A 200g block on a 50-cm-long string swings in a circle on a horizontal, frictionless table at 75rpm.

a. What is the speed of the block?

b. What is the tension in the string?

Short Answer

Expert verified

a). The speed of the block is 3.9m/s.

b). The tension in the string is6.2N.

Step by step solution

01

Step 1:Given Information (Part a)

A200g block on a 50-cm-long string swings in a circle on a horizontal, frictionless table at 75rpm.

02

Explanation (Part a)

The block 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. The block has angular velocity ωand a tangential speed vTand they are related to each other by

vT=rω

Where ris the radius of the circle. The angular velocity is given by ω=75rpm=75rev/min. InSI units, the angular frequency has unit radians per second ( rad/s). So, let us convert the unit from and 1min=60seconds

ω=75revmin2πrad1rev1min60s=7.85rad/s

Now, plug the values for ωand rinto equation (1) to get vT

vT=ωr=(7.85rad/s)(0.50m)=3.925m/s3.9m/s
03

Final Answer (Part a)

The speed of the block is3.9m/s.

04

Given Information (Part b)

A 200g block on a 50-cm-long string swings in a circle on a horizontal, frictionless table at 75rpm.

05

Explanation (Part b)

This acceleration of the block 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 block that we are calculated in part (a). 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 block 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 (2), we get the net force by

Fnet=ma=mv2r
06

Explanation (Part b)

This force has a direction toward the center of the circular motion. The block moves in a circle with a radius r=0.50m. Now, we plug the values for m,vand rinto equation (3) to get Fnet

Fnet=mv2r

=(0.200kg)(3.925m/s)20.50m

=6.16N

6.2N

This force is an identifiable agent and it is one of our familiar forces such as friction, normal or tension forces. The block moves on the table in a circular motion, there is no friction force because the table is frictionless, also there is no horizontally normal force, so the tension force in the wire causes this force. Therefore, the tension force will be

T=6.2N
07

Final Answer (Part b)

The tension force will be6.2N.

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

While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 15m, and you use your watch to find that each loop around takes 25s.

a. What are your speed and the magnitude of your acceleration?

b. What is the ratio of your weight at the top of the ride to your weight while standing on the ground?

c. What is the ratio of your weight at the bottom of the ride to your weight while standing on the ground?

A 2.0kgprojectile with initial velocity v=8.0ı^m/sexperiences the variable force F=-2.0tl^+4.0t2ȷ^N, where t is in s.

a. What is the projectile’s speed at t=2.0s?

b. At what instant of time is the projectile moving parallel to they-axis?

A small bead slides around a horizontal circle at height y inside the cone shown in FIGURE CP8.69. Find an expression for the bead’s speed in terms of a, h, y, and g.

A 4.0*1010kg asteroid is heading directly toward the center of the earth at a steady 20km/s. To save the planet, astronauts strap a giant rocket to the asteroid perpendicular to its direction of travel. The rocket generates 5.0*109N of thrust. The rocket is fired when the asteroid is 4.0*106km away from earth. You can ignore the earth’s gravitational force on the asteroid and their rotation about the sun.

a. If the mission fails, how many hours is it until the asteroid impacts the earth?

b. The radius of the earth is 6400km. By what minimum angle must the asteroid be deflected to just miss the earth?

c. What is the actual angle of deflection if the rocket fires at full thrust for300s before running out of fuel?

A 4.4-cm-diameter, 24 g plastic ball is attached to a 1.2-m-long string and swung in a vertical circle. The ball’s speed is 6.1 m/s at the point where it is moving straight up. What is the magnitude of the net force on the ball? Air resistance is not negligible.

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