A 1500kg car drives around a flat 200-m-diameter circular track at 25m/s. What are the magnitude and direction of the net force on the car? What causes this force?

Short Answer

Expert verified

The net force is Fnet=9.4×103N and its direction is toward the center.

The static friction causes this force.

Step by step solution

01

Given Information

A 1500kg car drives around a flat 200-m-diameter circular track at 25m/s.

02

Explanation

The car's velocity is tangent to the circle of motion, and its acceleration is called the centripetal acceleration and it points toward the center of the circle. This acceleration is given by equation (8.4) in the form

a=v2r

Where r is the radius of the circle and v is the velocity of the car. At a uniform circular motion, the velocity vector has a tangential component and the acceleration vector has a radial component. From Newton 's first law, the car has a net force exerted on it as it doesn't move with a constant velocity. So, using the expression of the acceleration from equation (1), we get the net force by

Fnet=ma=mv2r
03

Explanation

This force has a direction toward the center of the circular motion. The car moves in a circle with a diameter d=200m, so we can find its radius rby

r=d2=200m2=100m

Now, we plug the values for m,vand rinto equation (2) to get Fnet

Fnet=mv2r

=(1500kg)(25m/s)2100m

=9.375×103N

=9.4×103N

The car moves on a road in a circle motion, there is no tension force between the car and the center of the circle, also there is no normal force, so the static friction causes this force.

04

Final Answer

The net force is Fnet=9.4×103N and its direction is toward the center.

The static friction causes this force

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 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

The ultracentrifuge is an important tool for separating and analyzing proteins. Because of the enormous centripetal accelerations, the centrifuge must be carefully balanced, with each sample matched by a sample of identical mass on the opposite side. Any difference in the masses of opposing samples creates a net force on the shaft of the rotor, potentially leading to a catastrophic failure of the apparatus. Suppose a scientist makes a slight error in sample preparation and one sample has a mass 10 mg larger than the opposing sample. If the samples are 12 cm from the axis of the rotor and the ultracentrifuge spins at 70,000 rpm, what is the magnitude of the net force on the rotor due to the unbalanced samples?

A 500 g steel block rotates on a steel table while attached to a 2.0-m-long massless rod. Compressed air fed through the rod is ejected from a nozzle on the back of the block, exerting a thrust force of 3.5 N. The nozzle is 70o­ from the radial line, as shown in FIGURE P8.62. The block starts from rest.
a. What is the block’s angular velocity after 10 rev?
b. What is the tension in the rod after 10 rev?

Two wires are tied to the 2.0 kg sphere shown in FIGURE P8.45. The sphere revolves in a horizontal circle at constant speed.
a. For what speed is the tension the same in both wires?
b. What is the tension?

A 500g ball swings in a vertical circle at the end of a 1.5-m-long string. When the ball is at the bottom of the circle, the tension in the string is 15N. What is the speed of the ball at that point?

Suppose you swing a ball of mass m in a vertical circle on a string of length L. As you probably know from experience, there is a minimum angular velocity ωmin you must maintain if you want the ball to complete the full circle without the string going slack at the top.
a. Find an expression for ωmin

b. Evaluateωminin rpm for a 65 g ball tied to a 1.0-m-long string.

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.

Sign-up for free