/*! 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} Q15DQ A ball is rolling along at speed... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A ball is rolling along at speed vwithout slipping on a horizontal surface when it comes to a hill that rises at a constant angle above the horizontal. In which case will it go higher up the hill: if the hill has enough friction to prevent slipping or if the hill is perfectly smooth? Justify your answers in both cases in terms of energy conservation and in terms of Newton’s second law.

Short Answer

Expert verified

Only when there is a frictional force will the ball roll up.

Step by step solution

01

Newton’s second law of motion

When an external force acts on an item, the object accelerates in the same direction and at a rate proportional to the force. Which is given by

F=ma (1)

WhereF is external force, m is the mass of object, and a is acceleration of object.

02

Identification of the data

A ball is rolling along at speed without slipping on a horizontal surface when it comes to a hill that rises at a constant angle above the horizontal.

03

Finding the ball roll up the hill or not using Newton’s Second law

The ball would simply not go up the hill and instead continue to roll with the same angular velocity if there was no friction at all. The ball needs a little bit of friction to roll up the hill.

04

Finding the ball roll up the hill or not using Energy conservation

The initial kinetic energies of the ball's linear and rotational motion can be used to create gravitational potential energy. The ball will move farther as a result of the slope's increased friction.

The reason for this is because if a force is applied via the centre of mass of a ball that is at rest on a frictionless surface, the ball will glide over the surface without spinning. Given a force, the stationary ball will experience frictional force acting on its ground-contact edge, but in the opposite direction of the supplied force. The ball starts to rotate as a result of the torque produced by the friction.

A ball cannot start rolling without static friction as a result. The bottom edge of the ball will swiftly come to rest because the ball moves from top to bottom at a speed ofand its centre of mass moves at a speed of v. The ground's edge doesn't produce friction because it is immovable.

As a result, for a ball to begin rolling, friction is required; but, once the rolling condition has been met, friction no longer exists.

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!

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

The driver of a car wishes to pass a truck that is traveling at a constant speed of20.0m/s(about41mil/h). Initially, the car is also traveling at20.0m/s, and its front bumper is24.0mbehind the truck’s rear bumper. The car accelerates at a constant 0.600m/s2, then pulls back into the truck’s lane when the rear of the car is26.0mahead of the front of the truck. The car islong, and the truck is 21.0m long. (a) How much time is required for the car to pass the truck? (b) What distance does the car travel during this time? (c) What is the final speed of the car?

Question- Neptunium. In the fall of 2002, scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a nuclear chain reaction. Neptunium-237 has a density of 19.5 g/cm3. What would be the radius of a sphere of this material that has a critical mass?

For a spherical planet with mass M, volume V, and radius R,derive an expression for the acceleration due to gravity at the planet’s surface, g, in terms of the average density of the planet, ÒÏ=M/V, and the planet’s diameter, D=2R. The table gives the values of Dand gfor the eight major planets:

(a) Treat the planets as spheres. Your equation for as a function of and shows that if the average density of the planets is constant, a graph of versus will be well represented by a straight line. Graph as a function of for the eight major planets. What does the graph tell you about the variation in average density? (b) Calculate the average density for each major planet. List the planets in order of decreasing density, and give the calculated average density of each. (c) The earth is not a uniform sphere and has greater density near its center. It is reasonable to assume this might be true for the other planets. Discuss the effect this nonuniformity has on your analysis. (d) If Saturn had the same average density as the earth, what would be the value of at Saturn’s surface?

A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by the equationxt=αt2-βt3 where role="math" localid="1655226337795" α=1.50 m/s2androle="math" localid="1655226362269" β=0.0500m/s2 . Calculate the average velocity of the car for each time interval: (a) t = 0 to t = 2.00 s; (b) t = 0 to t = 4.00 s; (c) t = 2.00 s to t = 4.00 s.

Question-. (a) Does it make sense to say that a vector is negative? Why? (b) Does it make sense to say that one vector is the negative of another? Why? Does your answer here contradict what you said in part (a)?

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.