/*! 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 17 A solid disk of radius 8.50 \(\m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A solid disk of radius 8.50 \(\mathrm{cm}\) and mass \(1.25 \mathrm{kg},\) which is rolling at a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) , begins rolling without slipping up a \(10.0^{\circ}\) slope. How long will it take for the disk to come to a stop?

Short Answer

Expert verified
The disk takes approximately 1.47 seconds to come to a stop.

Step by step solution

01

Determine the Total Mechanical Energy

The total mechanical energy at the start is the sum of the translational and rotational kinetic energies. The translational kinetic energy (TKE) is given by \( \frac{1}{2}mv^{2} \), where \( m=1.25\,\text{kg} \) and \( v=2.50\,\text{m/s} \). The rotational kinetic energy (RKE) for a rolling solid disk is \( \frac{1}{2}I\omega^{2} \), where \( I=\frac{1}{2}mR^2 \) and \( \omega=\frac{v}{R} \). Calculate both.
02

Calculate Translational Kinetic Energy

Use the formula \(TKE = \frac{1}{2}mv^{2}\) where \(m = 1.25\,\text{kg}\) and \(v = 2.50\,\text{m/s}\). This gives:\[TKE = \frac{1}{2} \times 1.25 \times (2.50)^2 = 3.90625\,\text{Joules}\]
03

Calculate Rotational Kinetic Energy

First, find the moment of inertia \(I\) using \(I = \frac{1}{2} m R^2\). With \(m = 1.25\,\text{kg}\) and \(R = 0.085\,\text{m}\), we have:\[I = \frac{1}{2} \times 1.25 \times (0.085)^2 = 0.00453125\,\text{kg}\cdot\text{m}^2\]Now, \(\omega = \frac{v}{R} = \frac{2.50}{0.085} = 29.41\,\text{rad/s}\). Hence the RKE:\[RKE = \frac{1}{2} \times 0.00453125 \times (29.41)^2 = 1.9609375\,\text{Joules}\]
04

Calculate Total Initial Mechanical Energy

Add the translational and rotational kinetic energies together to find the total mechanical energy:\[E_{\text{total}} = TKE + RKE = 3.90625 + 1.9609375 = 5.8671875\,\text{Joules}\]
05

Relate Potential Energy Gain to Kinetic Energy Loss

As the disk rolls up the ramp, the kinetic energy converts to potential energy until all kinetic energy is converted, at which point the disk stops. The potential energy at height \(h\) is given by \(PE = mgh\) and is set equal to the total initial mechanical energy calculated earlier. Use this to find the height \(h\):\[mgh = E_{\text{total}}\]\[1.25 \times 9.81 \times h = 5.8671875\,\text{Joules}\]
06

Calculate Height of the Ramp

Rearrange the equation from Step 5 to solve for \(h\):\[h = \frac{5.8671875}{1.25 \times 9.81} \approx 0.48\,\text{m}\]
07

Calculate Distance Along the Ramp

The height \(h = 0.48\,\text{m}\) is related to the distance \(d\) up the incline by the equation \( h = d \sin(\theta) \). Given \(\theta = 10^{\circ}\), solve for \(d\):\[0.48 = d \sin(10^{\circ})\]\[d = \frac{0.48}{\sin(10^{\circ})} \approx 2.76\,\text{m}\]
08

Calculate Time to Stop

The time \(t\) taken can be found using the kinematic equation for motion: \( v^2 = u^2 + 2as \), where \( u = 2.50\,\text{m/s} \) (initial velocity), \( v = 0 \) (final velocity), and \( s = 2.76\,\text{m} \). Find \( a \), the deceleration (which is due to gravity along the incline):\[a = g \sin(\theta) = 9.81 \cdot \sin(10^{\circ}) \approx 1.70\,\text{m/s}^2\]Using \( v = u + at \), solve:\[0 = 2.50 + (-1.70)t\]\[t = \frac{2.50}{1.70} \approx 1.47\,\text{seconds}\]

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mechanical Energy
Mechanical energy is the sum of kinetic and potential energies of a system. In the case of a rolling disk, both translational and rotational kinetic energies contribute to the total mechanical energy. Understanding this energy helps predict how the disk behaves under different conditions. When the disk rolls up the incline, its kinetic energy decreases as it is converted to potential energy. This energy conservation helps determine how far the disk will travel up the slope before stopping. The total mechanical energy at the beginning is the initial translational and rotational kinetic energies. In this problem, these energies sum to provide the total mechanical energy, which remains constant if we ignore friction and air resistance.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. For an object rolling without slipping like our disk, it has two components of kinetic energy: translational kinetic energy and rotational kinetic energy.
  • Translational kinetic energy is given by the formula: \( rac{1}{2}mv^2 \).
  • Rotational kinetic energy is calculated as \( rac{1}{2}I\omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is angular velocity.
Together, both forms of kinetic energy determine how fast and energetically the disk is moving.
Moment of Inertia
The moment of inertia, often referred to as rotational inertia, quantifies how difficult it is to change an object's rotational motion. It depends on the mass of the object and its distribution relative to the axis of rotation. For a solid disk, the moment of inertia is calculated using \( I = \frac{1}{2} m R^2 \).
  • Here, \( m \) is the mass of the disk.
  • \( R \) is the radius of the disk.
The moment of inertia plays a key role in calculating the rotational kinetic energy, crucial for understanding the dynamics of rolling motion.
Potential Energy
Potential energy refers to the energy stored within an object due to its position in a force field, commonly a gravitational field. As the disk climbs the slope, its kinetic energy transforms into gravitational potential energy, given by \( PE = mgh \), where:
  • \( m \) is the mass,
  • \( g \) is the acceleration due to gravity,
  • \( h \) is the height the object has risen to.
When the disk stops at a height, all its initial kinetic energy is converted to potential energy, allowing us to determine the height it reaches and the distance it travels.
Inclined Plane Motion
Inclined plane motion involves objects moving along a slope, where the gravitational force acts along and perpendicular to the incline. The component of gravitational force parallel to the incline causes the object to accelerate or decelerate. This is calculated using: \( a = g \sin(\theta) \).
  • \( \theta \) is the incline's angle.
  • \( a \) is the acceleration parallel to the slope.
This acceleration determines how quickly the disk slows as it ascends the slope, influencing how far up the plane it will travel before stopping. Understanding how these forces work is crucial for solving problems involving inclined planes.

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

Disks \(A\) and \(B\) are mounted on shaft \(S S\) and may be connected or dis- connected by clutch \(C .\) (See Figure \(10.82 . )\) Disk \(A\) is made of a lighter material than disk \(B,\) so the moment of inertia of disk \(A\) about the shaft is one-third that of disk \(B\) . The moments of inertia of the shaft and clutch are negligible. With the clutch disconnected, \(A\) is brought up to an angular speed \(\omega_{0 . \text { . The }}\) accelerating torque is then removed from \(A,\) and \(A\) , and \(A\) is coupled to disk \(B\) by the clutch. (You can ignore bearing friction.) It is found that 2400 \(\mathrm{J}\) of thermal energy is developed in the clutch when the connection is made. What was the original kinetic energy of disk \(A\) ?

Back pains during pregnancy. Women often suffer from back pains during pregnancy. Let us investigate the cause of these pains, assuming that the woman's mass is 60 kg before pregnancy. Typically, women gain about 10 kg during pregnancy, due to the weight of the fetus, placenta, amniotic fluid, etc. To make the calculations easy, but still realistic, we shall model the unpregnant woman as a uniform cylinder of diameter 30 \(\mathrm{cm} .\) We can model the added mass due to the fetus as a 10 kg sphere 25 \(\mathrm{cm}\) in diameter and centered about 5 \(\mathrm{cm}\) outside the woman's original front surface. (a) By how much does her pregnancy change the horizontal location of the woman's center of mass? (b) How does the change in part (a) affect the way the pregnant woman must stand and walk? In other words, what must she do to her posture to make up for her shifted center of mass? (c) Can you now explain why she might have backaches?

\bullet A small block on a frictionless horizontal surface has a mass of 0.0250 \(\mathrm{kg}\) . It is attached to a massless cord passing through a hole in the surface. (See Figure \(10.54 .\) The block is originally revolving at a distance of 0.300 \(\mathrm{m}\) from the hole with an angular speed of 1.75 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 \(\mathrm{m} .\) You may treat the block as a particle. (a) Is angular momentum conserved? Why or why not? (b) What is the new angular speed? (c) Find the change in kinetic energy of the block. (d) How much work was done in pulling the cord?

A solid, uniform cylinder with mass 8.25 \(\mathrm{kg}\) and diameter 15.0 \(\mathrm{cm}\) is spinning at 220 \(\mathrm{rpm}\) on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is \(0.333 .\) What must the applied normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?

What would be the radial (centripetal) acceleration of an astronaut standing on the inner surface of the space station, which is at a distance of 240 \(\mathrm{m}\) from the axis of rotation, once the space station reaches its final angular velocity of 0.20 \(\mathrm{rad} / \mathrm{s}\) ? A. 0 \(\mathrm{m} / \mathrm{s}^{2}\) B. 4.8 \(\mathrm{m} / \mathrm{s}^{2}\) \(\mathrm{C}, 9.6 \mathrm{m} / \mathrm{s}^{2}\) D. 48 \(\mathrm{m} / \mathrm{s}^{2}\)

See all solutions

Recommended explanations on Physics Textbooks

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.