91Ó°ÊÓ

1. The angular speed of turntable $\mathrm{Ó\neg }$ is, is,$33\frac{1}{3}\raisebox{1ex}{${\displaystyle \mathrm{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \min }$}\right.$
2. The distance from the rotation axis r is,$6.0×{10}^{-2} \mathrm{m}$ ,
3. Time t is, $0.25 \mathrm{s}$.

By usingtheformula for radial acceleration and coefficient of friction, we can find the seed’s acceleration and theleast coefficient of static friction to avoid slippage. By calculating radial acceleration and tangential acceleration $a_t$, we can calculate the magnitude of the acceleration , $\left|\stackrel{→}{a}\right|$and using this value, we can find theleast coefficient to avoid slippage when the turntable had undergone constant angular acceleration from rest in 0.25 s.

1. The radial acceleration is $a={\mathrm{Ó\neg }}^{2}r$$a={\mathrm{Ó\neg }}^{2}r$
2. The coefficient of friction is ${\mathrm{μ}}_{s,min}=\frac{a}{g}$
3. The tangential acceleration is $a_t=\mathrm{Î\pm }r$
4. The magnitude of acceleration is $\left|\stackrel{→}{a}\right|=\sqrt{{a_r}^2+{a_t}^2}$

The angular speed in is,

$\mathrm{Ó\neg }=\left(33\frac{1}{3}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.\right)×\left(\frac{2\mathrm{Ï€}{\displaystyle \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{rev}}$}\right.}}{60s/min}\right)$

Consequently, the radial acceleration is given by $a={\mathrm{Ó\neg }}^{2}r$

role="math" localid="1660971880234" $$\begin{gathered} a=\left(3.49\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)×6.0×{10}^{-2} \text{m} \\ =0.73\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right. \end{gathered}$$

Hence the seed acceleration is, $=0.73\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$.

Step 3: (b) Calculation for the least coefficient of static friction to avoid slippage

By using the methods in chapter 6, we have

$ma=f_sâ\copyright ½f_{s,\max }={\mathrm{μ}}_{s}mg$

Thus, the coefficient of friction is given by

${\mathrm{Ï\dots }}_{s,min}=\frac{a}{g}$

Substitute all the value in the above equation.

$$\begin{gathered} {\mathrm{μ}}_{s,min}=\frac{0.73{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}}{9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.} \\ =0.075 \end{gathered}$$

Hence the value of least coefficient of static fraction is, $0.075$.

Step 3: (c) Calculation for the least coefficient of static friction to avoid slippage when the turntable had undergone constant angular acceleration from rest in 0.25 s

The radial acceleration of the object is given by

$a_r={\mathrm{Ó\neg }}^{2}r$

While the tangential acceleration is given by

$a_t=\mathrm{δ}r$

Thus, the magnitude of acceleration is,

$$\begin{gathered} \left|\stackrel{â‡\P Ä}{a}\right|=\sqrt{a_r^2}+a_t^2 \\ =\sqrt{{\left({\mathrm{Ó\neg }}^{2}r\right)}^2+{\left(ar\right)}^2} \\ =r\sqrt{{\mathrm{Ó\neg }}^4+{\mathrm{δ}}^2} \end{gathered}$$

If the object is not to slip at any time, we require

$$\begin{gathered} f_{s,max}={\mathrm{μ}}_{s}mg \\ ={ma}_{max} \\ =mr\sqrt{{\mathrm{Ó\neg }}_{max}^4}+{\mathrm{δ}}^2 \end{gathered}$$

Thus, since $\mathrm{Î\pm }=\frac{\mathrm{Ó\neg }}{t}$, we get

$\begin{array}{l}\underset{m→n}{\lim }\frac{a}{g}\\ \sqrt[=0]{\frac{{\mathrm{Ó\neg }}_{max}^4+{\mathrm{δ}}^2}{g}}\\ \sqrt[=r]{\frac{{\mathrm{Ó\neg }}_{max}^4+{\left({\displaystyle \frac{{\mathrm{Ó\neg }}_{max}}{t}}\right)}^2}{g}}\\ =\frac{0.060m×\sqrt{{\left(3.49{\displaystyle \raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}\right)}^4}+{\left({\displaystyle \frac{3.4{\displaystyle \raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{0.25s}}\right)}^2}{9.8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}}\\ =0.11\end{array}$

Hence the value of least coefficient of static fraction is, 0.11