Calculating Angular and Linear Speeds of a Rotating Disk

[whoopie88]

Homework Statement

A 3.5 inch floppy disk in a computer rotates with a period of 1.45 multiplied by 10-1 s. (Note: A 3.5 inch floppy disk is 3.5 inches in diameter.)
(a) What is the angular speed of the disk?

(b) What is the linear speed of a point on the rim of the disk?

(c) Does a point near the center of the disk have an angular speed that is greater than, less than, or the same as the angular speed found in part (a)?

Homework Equations

All Rotational Kinematics Formulas

The Attempt at a Solution

I don't know how to find the angular speed with the givens provided...can somebody help me get started with this problem??

Thanks in advance for any help.

 

[SEngstrom]

It may be that the definition of the rotation is unclear? From what you say I interpret the period (T, the time for the disk to rotate 360 degrees) is 1.45*10-1s = 0.145 s. Angular speed is angle/unit time. How many degrees does the disk rotate in a second? Consider what units you are expected to respond in degrees or radians.

 

[whoopie88]

Actually, I forgot about one of the formulas and what the Period of a circle represents. I found the right answer. Thanks for your help!

But I have another question related to Circular Motion:

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a car can safely travel if the radius of the track is 79.0 m and the coefficient of friction is 0.38?

I feel like I should use the formulas associated with Centripetal Force, but I'm not sure which force represents Centripetal Force...so I'm not sure where to start. Help?

 

[SEngstrom]

The centrifugal force is the outward inertial force because of the rotation. The centripetal force is the balancing force that prevents the car in this case to fly off the track. There is a maximal centripetal force the tires can provide (friction coeff)*(normal force to the surface). To find the balancing point you need to find what the maximal centrifugal force the setup can handle is.

 

[whoopie88]

How do I go about finding this centrifugal force? None of my formulas even make mention of the centrifugal force..

 

[SEngstrom]

Take a look at http://en.wikipedia.org/wiki/Centripetal_force
The centripetal force required for motion along a path with curvature radius r is mv^2/r, or (equivalently) mw^2r, where w is the angular velocity. This is the friction force the tires need to supply to keep the car on a circular path.

 

[whoopie88]

I plugged the given values into the formula:

mv^2/(79) = 0.38(Fn)

But there are still 3 unknowns. This is why I'm confused.

 

[SEngstrom]

Fn=f*m*g, where f is the friction coefficient. The mass of the car does matter.

 

[whoopie88]

If the mass of the car does matter and it's not given, how do I solve?

 

[SEngstrom]

If the mass of the car does matter and it's not given, how do I solve?

write the equation you need - the mass of the car is a factor on either side of it.

 

[whoopie88]

Ah, I see. I found the correct answer. Thank you. I just have one question about the process I went through. Why does Fn = f*m*g where f is the coefficient of Friction? I thought Fg was the only force that was equal to mg. Am I missing something conceptually?

 

[SEngstrom]

My mistake - the normal force (Fn) at rest (in the z-direction) is the force the road exerts on the tires to balance the gravitational pull (mg) on the car. The maximal lateral friction force is f*Fn. Sorry for the confusion - seems you have the concepts clear :-)

 

[whoopie88]

Oh, I understand now. Thanks for your help!