# Biophysics Problem 33

If the person in Question 31 is in a skiing accident where one end of the bone described is twisted 10 further than the other, what torque must have been applied? Assume the shaft of the bone is \(0.30\;m\) long.

The answer given assumes \(G = 1 \times 10^{10}\; N\; m^2\) (found on page 104 of your textbook).

The equation for twisting a hollow bone is:

\(\tau = \frac{2G\pi r^2 t \theta}{\ell}\)

The three most common mistakes in this type of problem are:

- Forgetting to express the twist angle in radians.
*Remember that there are \(2\pi\) radians in one circle of \(360^\circ.\)* - Calculating the average radius.
- Calculating the thickness \((t)\) of the shell bone.

Calculate the average radius now.

\(\tau = \frac{2(1\times 10^{10}\pi (0.015)^3(0.005)(10)}{0.3(57.3^\circ /rad)}\\ = 620\)

The inner and outer radii are shown in the diagram above. The average radius is just half the sum of these two radii, i.e. \(r_{av} = 0.015\; m.\)

Now you should be able to calculate the thickness \('t'.\)

The thickness of the layer of bone is the difference between the two radii. That is:

\(t = 0.0175 - 0.0125 = 0.005\; m\)

Remember that \('\ell'\) in the denominator is the length of the bone in metres. You should now be able to find the moment of torque required. Do this now.

Using the equation given at the beginning of this solution, we can subsititute in the numbers, and get:

\(\tau = \frac{2(1\times 10^{10})\pi (0.015)^3(0.005)(10)}{0.3(57.3^\circ/rad)}\\ = 620\)

(I leave it to you to figure out what units moment \(\tau\) would have.)