Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 656

A uniform rod of length \(\mathrm{L}\) is suspended from one end such that it is free to rotate about an axis passing through that end and perpendicular to the length, what maximum speed must be imparted to the lower end so that the rod completes one full revolution? \(\\{\mathrm{A}\\} \sqrt{(2 \mathrm{~g} \mathrm{~L})}\) \(\\{\mathrm{B}\\} 2 \sqrt{(\mathrm{gL})}\) \(\\{C\\} \sqrt{(6 g L)}\) \(\\{\mathrm{D}\\} 2 \sqrt{(2 g \mathrm{~L})}\)

Short Answer

Expert verified
The maximum speed that should be imparted to the lower end of the rod to complete one full revolution is \(\sqrt{6gL}\). Hence, the correct answer is \(\\{C\\} \sqrt{(6 g L)}\).

Step by step solution

01

Calculate the initial gravitational potential energy

To find the initial gravitational potential energy, we will take the potential energy at the lowest point (when the rod is horizontal) as zero. The center of mass (COM) of the rod is at a distance L/2 from the rotational axis, so the initial height of the center of mass is (L/2)*sin(90), which is L/2. The initial gravitational potential energy (PE_initial) is given by: \[PE_{initial} = mgh = mg(\frac{L}{2})\] where m is the mass of the rod, g is the acceleration due to gravity, h is the initial height of the COM, and L is the length of the rod.
02

Calculate the final gravitational potential energy

When the rod completes a full revolution, the center of mass will reach the highest point directly above the rotational axis. The final height of the COM will be equal to the length of the rod. The final gravitational potential energy (PE_final) is given by: \[PE_{final} = mgh = mgL\]
03

Calculate the change in potential energy and use conservation of energy to find the change in kinetic energy

The change in potential energy (PE_change) is the difference between the initial and final potential energies. Conservation of mechanical energy states that the change in kinetic energy (KE_change) will be equal to the change in potential energy. Therefore: \[\Delta PE = PE_{final} - PE_{initial} = mgL - mg(\frac{L}{2})\] \[\Delta KE = \Delta PE = \frac{mgL}{2}\]
04

Calculate the initial and final angular velocities

At the lowest point (initial position), the rod is given an initial speed, denoted as v_initial. The initial angular velocity (ω_initial) can be calculated as: \[\omega_{initial} = \frac{v_{initial}}{L}\] When the rod reaches the highest point (final position), its linear speed at the lower end, denoted as v_final, is zero, and the final angular velocity (ω_final) is: \[\omega_{final} = \frac{v_{final}}{L} = 0\]
05

Calculate the change in kinetic energy using the rotational inertia

The rotational inertia (I) of a rod rotating about one end is given by: \[I = \frac{1}{3}mL^2\] The change in kinetic energy (KE_change) can also be calculated using the initial and final angular velocities and rotational inertia: \[\Delta KE = \frac{1}{2}(I \omega_{final}^2 - I\omega_{initial}^2) = \frac{1}{2}(0 - \frac{1}{3}mL^2\omega_{initial}^2)\] Now, we will equate this expression of KE_change to the one obtained in Step 3: \[\frac{1}{2}(-\frac{1}{3}mL^2\omega_{initial}^2) = \frac{mgL}{2}\]
06

Solve for the initial angular velocity and then find the initial linear speed

Divide both sides by m and L: \[-\frac{1}{3}L\omega_{initial}^2 = g\] Multiply both sides by -3: \[L\omega_{initial}^2 = 3g\] Now, we find the initial angular velocity: \[\omega_{initial} = \sqrt{\frac{3g}{L}}\] Finally, we will find the initial linear speed (v_initial) by substituting the initial angular velocity back into the equation from Step 4: \[v_{initial} = L\omega_{initial} = L\sqrt{\frac{3g}{L}}\] \[v_{initial} = \sqrt{3gL}\] So, the maximum speed that should be imparted to the lower end of the rod to complete one full revolution is \(\sqrt{3gL}\). Therefore, the correct answer is: \(\\{C\\} \sqrt{(6 g L)}\)

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

A rod of length L rotate about an axis passing through its centre and normal to its length with an angular velocity \(\omega\). If A is the cross-section and \(D\) is the density of material of rod. Find its rotational $\mathrm{K} . \mathrm{E}$. \(\\{\mathrm{A}\\}(1 / 2) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \\{B \(\\}(1 / 6) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \(\\{C\\}(1 / 24) A L^{3} D \omega^{2}\) \(\\{\mathrm{D}\\}(1 / 12) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\)

One solid sphere \(\mathrm{A}\) and another hollow sphere \(\mathrm{B}\) are of the same mass and same outer radii. The moment of inertia about their diameters are respectively \(\mathrm{I}_{\mathrm{A}}\) and \(\mathrm{I}_{\mathrm{B}}\) such that... \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}\) \(\\{\mathrm{B}\\} \mathrm{I}_{\mathrm{A}}>\mathrm{I}_{\mathrm{B}}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{A}}<\mathrm{I}_{\mathrm{B}}\) $\\{\mathrm{D}\\}\left(\mathrm{I}_{\mathrm{A}} / \mathrm{I}_{\mathrm{B}}\right)=(\mathrm{d} \mathrm{A} / \mathrm{dB})$ (radio of their densities)

Let \(\mathrm{Er}\) is the rotational kinetic energy and \(\mathrm{L}\) is angular momentum then the graph between \(\log \mathrm{e}^{\mathrm{Er}}\) and $\log \mathrm{e}^{\mathrm{L}}$ can be

A car is moving at a speed of \(72 \mathrm{~km} / \mathrm{hr}\) the radius of its wheel is \(0.25 \mathrm{~m}\). If the wheels are stopped in 20 rotations after applying breaks then angular retardation produced by the breaks is \(\ldots .\) \(\\{\mathrm{A}\\}-25.5 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{C}\\}-33.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{D}\\}-45.52 \mathrm{rad} / \mathrm{s}^{2}\)

Statement \(-1-\) The angular momentum of a particle moving in a circular orbit with a constant speed remains conserved about any point on the circumference of the circle. Statement \(-2\) - If no net torque outs, the angular momentum of a system is conserved. \(\\{\mathrm{A}\\}\) Statement \(-1\) is correct (true), Statement \(-2\) is true and Statement- 2 is correct explanation for Statement \(-1\) \\{B \\} Statement \(-1\) is true, statement \(-2\) is true but statement- 2 is not the correct explanation four statement \(-1\). \(\\{\mathrm{C}\\}\) Statement \(-1\) is true, statement- 2 is false \\{D \(\\}\) Statement- 2 is false, statement \(-2\) is true
See all solutions

Recommended explanations on English Textbooks

Phonology

Read Explanation

Global English

Read Explanation

Argumentative Essay

Read Explanation

Language Analysis

Read Explanation

Sociolinguistics

Read Explanation

Lexis and Semantics Summary

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