Question

A thin, 100g disk with a diameter of 8.0 cm rotates about an axis through its center with 0.15 J of kinetic energy. What is the speed of a point on the rim?

Short answer

The speed of a point on the rim is$2.4\text{m}/\text{s}$

Step-by-step solution

Step 1. Given information

$\begin{array}{l}\text{Radius\hspace{0.17em}of\hspace{0.17em}rim,\hspace{0.17em}}R=0.040\text{m}\\ \text{Mass\hspace{0.17em}of\hspace{0.17em}rim,\hspace{0.17em}}M=0.10\text{kg}\\ \text{Rotational\hspace{0.17em}kinetic\hspace{0.17em}energy,\hspace{0.17em}}K=0.15\text{J}\end{array}$

Step 2. Explanation

The moment of inertia of the disk

$I=\frac{1}{2}M{R}^2=\frac{1}{2}(0.10\text{kg}){(0.040\text{m})}^2=8.0\times {10}^{-5}\text{kg}{\text{m}}^2$

The rotational kinetic energy

$\begin{array}{l}K=\frac{1}{2}I{\omega }^2=0.15\text{J}\\ {\omega }^2=\frac{2(0.15\text{J})}{I}=\frac{0.30\text{J}}{8.0\times {10}^{-5}\text{kg}{\text{m}}^2}\\ \omega =61.237\text{rad}/\text{s}\end{array}$

The speed of a point on the rim

.$v_{\text{rim}}=R\omega =(0.040\text{m})(61.237\text{rad}/\text{s})=2.4\text{m}/\text{s}$$v_{\text{rim}}=R\omega =(0.040\text{m})(61.237\text{rad}/\text{s})=2.4\text{m}/\text{s}$