Angular Velocity Of A Cylinder

Terminal on our Physics questions for today:

Futurity infinite stations and spacecraft may create "artificial gravity" by rotating. Consider a cylindrical space station with a radius of 894 m that rotates virtually its cardinal axis. The within surface of the cylinder is the floor of the space station upon which the astronauts live and work. What period of rotation (in s) is necessary for a 70 kg astronaut to have the aforementioned weight as he has on Earth?

Now let'southward place what we have:

Radius = 894 m
Astronaut Mass = 70 kg
Gravitational Acceleration (what nosotros desire it to be anyhow) = nine.8 m/s^2

The important thing to note is that nosotros demand to find the Normal Force being exerted on the astronaut as a effect of the angular acceleration. That force should exist the equivalent of the Normal Forcefulness on World. That pretty much tells united states of america everything we demand to get started. Now, by this, we tin figure the force intended by simple F = ma which gives us 686 N of force. This is the Normal Force we want to attain.

Now, we need to detect out at what velocity the cylinder needs to be spinning in order to obtain 686 Due north of force. Nosotros can do this past using our angular velocity formula:

$v = \sqrt{\frac{rF_n}{m}} = \sqrt{\frac{(894 m)(686 N)}{70 kg}} = 93.6 m/s$

Easy as pie correct? At present that we take the velocity, we need to discover the total circumference, this will give united states the distance that the cylinder has to rotate through in order to complete 1 revolution.

$c = 2\pi(r) = 2\pi(894) = 56.17.2 m$

Now that we have the total distance and we have velocity, our handy-dandy formula for distance, velocity and time tin be used.

$d = vt \equiv t = \frac{d}{v} = \frac{5617.2 m}{93.6 m/s} = 60 s$

And in that location you lot have it, it would take the infinite station sixty seconds to make ane complete revolution in order to provide the same force on the astronaut as if on Earth.

Equally always, feel free to correct anything you come across wrong with this.

A question was posed in Physics class:

A toy "rocket-car" is attached to a metal rod that keeps the rocket moving in a horizontal circle on a frictionless tabular array. The radius of the circle is ane.33 m, the mass of the rocket is 0.87 kg, and the rocket engine provides a thrust of 5.seven N. If the rocket starts from residual, how many revolutions volition the rocket make before the tension in the rod equals 215 Northward?

Let'due south pick out the important information from the equation:

Attached to a rod moving in a circle.
Radius (of the circle) = i.33 m
Mass (of the rocket) = 0.87 kg
Thrust (of the engine) = 5.vii Northward
Final Tension Force = 215 North

Since it is going around in a circle on the table that is frictionless we can ignore gravity for this since information technology'southward only 1 plane of motion. The important thing to realize is that we demand to find the human relationship between tension on the arm and thrust of the engine which produces dispatch. We know that as the object accelerates the tension increases. We also know that if we can find the total altitude traveled and divide it by the distance in ane revolution, nosotros'll get the number of revolutions.

First, we volition demand to discover out exactly how much the object is accelerating. To get acceleration, nosotros can piece of work backwards since we have the force of thrust as 5.seven Due north.

$F = ma \equiv a = \frac{F}{m} \equiv a = \frac{5.7 N}{0.87 kg} = 6.55 m/s^2$

Now, we need to figure out what the velocity of the object is when the tension force is 215 N. To exercise this, we use the formula for athwart forcefulness which tin can exist changed to angular velocity:

$F = \frac{mv^2}{r} \equiv rF = mv^2 \equiv v = \sqrt{\frac{rF}{m}}$

This volition give us our velocity. We already know the variables needed, r is 1.33 1000, F is the final tension forcefulness, 215 Northward and our mass is 0.87 kg.

$v = \sqrt{\frac{(1.33 m)(215 N)}{0.87 kg}} = 18.13 m/s$

Now we have the velocity. Getting the circumference is the piece of cake role:

$c = 2\pi(r) \equiv 2\pi(1.33 m) = 8.36 m$

Great! Now we have two more steps. The showtime is to utilise our handy formula that deals specifically with velocity, dispatch and distance values and obtain the full distance:

$v_f^2 = v_i^2 + 2a\Delta(s) \equiv \Delta s = \frac{v_f^2}{v_i^2 + 2a} = \frac{(18.13 m/s^2)}{(0)^2 + 2(6.55 m/s^2)} = 25.09 m$

Nosotros're almost done. Now that we take the full change in distance we can do unproblematic division and we'll get our respond:

$revolutions = \frac{\Delta s}{c} = \frac{25.09 m}{8.36 m} = 3.00 revolutions$

And at that place you have it. So the total times the object goes around the runway before reaching 215 North of force on the arm is 3.

Equally e'er, please correct me if y'all encounter whatever bug. Thanks!

A question posed in Physics class (much help to Judy for the assistance on this one):

1 of the rides constitute at carnivals is the rotating cylinder. The riders footstep inside the vertical cylinder and stand up with their backs against the curved wall. The cylinder spins very rapidly, and at some athwart velocity, the floor is pulled abroad. The thrill-seekers now hang like flies on the wall. If the radius of the cylinder is two.1 m and the coefficient of static friction betwixt the people and the wall is thou _s = 0.390, what is the maximum menstruation of rotation (in south) of the cylinder for the floor to exist removed safely?

Now, what'due south important to note out of all this is that yous're looking to get the angular velocity that volition counter the forcefulness of gravity plenty that static friction is effective in holding the people against the wall.

We know:

Coefficient of static friction = 0.390
Radius = 2.1 meters
Gravitational Acceleration = 9.8 k/due south^2

We as well know that static friction must exist greater than or equal to the strength and that must be greater than or equal to 0 in order to remain in place. Annihilation less than 0 and the object moves. So, we need 2 equations for this, i for the strength related to gravity and one related to the normal force related to friction:

$F_s = \mu_s(n)$

$F_g = ma \equiv F_g = mg$

At present, the key to this is realizing that the normal force is related, or rather, equal to, the amount of force needed to keep the person in identify. The formula for that is:

$n = \frac{mv^2}{r}$

These can be combined equally follows:

$\mu_s\frac{mv^2}{r} = mg \equiv \mu_s\frac{v^2}{r} = g \equiv v = \sqrt{\frac{gr}{\mu}}$

Now you tin plug in numbers for the variables and get an answer for velocity.

$v = \sqrt{\frac{(9.8 m/s^2)(2.1 m)}{0.390}} = 7.26 m/s$

At present the second key to this is remembering that the distance around a circle is its circumference, that is the value that y'all'll need to put in to get your velocity value spitting out seconds around.

$\frac{2\pi(r)}{v} \equiv \frac{2\pi(2.1 m)}{7.26 m/s} = 1.8 seconds$

And there you have it. The minimum velocity that the cylinder needs to be spinning is 7.26 g/s which volition crusade it to rotate in one case every 1.8 seconds keeping the riders firmly in identify.

Equally always, delight correct me if y'all find any errors.

Angular Velocity Of A Cylinder,

Source: http://www.collegeastronomy.com/tag/angular-velocity/

Posted by: mossstrater.blogspot.com

Angular Velocity Of A Cylinder

Angular Velocity Of A Cylinder,

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