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Chapter 11: Problem 27

A potter's wheel with rotational inertia \(6.40 \mathrm{kg} \cdot \mathrm{m}^{2}\) is spinning freely at 19.0 rpm. The potter drops a 2.70 -kg lump of clay onto the wheel, where it sticks \(46.0 \mathrm{cm}\) from the rotation axis. What's the wheel's subsequent angular speed?

Short Answer

Expert verified
The wheel's subsequent angular speed is approximately \(1.72 \text{ rad/s}\).

Step by step solution

01

Convert given quantities to SI units

First, we have to convert all given quantities to the standard units used in physics. The rotational inertia is already given in \(kg \cdot m^{2}\), the standard unit. The initial angular speed is given in revolutions per minute. We need to convert it to radians per second. The conversion is as follows: 19 rpm \(\times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}}\) = \(1.99 \text{ rad/s} = \omega_{i}\). The radial distance of clay from the rotation axis is given in centimeters, we convert it to meters: \(46.0 \text{ cm} = 0.46 \text{ m} = r\).
02

Calculate initial angular momentum

Before the clay hits the wheel, the system contains only the wheel, and its angular momentum can be calculated as the rotational inertia of the wheel times its initial angular speed: \(L_{i} = I_{wheel}\omega_{i} = 6.4 \text{ kg}\cdot \text{m}^{2}\times 1.99 \text{ rad/s} = 12.736 \text{ kg}\cdot\text{m}^{2}/\text{s}\)
03

Calculate final angular momentum

After the clay hits the wheel and sticks to it, the system contains the wheel and the clay. The angular momentum after sticking the clay can be calculated as: \(L_{f} = I_{wheel}\omega_{f} + I_{clay}\omega_{f}\). Where \(I_{wheel}\) and \(I_{clay}\) are the inertia of the wheel and clay respectively, and \(\omega_{f}\) is the final angular speed. We already have the value of \(I_{wheel}\), but the inertia of clay can be calculated as \(I_{clay} = m \cdot r^{2} = 2.7 \text{ kg} \times (0.46 \text{ m})^{2} = 0.57 \text{ kg} \cdot \text{m}^{2}\)
04

Use conservation of angular momentum

As there is no external torque acting on the system, the angular momentum of the system should remain the same before and after the clay hits the wheel. Therefore: \(L_{i}=L_{f}\). Substituting the known values and solving for the final angular speed, we obtain: \(12.736 \text{ kg}\cdot\text{m}^{2}/s = (6.4 \text{ kg}\cdot\text{m}^{2} + 0.57 \text{ kg}\cdot\text{m}^{2})\omega_{f}\) or \(\omega_{f} = \frac{12.736 \text{ kg}\cdot\text{m}^{2}/s}{(6.4 \text{ kg}\cdot\text{m}^{2} + 0.57 \text{ kg}\cdot\text{m}^{2})} = 1.72 \text{ rad/s}\)

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