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Q16P

Expert-verifiedFound in: Page 321

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

*Non-uniform cylindrical object.* InFigure, a cylindrical object of mass*M*** and radius** *R** *rolls smoothly from rest down a ramp and onto a horizontal section. From there it rolls off the ramp and onto the floor, landing a horizontal distance *d* = 0.506m from the end of the ramp. The initial height of the object is *H* = 0.90m**; the end of the ramp is at height h = 0.10m**

Value of $\beta $ is 0.25

- The horizontal distance
*d*= 0.506m - Initial height of ramp
*H*= 0.90m , - New height of the ramp
*h*= 0.10m

**This is based on the concept of energy conservation and projectile motion. Here, the body is made to freely along an inclined plane then, it gets projected from a height, h to drop at a distance from the ramp. Thus, the rolling of the cylindrical shell gives us the involvement of angular velocity to get the moment of inertia of the shell along its axis. Thus, the required term that is ${\mathit{\beta}}$ ****.**

Formulae:

The distance travelled by the body according to second law of kinematics,

$s=Vt+\frac{1}{2}g{t}^{2}$

where, *V* is the initial velocity of the body, *g* is the acceleration due to gravity, *t* is the time taken.

The transverse velocity of the body,

${V}_{com}={R}_{\omega}$

Where, *R* is the radius of the circular path, $\omega $ is the angular velocity of the body.

The time of flight of the projected body,

$t=\sqrt{\frac{2h}{g}}$

where, *h* is the height of the flight

By using energy conservation, we can write

$MgH=\frac{1}{2}l{\omega}^{2}+\frac{1}{2}m{v}_{0}^{2}+mgh$

As the cylinder would have potential energy only at height H. However, at new height h, part of the energy would be rotational kinetic energy, translational kinetic energy and potential energy.

Using, $v=r\omega $

$\begin{array}{rcl}Mg\left(H-h\right)& =& \frac{1}{2}l\left(\frac{{v}^{2}}{{R}^{2}}\right)+\frac{1}{2}m{v}_{0}^{2}\\ & =& \frac{1}{2}{v}_{0}^{2}\left(\frac{1}{{R}^{2}}+M\right)\\ \frac{2Mg\left(H-h\right)}{{v}_{0}^{2}}& =& \left(\frac{1}{{R}^{2}}+M\right)\\ \frac{2Mg\left(H-h\right)}{{v}_{0}^{2}}-M& =& \frac{1}{{R}^{2}}\\ M{R}^{2}\left(\frac{2g\left(H-h\right)}{{v}_{0}^{2}}-1\right)& =& {l}_{com}\\ {l}_{com}& =& M{R}^{2}\left[\frac{2g\left(H-h\right)}{{v}_{0}^{2}}-1\right]\end{array}$

In addition, we have

${l}_{com}=\beta M{R}^{2}$

We can say that

$\beta =\begin{array}{rcl}& & \frac{2g\left(H-h\right)}{{v}_{0}^{2}}\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 1\end{array}$

We know the vertical component of velocity is zero, we can say that V_{0y} = 0 and by using kinematic equation, we have

$\begin{array}{rcl}\u2206y& =& {v}_{i}t+\frac{1}{2}a{t}^{2}\\ h& =& \frac{1}{2}g{t}^{2}\\ t& =& \sqrt{\frac{2h}{g}}\end{array}$

Now,

$\begin{array}{rcl}s& =& {v}_{o}t+\frac{1}{2}g{t}^{2}\\ d& =& {v}_{o}\times \sqrt{\frac{2h}{g}}\\ {v}_{0}^{2}& =& \frac{g{d}^{2}}{2h}\end{array}$

After plugging the value of velocity in equation of $\beta $ we get,

$\begin{array}{rcl}\beta & =& \frac{2g\left(H-h\right)}{{v}_{0}^{2}}-1\\ & =& \frac{2g\left(H-h\right)}{\frac{g{d}^{2}}{2h}}-1\\ & =& \frac{4h\left(H-h\right)}{{d}^{2}}-1\\ & =& \frac{4\left(0.10m\right)\left(0.80m\right)}{0.256{m}^{2}}-1\\ & =& 0.25\end{array}$

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