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Chapter 8: Problem 63

A 1.10 -kg bucket is tied to a rope that is wrapped around a pole mounted horizontally on friction-less bearings. The cylindrical pole has a diameter of \(0.340 \mathrm{m}\) and a mass of \(2.60 \mathrm{kg} .\) When the bucket is released from rest, how long will it take to fall to the bottom of the well, a distance of \(17.0 \mathrm{m} ?\)

Short Answer

Expert verified
Answer: The bucket will take approximately 3.32 seconds to fall to the bottom of the well.

Step by step solution

01

1. Analyzing the forces on the bucket and the pole

First, let's analyze the forces acting on the bucket and the pole separately. For the bucket, there is a gravitational force (\(mg\)) acting downward and tension force (T) acting upward. For the pole, the tension force acts tangentially.
02

2. Using Newton's Second Law on the bucket

To find the acceleration of the bucket, we can apply Newton's Second Law on the bucket: \(F_{net} = m_{bucket}a_{bucket}\) As the bucket is moving downwards, we consider its downward motion positive. So, we have: \(mg - T = m_{bucket}a_{bucket}\) (1)
03

3. Torque and angular acceleration of the pole

Since the falling bucket unwinds the rope, the tension force acts on the pole and causes an angular acceleration. The torque due to tension force on the pole can be calculated as follows: \(\tau = -rT\), where \(r\) is the radius of the pole. The minus sign comes from considering clockwise rotation negative. The angular acceleration can be obtained by dividing the torque by the moment of inertia of the pole (\(I_{pole}\)), which for a solid cylinder is \(\frac{1}{2}m_{pole}r^2\): \(-\frac{I_{pole}\alpha_{pole}}{r} = -rT\)
04

4. Relating linear acceleration and angular acceleration

Since the rope has to unwind from the pole for the bucket to fall, the linear acceleration of the bucket and the angular acceleration of the pole are related by: \(a_{bucket}=r\alpha_{pole}\) (2)
05

5. Solving for bucket acceleration

Substitute the value of torque from step 3 and angular acceleration from step 4 into equation (2): \(T = I_{pole}\alpha_{pole} = I_{pole}\frac{a_{bucket}}{r} = \frac{1}{2}m_{pole}r^2\frac{a_{bucket}}{r} = \frac{1}{2}m_{pole}ra_{bucket}\) Substitute this equation for T into equation (1): \(mg - \frac{1}{2}m_{pole}ra_{bucket} = m_{bucket}a_{bucket}\) Solve for \(a_{bucket}\): \(a_{bucket} = \frac{mg}{m_{bucket} + \frac{1}{2}m_{pole}r}\)
06

6. Calculating the bucket acceleration

Plug the known values for mass and radius, and calculate the acceleration: \(a_{bucket} = \frac{(1.10kg)(9.8m/s^2)}{1.10kg + \frac{1}{2}(2.60kg)(0.17m)}\) \(a_{bucket} \approx 3.16 m/s^2\)
07

7. Finding the time it takes for the bucket to fall 17m

Now that we have the acceleration of the bucket, we can find the time it takes for the bucket to fall 17m using the kinematic equation: \(h = \frac{1}{2}at^2\) \(17m = \frac{1}{2}(3.16m/s^2)t^2\) Solve for \(t\): \(t \approx 3.32 s\) So, it will take approximately 3.32 seconds for the bucket to fall to the bottom of the well.

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