A centrifuge accelerates uniformly from rest to 15,000 rpm in 240 s. Through how many revolutions did it turn in this time?

Short Answer

Expert verified

The number of revolutions made by the centrifuge is \(3 \times {10^4}\;{\rm{rev}}\).

Step by step solution

01

Given data

The revolutions per minute are\({\omega _2} = 15000\,{\rm{rpm}}\).

The time period is \(t = 240\;{\rm{s}}\).

02

Understanding angular displacement

In this problem, the angular displacement is determined by using the equation of kinematics for the rotational motion of the object. Also, the initial angular speed of the centrifuge is zero because it is at rest.

03

Determine the angular displacement

The relation to find the angular displacement is given by:

\(\theta = \frac{1}{2}\left( {{\omega _1} + {\omega _2}} \right)t\)

Here, \({\omega _1}\) is the initial angular speed, whose value is zero.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}\theta &= \frac{1}{2}\left( {0 + 15000\,{\rm{rpm}}} \right)\left( {240\;{\rm{s}} \times \frac{{1\,{\rm{min}}}}{{60\;{\rm{s}}}}} \right)\\\theta &= 3 \times {10^4}\;{\rm{rev}}\end{aligned}\)

Thus, \(\theta = 3 \times {10^4}\;{\rm{rev}}\) is the required angular displacement.

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

Figure 8–59 illustrates an \({{\bf{H}}_{\bf{2}}}{\bf{O}}\) molecule. The \({\bf{O - H}}\) bond length is 0.096 nm and the \({\bf{H - O - H}}\) bonds make an angle of 104°. Calculate the moment of inertia of the \({{\bf{H}}_{\bf{2}}}{\bf{O}}\)molecule (assume the atoms are points) about an axis passing through the center of the oxygen atom (a) perpendicular to the plane of the molecule, and (b) in the plane of the molecule, bisecting the \({\bf{H - O - H}}\) bonds.

FIGURE 8-59 Problem 82

Bonnie sits on the outer rim of a merry-go-round, and Jill sits midway between the centre and the rim. The merry-go-round makes one complete revolution every 2 seconds. Jill’s linear velocity is:

(a) the same as Bonnie’s.

(b) twice Bonnie’s.

(c) half of Bonnie’s.

(d) one-quarter of Bonnie’s.

(e) four times Bonnie’s.

Calculate the moment of inertia of an align of point objects, as shown in Fig. 8–47 about (a) the y axis and (b) the x-axis. Assume two masses, \(m = 2.2\;{\rm{kg}}\)and \(M = 3.4\;{\rm{kg}}\), and the objects are wired together by very light, rigid pieces of wire. The align is rectangular and split through the middle by the x-axis. (c) About which axis would it be harder to accelerate this align?

FIGURE 8-47

Problem 39

A hammer thrower accelerates the hammer\(\left( {{\bf{mass}} = {\bf{7}}{\bf{.30 kg}}} \right)\)from rest within four full turns (revolutions) and releases it at a speed of\({\bf{26}}{\bf{.5 m/s}}\).Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.20 m, calculate (a) the angular acceleration, (b) the (linear) tangential acceleration, (c) the centripetal acceleration just before release, (d) the net force being exerted on the hammer by the athlete just before release, and (e) the angle of this force with respect to the radius of the circular motion. Ignore gravity.

A grinding wheel is a uniform cylinder with a radius of 8.50 cm and a mass of 0.380 kg. Calculate (a) its moment of inertia about its center and (b) the applied torque needed to accelerate it from rest to 1750 rpm in 5.00 s. Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 55.0 s.

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