/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 A marble of mass \(m\) and radiu... [FREE SOLUTION] | 91Ó°ÊÓ

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A marble of mass \(m\) and radius \(r\) rolls along the looped rough track of Fig. \(10-67 .\) What is the minimum value of the vertical height \(h\) that the marble must drop if it is to reach the highest point of the loop without leaving the track? (a) Assume \(r \ll R ;\) (b) do not make this assumption. Ignore frictional losses.

Short Answer

Expert verified
(a) \( h = \frac{5}{2}R \); (b) \( h = \frac{7}{2}R \).

Step by step solution

01

Understand the Problem

We need to calculate the minimum vertical height \( h \) from which a marble must be released to just barely complete a loop of radius \( R \). This revolves around converting potential energy into kinetic energy and ensuring enough radial force at the top of the loop.
02

Analyzing Forces at the Top of the Loop

At the top of the loop, the marble must have enough speed to maintain contact with the track. The gravitational force must provide the necessary centripetal force. This condition is given by \[ mg = \frac{mv^2}{R} \] where \( v \) is the velocity at the top of the loop.
03

Relating Velocity and Energy

The potential energy at height \( h \) converts to kinetic energy. The total mechanical energy at the starting height must equal at least the sum of the kinetic energy and potential energy at the top of the loop, \[ mgh = mg(2R) + \frac{1}{2}mv^2. \]
04

Solving for Velocity at the Top

From the equation \( mg = \frac{mv^2}{R} \), we find \( v^2 = gR \).
05

Solving the Energy Equation

Substitute the value of \( v^2 \) from Step 4 into the energy equation: \[ mgh = mg(2R) + \frac{1}{2}m(gR). \] Simplify to \( h = \frac{5}{2}R \).
06

Considering the Radius Assumptions Part(a)

Under the assumption \( r \ll R \), the rotational energy of the marble can be ignored compared to the translational energy, and the height calculated is \( h = \frac{5}{2}R \).
07

Considering the Rotational Energy Part(b)

Without assuming \( r \ll R \), we must include rotational energy. The moment of inertia \( I \) of a sphere is \( \frac{2}{5}mr^2 \), and the rotational kinetic energy is \( \frac{1}{2}I \omega^2 = \frac{1}{2}\frac{2}{5}mr^2\left(\frac{v^2}{r^2}\right) = \frac{1}{5}mv^2 \).
08

Setting Up Energy Equation Considering Rotation

Accounting for both translational and rotational kinetic energy at the top of the loop,\[ mgh = mg(2R) + \frac{1}{2}mv^2 + \frac{1}{5}mv^2 \]. Solving this equation leads to \( h = \frac{7}{2}R \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
Centripetal force is the invisible hand that keeps objects moving in a circular path. It always points towards the center of the circle and is essential for objects to maintain circular motion. For a marble rolling through a loop, the centripetal force is the net force required to keep it from deviating from its circular track.

At the top of the loop, gravity acts as the centripetal force. This means that the force of gravity is just enough to keep the marble in contact with the track. The condition for this is given by the equation: \[ mg = \frac{mv^2}{R} \]Here, \( m \) is the mass of the marble, \( g \) is the acceleration due to gravity, \( v \) is the velocity, and \( R \) is the loop radius. This relationship shows how the marble needs to have a certain speed, determined by gravity and the loop's radius, to stay on its circular path.

If the marble travels too slowly, it won't generate enough centripetal force and will fall from the track. If it goes fast enough to satisfy this equation, it will make it to the top successfully.
Rotational Kinetic Energy
Rotational kinetic energy is the energy possessed by a rotating object due to its rotation. Just as objects in linear motion have kinetic energy, rotating objects have a similar form of energy. The marble not only translates (moves along a path) but also rotates as it rolls along the looped track.

The rotational kinetic energy is expressed as:\[ \frac{1}{2}I \omega^2 \]Where \( I \) is the moment of inertia, and \( \omega \) is the angular velocity. For a sphere such as a marble, the moment of inertia is \( \frac{2}{5}mr^2 \), where \( r \) is the radius of the marble. This factor shows that smaller or larger marbles will have different rotational energies when they spin.

When solving for the minimum height \( h \) required for the marble to complete a loop, rotational kinetic energy must be considered, especially if the marble's radius is significant compared to the loop's radius. Incorporating rotational energy into the energy conservation equation gives a more precise calculation of \( h \), ensuring the marble completes the loop.
Moment of Inertia
Moment of inertia is a crucial concept in rotational dynamics, analogous to mass in linear motion. It determines how much torque is required for a given angular acceleration around an axis. For different shapes and distributions of mass, the moment of inertia varies.

For the marble in the scenario, the moment of inertia is given by:\[ I = \frac{2}{5}mr^2 \]This expression signifies that how much mass is distributed around the axis of rotation impacts rotational motion. When accounting for the marble's rotation as it moves through the loop, the moment of inertia helps determine its rotational kinetic energy.

Understanding moment of inertia aids in solving problems where rotation is involved, particularly when the object's spinning nature affects its motion, as it does in the marble's loop scenario. Considering the moment of inertia in energy calculations ensures accurate predictions of the initial height required to complete the loop.

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Most popular questions from this chapter

One possibility for a low-pollution automobile is for it to use energy stored in a heavy rotating flywheel. Suppose such a car has a total mass of \(1100 \mathrm{~kg}\), uses a uniform cylindrical flywheel of diameter \(1.50 \mathrm{~m}\) and mass \(240 \mathrm{~kg},\) and should be able to travel \(350 \mathrm{~km}\) without needing a flywheel "spinup." (a) Make reasonable assumptions (average frictional retarding force \(=450 \mathrm{~N},\) twenty acceleration periods from rest to \(95 \mathrm{~km} / \mathrm{h},\) equal uphill and downhill, and that energy can be put back into the flywheel as the car goes downhill), and estimate what total energy needs to be stored in the flywheel. ( \(b\) ) What is the angular velocity of the flywheel when it has a full "energy charge"? (c) About how long would it take a 150 -hp motor to give the flywheel a full energy charge before a trip?

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