Chapter 8, Problem 77GP

(a)

To determine

To Find: The distance up the inclined plane that a hollow cylinder covers.

Answer to Problem 77GP

  $4.3\text{m}$

Explanation of Solution

Given data:

Speed, $v=3.3\text{m/s}$

Angle, $\theta ={15}^0$

Formula used:

  $\begin{array}{l}{v}^2-u^2=2as\\ a=\frac{g\sin \theta }{1+\left(\frac{k^2}{r^2}\right)}\end{array}$

Here,

  $k$ is the radius of gyration,

  $\theta$ is the angle of inclination,

  $a$ is the acceleration,

  $v$ is the final velocity,

  $u$ is the initial velocity,

  $s$ is the displacement.

Calculation:

On an inclined plane, for rolling up or down the body may accelerate or decelerates.

  $\begin{array}{l}{v}^2-u^2=2as\\ a=\frac{g\sin \theta }{1+\left(\frac{k^2}{r^2}\right)}\end{array}$

For hollow cylinder $k=r$ (Hoop)

  $a=\frac{g\sin 15°}{1+1}$

  $\text{a=1}{\text{.268 m/s}}^2$

Now substitute the value of a ,

  $\begin{array}{l}{v}^2=u^2-2as\\ 0={3.3}^2-(2\times 1.268\times s)\\ \frac{{3.3}^2}{2\times 1.268}=s\\ s=4.29\text{ m}\end{array}$

Conclusion: 

The height covered by the hollow cylinder is 4.29 m.

(b)

To determine

To Find: The time that the hollow cylinder should be on the incline before it arrives back at the bottom.

Answer to Problem 77GP

  $t=2.6\text{s}$

Explanation of Solution

Given data:

Speed $v=3.3\text{m/s}$

Formula used:

  $v=u+at$

Here,

  $a$ is the acceleration,

  $v$ is the final velocity,

  $u$ is the initial velocity,

  $t$ is the time.

Calculation:

The following equation used to determine the time,

  $\begin{array}{l}v=u+at\\ v\text{=0}\end{array}$

and moving upwards gives negative sign

  $\begin{array}{l}0=3.3-at\\ 0=3.3-1.268\times t\\ 3.3=1.268t\\ t=2.6\text{ s}\end{array}$

Conclusion:

For the period of $2.6\text{s}$ , the hollow cylinder should be on the incline before it arrives back at the bottom.