Chapter 8, Problem 48P

a.

To determine

To find: The translational speeds of the sphere.

Answer to Problem 48P

The translational speed of the speed is $8.367\text{m/s}$ and the rotational speed is $24.25\text{rad/s}$ .

Explanation of Solution

Given Information: Radius of the sphere $r=0.2\text{m}$

Mass of sphere $m=1.80\text{kg}$

Angle of Inclination $\theta =30°$

  $L=10.0\text{m}$

Concept used:

1. $I_{sphere}=\frac{2m{r}^2}{5}$

   Here,

     $m$ is the mass of sphere,

     $I_{sphere}$ is the moment of inertia of sphere,

     $r$ is the radius of sphere.

2. $\omega =\frac{v}{r}$

Here,

  $\omega$ is the angular velocity,

  $v$ is the linear veloity,

  $r$ is the radius.

3. By-law of conservation of energy

  $\begin{array}{l}{E}_I=E_F\\ PE=K{E}_{translational}+K{E}_{rotation}\\ mgL\sin \theta =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\\ mgL\sin \theta =\frac{1}{2}m{v}^2+\frac{1}{2}\left({\frac{2mr}{5}}^2\right){\left(\frac{v}{r}\right)}^2\\ mgL\sin \theta =\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2\\ mgL\sin \theta =\frac{7}{10}m{v}^2\\ v={\left(\frac{10Lg\times \sin \theta }{7}\right)}^{\frac{1}{2}}\end{array}$

Here,

  $v$ is the linear velocity,

  $L$ is the length.

Calculation:

As $v={\left(\frac{10Lg\times \sin \theta }{7}\right)}^{\frac{1}{2}}$ then put all the known values from the given

  $\begin{array}{l}v={\left(\frac{(10)(10)(9.8)\sin 30°}{7}\right)}^{1/2}\\ =8.367\text{m/s}\end{array}$

and,

  $\begin{array}{l}\omega =\frac{v}{r}\\ \omega =\frac{8.367}{0.347}\\ \omega =24.251\text{rad/s}\end{array}$

Conclusion: The translational speed of sphere is $8.367\text{m/s}$ and the rotational speed is $24.251\text{rad/s}$ .

b.

To determine

To find: The ratio of translational KE to rotational KE.

Answer to Problem 48P

The ratio of translational KE to rotational KE is $\frac{5}{2}$.

Explanation of Solution

Concept used:

Translational KE = $\frac{1}{2}{\text{mv}}^{\text{2}}$

Rotational KE= $\frac{1}{5}{\text{mv}}^{\text{2}}$

Here,

  $v$ is the linear veloity,

  $m$ is the mass.

Calculation:

As given in question

  $\begin{array}{l}\frac{K{E}_{trans}}{K{E}_{rot}}=\frac{\frac{1}{2}{\text{mv}}^{\text{2}}}{\frac{1}{5}\text{mv}}\\ \frac{K{E}_{trans}}{K{E}_{rot}}=\frac{5}{2}\end{array}$

Conclusion:

Ratio of translational KE to rotational KE is $\frac{5}{2}$ .

c.

To determine

To check: Dependence on radius of sphere and its mass.

Answer to Problem 48P

Only angular speed depends on the radius and none of the result depends upon the mass.

Explanation of Solution

Given Information: $\omega =\frac{v}{r}$ and, $\frac{K{E}_{trans}}{K{E}_{rot}}=\frac{5}{2}$

Calculation: V does not depends upon m or r.

  $\omega =\frac{v}{r}$ also, dependent on radius not mass.

  $\frac{K{E}_{trans}}{K{E}_{rot}}=\frac{5}{2}$ so ,not dependent on mass or radius.

Conclusion: Only angular speed depends on the radius.