped around a pulley. It will take energy to spin the pulley, which is not in our rough estimate. Nonetheless, 30m is a good starting estimate.

To tackle this problem (with the pulley) we’ll need to employ Newton’s Second law and kinematic equations. (Why this route, instead of energy methods? We are given a time and time is easily related to acceleration via a kinematic equation. It is difficult, although not impossible, to connect time to energy.)

• We’ll assume that there is no friction in the pulley’s axle, but there is sufficient friction between the rope and pulley such that the rope does not slip.

• We’ll approximate the pulley by a simple ring and ignore the “light spokes.”

• We’ll go with the equal mass assumption.

We can apply the lienar version of Newton’s Second Law to the bucket.

T - mg = -ma

Solve this for the rope’s tension.

T = m(g - a)

We can also use Newton’s Second Law (rotational version) on the pulley, with the pivot at the center of the pulley.

TR = I*

Now we’ll make use of the pulley’s shape and substitute in the moment of inertia for a simple hoop. TR = mR2

Next, make use of the no slip condition to relate the angular acceleration of the pulley to the linear acceleration of the rope & bucket.

  TR = mR^2 (a /R) ; T = ma

This can be substituted into the equation for the bucket, so we can solve for acceleration.

ma = m(g - a)

a = g/ 2

Now we’re ready to return to the same kinematic equation that we used in our rough estimate to find the distance that the bucket fell during 2.0seconds.

y = 1 /2* g/ 2*t ^2 = 1/ 2 *4.9*(2.0s)^ 2 =9.8m

So, our initial estimate was a factor of two too large. This illustrates the importance of considering rotational motion along with linear motion

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