Chapter 6, Problem 45P

(a)

To determine

To Find:The acceleration of each mass.

Explanation of Solution

Given:

Smaller mass, $m=50.0\text{g = 0}\text{.05kg}$ .

Larger mass, $M=2.00\text{kg}$

Distance between two smaller masses, $d_m=6.00\text{cm = 0}\text{.06 m}$

Distance between smaller mass and larger mass, $d_{mM}=5.00\text{cm = 0}\text{.05 m}$

Formula used:

Gravitational force between two bodies having mass m and M , d distance away is:

  $F=\text{G}\frac{mM}{d^2}$

G is the gravitational constant.

Newton’s second law of motion:

  $F=ma$

Calculation:

Acceleration of smaller mass (m):

  $\begin{array}{l}m{a}_m=\frac{GmM}{d_{mM}{}^2}\\ \Rightarrow a_m=6.67\times {10}^{-11}\times \frac{2.00}{{\left(0.05\right)}^2}{\text{m/s}}^{\text{2}}\\ \Rightarrow a_m=5.34\times {10}^{-8}{\text{m/s}}^{\text{2}}\end{array}$

Acceleration of larger mass (m):

  $\begin{array}{l}M{a}_M=\frac{GmM}{d_{mM}{}^2}\\ \Rightarrow a_M=6.67\times {10}^{-11}\times \frac{0.05}{{\left(0.05\right)}^2}{\text{m/s}}^{\text{2}}\\ \Rightarrow a_M=1.33\times {10}^{-9}{\text{m/s}}^{\text{2}}\end{array}$

Conclusion:

Thus, mass m accelerates with $5.34\times {10}^{-8}{\text{m/s}}^{\text{2}}$ and mass M accelerates with $1.33\times {10}^{-9}{\text{m/s}}^{\text{2}}$ .

(b)

To determine

To Find: The time required by each mass to move by $\text{1 mm}\text{.}$

Explanation of Solution

Given:

Smaller mass, $m=50.0\text{g = 0}\text{.05kg}$ .

Larger mass, $M=2.00\text{kg}$

Distance between two smaller masses, $d_m=6.00\text{cm = 0}\text{.06 m}$

Distance between smaller mass and larger mass, $d_{mM}=5.00\text{cm = 0}\text{.05 m}$

Distance moved by each mass, $s=1\text{mm = 1}\text{.0}\times {\text{10}}^{-3}\text{m}$

mass m accelerates with $5.34\times {10}^{-8}{\text{m/s}}^{\text{2}}$

mass M accelerates with $1.33\times {10}^{-9}{\text{m/s}}^{\text{2}}$ .

Formula used:

Second equation of motion:

  $s=ut+\frac{1}{2}a{t}^2$

Where, u is the initial velocity, t is the time and a is the acceleration.

Calculation:

Consider that each mass is initially at rest. Then, $u=0$ .

For mass m:

  $\begin{array}{l}s=ut+\frac{1}{2}{a}_m{t}^2\\ \Rightarrow 1.0\times {10}^{-3}=0+\frac{1}{2}\times \left(5.34\times {10}^{-8}\right)t^2\\ \Rightarrow t=\sqrt{\frac{2\times {10}^{-3}}{5.34\times {10}^{-8}}}\\ \Rightarrow t=193.53\text{s}\end{array}$

For mass M:

  $\begin{array}{l}s=ut+\frac{1}{2}{a}_M{t}^2\\ \Rightarrow 1.0\times {10}^{-3}=0+\frac{1}{2}\times \left(1.33\times {10}^{-9}\right)t^2\\ \Rightarrow t=\sqrt{\frac{2\times {10}^{-3}}{1.33\times {10}^{-9}}}\\ \Rightarrow t=1226.28\text{s}\end{array}$

Conclusion:

Thus, mass mmoves in $193.53\text{s}$ and mass Mmoves in $1226.28\text{s}\text{.}$

(c)

To determine

To Find: The angle by which the rotates in the given time interval.

Explanation of Solution

Given:

Smaller mass, $m=50.0\text{g = 0}\text{.05kg}$ .

Larger mass, $M=2.00\text{kg}$

Distance between two smaller masses, $d_m=6.00\text{cm = 0}\text{.06 m}$

Distance between smaller mass and larger mass, $d_{mM}=5.00\text{cm = 0}\text{.05 m}$

Distance moved by each mass, $s=1\text{mm = 1}\text{.0}\times {\text{10}}^{-3}\text{m}$

Mass m accelerates with $5.34\times {10}^{-8}{\text{m/s}}^{\text{2}}.$

Mass M accelerates with $1.33\times {10}^{-9}{\text{m/s}}^{\text{2}}.$

Mass m moves in $193.53\text{s}$

Formula used:

Second equation of rotational motion:

  $\theta ={\omega }_{o}t+\frac{1}{2}\alpha t^2$

Where, ${\omega }_o$ is the initial angular speed, t is the time and $\alpha$ is the angular acceleration.

Angular acceleration is related to linear acceleration as:

  $\alpha =\frac{a}{r}$

Where, a is the linear acceleration and r is the radius.

Calculation:

Initially, the rod is at rest. So, ${\omega }_o=0$ .

  $\begin{array}{l}\theta ={\omega }_{o}t+\frac{1}{2}\alpha t^2\\ \Rightarrow \theta =0+\frac{1}{2}\times \frac{a}{r}\times t^2\\ \Rightarrow \theta =\frac{1}{2}\times \frac{5.34\times {10}^{-8}}{0.06}\times {\left(193.53\right)}^2\\ \Rightarrow \theta =1.67\times {10}^{-2}\text{rad}\\ \Rightarrow \theta =1.67\times {10}^{-2}\times \frac{180}{\pi }\mathrm{deg}\\ \Rightarrow \theta =0.96°\end{array}$

Conclusion:

Thus, the rod rotates by an angle of $1.67\times {10}^{-2}\text{rad}$ or $0.96°$ .