Question

A seed is on a turntable rotating at$\frac{\mathbf{1}}{\mathbf{3}}\raisebox{1ex}{$\mathbf{r}\mathbf{e}\mathbf{v}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{m}\mathbf{i}\mathbf{n}$}\right.\mathbf{,}\mathbf{}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$from the rotation axis. What are (a) the seed’s acceleration and (b) the least coefficient of static friction to avoid slippage? (c) If the turntable had undergone constant angular acceleration from rest in0.25s, what is the least coefficient to avoid slippage?

Short answer

1. The seed’s acceleration a is, $0.73\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\mathrm{s}}^2}$}\right.$.
2. The least coefficient of static friction to avoid slippage ${\mu }_{s,\min }$ is, $0.075$.
3. The least coefficient to avoid slippage when the turntable had undergone constant angular acceleration from rest in 0.25 s${\mu }_{s,\min }$ is , 0.11 .

Step-by-step solution

Understanding the given information

1. The angular speed of turntable $\omega$ 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,$6.0\times {10}^{-2} \mathrm{m}$ ,
3. Time t is, $0.25 \mathrm{s}$.

Concept and Formula used for the given question

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 , $\left|\overrightarrow{a}\right|$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={\omega }^{2}r$$a={\omega }^{2}r$
2. The coefficient of friction is ${\mu }_{s,min}=\frac{a}{g}$
3. The tangential acceleration is $a_t=\alpha r$
4. The magnitude of acceleration is $\left|\overrightarrow{a}\right|=\sqrt{{a_r}^2+{a_t}^2}$

(a) Calculation for the seed’s acceleration

The angular speed in is,

$\omega =\left(33\frac{1}{3}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.\right)\times \left(\frac{2\pi {\displaystyle \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{rev}}$}\right.}}{60s/min}\right)$

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

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

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

$ma=f_s⩽f_{s,\max }={\mu }_{s}mg$

Thus, the coefficient of friction is given by

${\upsilon }_{s,min}=\frac{a}{g}$

Substitute all the value in the above equation.

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

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

$a_r={\omega }^{2}r$

While the tangential acceleration is given by

$a_t=\delta r$

Thus, the magnitude of acceleration is,

$$\begin{gathered} \left|\stackrel{\rightharpoonup }{a}\right|=\sqrt{a_r^2}+a_t^2 \\ =\sqrt{{\left({\omega }^{2}r\right)}^2+{\left(ar\right)}^2} \\ =r\sqrt{{\omega }^4+{\delta }^2} \end{gathered}$$

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

$$\begin{gathered} f_{s,max}={\mu }_{s}mg \\ ={ma}_{max} \\ =mr\sqrt{{\omega }_{max}^4}+{\delta }^2 \end{gathered}$$

Thus, since $\alpha =\frac{\omega }{t}$, we get

$\begin{array}{l}\underset{m\to n}{\lim }\frac{a}{g}\\ \sqrt[=0]{\frac{{\omega }_{max}^4+{\delta }^2}{g}}\\ \sqrt[=r]{\frac{{\omega }_{max}^4+{\left({\displaystyle \frac{{\omega }_{max}}{t}}\right)}^2}{g}}\\ =\frac{0.060m\times \sqrt{{\left(3.49{\displaystyle \raisebox{1ex}{$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