# Homework Set 7

## Solution to recitation problem 5

a) $v=at$

$1=a\times0.5$
$a=2\mathrm{ms^{-2}}$

$Fr-Tr=\frac{1}{2}Mr^2\alpha$ → $F-T=\frac{1}{2}Ma$

$T-mg=ma$ → $T=mg+ma$

$F=\frac{1}{2}Ma+mg+ma$

$F=0.5\times 0.2\times 2+1\times(9.8+2)=12\mathrm{N}$

b)$\alpha=\frac{a}{r}=\frac{2}{0.1}=20\mathrm{rad\,s^{-2}}$

c) At constant speed $a=0$ so $F=mg=1\times9.8=9.8\mathrm{N}$

d) $\omega=\frac{v}{r}=\frac{1}{0.1}=10\mathrm{rad\,s^{-1}}=10\times\frac{60}{2\pi}=95.5\mathrm{rpm}$

## Problem 12.39

## Problem 12.79

An equilibrium problem. Find the height above the ground where the Tension/per unit area in the rope is greater than $500\times10^{6}\mathrm{N/m^{2}}$ (Tensile strength of nylon). Multiply the maximum tension per unit area by the area of the rope (calculated from the given diameter) to find the maximum tension allowed. You can then write an equation for the balance of the vertical forces in terms of the angle of the rope. Using the maximum tension you found before, you can either solve for the angle and then convert by geometry to height, or first convert to an expression in height and then solve.

## Problem 13.20

Use the balance of the torques around the lever to find the force per unit area on the small cylinder, which is the same as the force per unit area on the large one. The pressure on the sample is the force on the large cylinder divided by the sample area.

## A Submerged Ball

As the ball is in equilibrium the sum of the tension force, gravitational force and bouyant force should be zero.

## Crown of Gold?

The apparent weight is given by the true weight minus the bouyant force.

## Problem 13.56

An application of Bernoulli's theorem. This problem is similar to the one considered in Toricelli's theorem, the only difference is that the pressure at the top is not the same as atmospheric pressure.

## Venturi Meter with Two Tubes

Apply Bernoulli's theorem and the continuity equation to understand the Venturi Tube.

## Problem 13.65