Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 649

A solid cylinder of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls down an inclined plane of height \(\mathrm{h}\). The angular velocity of the cylinder when it reaches the bottom of the plane will be. \(\\{\mathrm{A}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh})}\) \(\\{\mathrm{B}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh} / 2)}\) \(\\{\mathrm{C}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh} / 3)}\) \(\\{\mathrm{D}\\}(1 / 2 \mathrm{R}) \sqrt{(\mathrm{gh})}\)

Short Answer

Expert verified
The short answer is: The angular velocity of the cylinder when it reaches the bottom of the plane is \(\frac{2}{R}\sqrt{\frac{gh}{3}}\).

Step by step solution

01

Find the total potential energy at the incline's top.

At the top of the inclined plane, the cylinder has a potential energy given by: \(PE = mgh\) Where: - \(PE\) = potential energy of the cylinder - \(m\) = mass of the cylinder (\(M\)) - \(g\) = acceleration due to gravity - \(h\) = height of the inclined plane
02

Find the rotational and translational kinetic energies at the bottom of the incline.

At the bottom of the inclined plane, the cylinder has both translational kinetic energy (due to its linear motion) and rotational kinetic energy (due to its rotation). These can be given as follows: \(KE_t = \frac{1}{2} mv^2\) \(KE_r = \frac{1}{2} I \omega^2\) Where: - \(KE_t\) = translational kinetic energy - \(KE_r\) = rotational kinetic energy - \(v\) = linear velocity at the bottom of the incline - \(I\) = moment of inertia of the cylinder - \(\omega\) = angular velocity of the cylinder at the bottom of the incline The moment of inertia for a solid cylinder is \(I = \frac{1}{2} MR^2\).
03

Apply the conservation of energy.

By the conservation of energy, the potential energy at the top of the incline is equal to the sum of the translational and rotational kinetic energies at the bottom. Thus, \(mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2\) Substituting the moment of inertia for the solid cylinder, we get: \(mgh = \frac{1}{2} mv^2 + \frac{1}{4} MR^2 \omega^2\)
04

Relate linear velocity and angular velocity.

For a rolling object without slipping, the linear velocity \(v\) and the angular velocity \(\omega\) are related as follows: \(v = \omega R\) We can substitute this into the conservation of energy equation: \(mgh = \frac{1}{2} m(\omega R)^2 + \frac{1}{4} MR^2 \omega^2\)
05

Solve for angular velocity.

Now we can solve for the angular velocity \(\omega\): \(\omega^2(R^2 + \frac{1}{2}R^2) = 2gh\) Solving for \(\omega\), we get: \(\omega = \sqrt{\frac{2gh}{\frac{3}{2}R^2}}\) \(\omega = \sqrt{\frac{4gh}{3R^2}}\) \(\omega = \frac{2}{R}\sqrt{\frac{gh}{3}}\) Thus, the angular velocity of the cylinder when it reaches the bottom of the plane is \(\frac{2}{R}\sqrt{\frac{gh}{3}}\), which corresponds to option C.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the moment of inertia of a solid sphere of density \(\rho\) and radius \(\mathrm{R}\) about its diameter? \(\\{\mathrm{A}\\}(105 / 176) \mathrm{R}^{5} \rho\) \(\\{\mathrm{B}\\}(176 / 105) \mathrm{R}^{5} \rho\) \(\\{C\\}(105 / 176) R^{2} \rho\) \(\\{\mathrm{D}\\}(176 / 105) \mathrm{R}^{2} \rho\)

A thin circular ring of mass \(\mathrm{M}\) and radius \(\mathrm{r}\) is rotating about its axis with a constant angular velocity \(\mathrm{w}\). Two objects each of mass \(\mathrm{m}\) are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity.... $\\{\mathrm{A}\\}[\\{\omega(\mathrm{M}-2 \mathrm{~m})\\} /\\{\mathrm{M}+2 \mathrm{~m}\\}]$ \(\\{\mathrm{B}\\}[\\{\omega \mathrm{M}\\} /\\{\mathrm{M}+2 \mathrm{~m}\\}]\) \(\\{C\\}[\\{\omega M)\\} /\\{M+m\\}]\) \(\\{\mathrm{D}\\}[\\{\omega(\mathrm{M}+2 \mathrm{~m})\\} / \mathrm{M}]\)

The moment of inertia of a uniform circular disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) about any of its diameter is \((1 / 4) \mathrm{MR}^{2}\), what is the moment of inertia of the disc about an axis passing through its centre and normal to the disc? \(\\{\mathrm{A}\\} \mathrm{MR}^{2}\) \\{B \(\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(3 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\} 2 \mathrm{MR}^{2}\)

Match list I with list II and select the correct answer $$ \begin{aligned} &\begin{array}{|l|l|} \hline \text { List-I } & \begin{array}{l} \text { List - II } \\ \text { System } \end{array} & \text { Moment of inertia } \\ \hline \text { (x) A ring about it axis } & \text { (1) }\left(\mathrm{MR}^{2} / 2\right) \\ \hline \text { (y) A uniform circular disc about it axis } & \text { (2) }(2 / 5) \mathrm{MR}^{2} \\ \hline \text { (z) A solid sphere about any diameter } & \text { (3) }(7 / 5) \mathrm{MR}^{2} \\ \hline \text { (w) A solid sphere about any tangent } & \text { (4) } \mathrm{MR}^{2} \\ \cline { 2 } & \text { (5) }(9 / 5) \mathrm{MR}^{2} \\ \hline \end{array}\\\ &\text { Select correct option }\\\ &\begin{array}{|l|l|l|l|l|} \hline \text { Option? } & \mathrm{X} & \mathrm{Y} & \mathrm{Z} & \mathrm{W} \\\ \hline\\{\mathrm{A}\\} & 2 & 1 & 3 & 4 \\ \hline\\{\mathrm{B}\\} & 4 & 3 & 2 & 5 \\ \hline\\{\mathrm{C}\\} & 1 & 5 & 4 & 3 \\ \hline\\{\mathrm{D}\\} & 4 & 1 & 2 & 3 \\ \hline \end{array} \end{aligned} $$

The moment of inertia of a meter scale of mass \(0.6 \mathrm{~kg}\) about an axis perpendicular to the scale and passing through \(30 \mathrm{~cm}\) position on the scale is given by (Breath of scale is negligible). \(\\{\mathrm{A}\\} 0.104 \mathrm{kgm}^{2}\) \\{B \(\\} 0.208 \mathrm{kgm}^{2}\) \(\\{\mathrm{C}\\} 0.074 \mathrm{kgm}^{2}\) \(\\{\mathrm{D}\\} 0.148 \mathrm{kgm}^{2}\)
See all solutions

Recommended explanations on English Textbooks

Essay Writing Skills

Read Explanation

Free Response Essay

Read Explanation

Phonetics

Read Explanation

Cues and Conventions

Read Explanation

Synthesis Essay

Read Explanation

Multiple Choice Questions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free