Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Q14P (page 377)

Agroup of particles of total mass 35kghas a total kinetic energy of 340J. The kinetic energy relative to the center of mass is 85J. What is the speed of the center of mass?

Short Answer

Expert verified

The speed of the center of mass is4.93m/s

Step by step solution

01

Identification of the given data

  • The total mass is m=35kg
  • Total kinetic energy is KT=340J
  • Kinetic energy relative to the center of mass is K=85J
02

Concept of the center of mass of the system

The center of mass of the system is calculated by considering the placement of the of all the objects acting in the system.

03

Determination of the speed of the center of mass

According to the conservation of energy principle,

KT+K=E

Where,

KT= Total kinetic energy

K = Kinetic energy relative to the center of mass

Potential Energy, localid="1653897038194" role="math" E=12mvc2

Substitute Equation (2) in (1),

KT+K=12mvc2vc2=2·KT+Km=2·340+8535vc2=24.29vc=4.93m/s

Hence,the speed of the center of mass is localid="1653897138661" 4.93m/s

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 solid uniform-density sphere is tied to a rope and moves in a circle with speed v. The distance from the center of the circle to the center of the sphere is d, the mass of the sphere is M, and the radius of the sphere is R. (a) What is the angular speed ω? (b) What is the rotational kinetic energy of the sphere? (c) What is the total kinetic energy of the sphere?

Tarzan, whose mass is 100kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets go, his center of mass is at a height 2.9m above the ground and the bottom of his dangling feet are at a height 2.1 above the ground. When he first hits the ground he has dropped a distance 2.1, so his center of mass is (2.9-2.1) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height above the ground. (a) Consider the point particle system. What is the speed v at the instant just before Tarzan's feet touch the ground? (b) Consider the extended system. What is the net change in internal energy for Tarzan from just before his feet touch the ground to when he is in the crouched position?

It is sometimes claimed that friction forces always slow an object down, but this is not true. If you place a box of mass Mon a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed vof the belt. The coefficient of friction between box and belt isμ. (a) What is the distance d(relative to the floor) that the box moves before reaching the final speed v? Use energy arguments, and explain your reasoning carefully. (b) How much time does it take for the box to reach its final speed? (c) The belt and box of course get hot. Is the effective distance through which the friction force acts on the box greater than or less than d? Give as quantitative an argument as possible. You can assume that the process is quick enough that you can neglect transfer of energyQ due to a temperature difference between the belt and the box. Do not attempt to use the results of the friction analysis in this chapter; rather, apply the methods of that analysis to this different situation. (d) Explain the result of part (c) qualitatively from a microscopic point of view, including physics diagrams.E

Under what conditions does the energy equation for the point particle system differ from the energy equation for the extended system? Give two examples of such a situation. Give one example of a situation where the two equations look exactly alike.

By calculating numerical quantities for a multiparticle system. One can get a concrete sense of the meaning of the relationships p→sys=Mtotv→CMand Ktot=Ktrans+Krel. Consider an object consisting of two balls connected by a spring, whose stiffness is 400 N/m. The object has been thrown through the air and is rotating and vibrating as it moves. At a particular instant, the spring is stretched 0.3m, and the two balls at the ends of the spring have the following masses and velocities: 1:5kg.(8,14,0)m/s2:3kg(-5,9,0)m/s

(a)For this system, calculate p→sys. (b) Calculate v→CM(c) Calculate Ktot3. (d) Calculate Ktrans. (e) Calculate Krel. (f) Here is a way to check your result for Krel. The velocity of a particle relative to the center of mass is calculated by subtracting v→CMfrom the particle’s velocity. To take a simple example, if you’re riding in a car that’s moving with v→CM,x=20m/sand you throw a ball with v→CM,x=35m/s, relative to the car, a bystander on the ground sees the ball moving with vx=55m/sSo v→=v→CM=v→reland therefore we have=v→relv→=v→CMfor each mass and calculate the correspondingKrel. Compare with the result you obtained in part (e).
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Measurements

Read Explanation

Electricity and Magnetism

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Electromagnetism

Read Explanation

Magnetism and Electromagnetic Induction

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