A helicopter blade spins at exactly 100 revolutions per minute. Its tip is5.00mfrom the centre of rotation.

(a) Calculate the average speed of the blade tip in the helicopter’s frame of reference.

(b) What is its average velocity over one revolution?

Short Answer

Expert verified
  1. The average speed is 52.38m/s.
  2. The average velocity is zero.

Step by step solution

01

Given Data

  • The number of revolutions in one minute = 100.
  • Distance of the tip from the center of rotation = 5.00 m.
02

Average speed and average velocity

Average speed is a scalar quantity. It is the ratio of the total distance covered to that of the total time taken to cover that distance. Hence the distance covered by the blade will be the circumference of the circle.

a) The distance can be calculated as:

D=2πRD=2×π×5D=31.42 m


Hence the total distance traveled is 31.4 meters in one rotation.

Such 100 revolutions are made by the blade.

So, the total distance can be calculated as:

=31.42×100=3142 m


Average Speed=Total DistanceTime


Substituting values in the above equation, we get,

Average speed=314260=52.38 m/s


Hence the average speed of the helicopter blade is52.38m/s.

Average velocity is the ratio of the displacement of the body to that of time when the displacement is the shortest distance from the initial and final position.


b.


Here in this, the body is in a circular motion. So the initial and final points will be the same. So the displacement of the body will be zero.


Average veloctiy=Total displacementTime=060=0 m/s



The average velocity is zero.

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

A swan on a lake gets airborne by flapping its wings and running on top of the water.

(a) If the swan must reach a velocity of6.00 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s2, how far will it travel before becoming airborne?

(b) How long does this take?

Find the following for path C in Figure 2.59:

(a) The distance travelled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish.

A student drove to the university from her home and noted that the odometer reading of her car increased by 12km . The trip took 18min.

(a) What was her average speed?

(b) If the straight-line distance from her home to the university is10.3km in a direction25.0º south of east, what was her average velocity?

(c) If she returned home by the same path7h30min after she left, what were her average speed and velocity for the entire trip?

Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Specifically, identify each quantity in your example

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/sand accelerates at the rate of 0.500for 7.00 s.

(a) What is his final velocity?

(b) The racer continues at this velocity to the finish line. If he was 300 mfrom the finish line when he started to accelerate, how much time did he save?

(c) One other racer was 5.00 mahead when the winner started to accelerate, but he was unable to accelerate, and travelled at 11.8 m/suntil the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

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