What is the torque of a cd after accelerating to 500 rev/min in 3.0 revolutions?

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Homework Help Overview

The problem involves a CD accelerating uniformly from rest to a speed of 500 revolutions per minute over a distance of 3.0 revolutions. The relevant parameters include the radius of the CD and its mass, with the goal of determining the torque applied during this acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between linear and rotational motion equations, questioning how to derive angular acceleration from the given revolutions and time. There is also a query about how torque relates to these equations.

Discussion Status

Some participants have provided analogies from linear to rotational motion and suggested equations that may be relevant. However, there is still uncertainty regarding the application of these concepts to find torque, and no consensus has been reached on the approach to take.

Contextual Notes

Participants note the need to convert the speed from revolutions per minute to radians per second and to consider the distance traveled in terms of angular displacement. There is an acknowledgment of the challenge in connecting these concepts to calculate torque.

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Homework Statement


When play is pressed, a cd accelerates uniformly from rest to 500 rev/min in 3.0 revolutions. If the Cd has a radius of 5.5 cm, and a mass of 20 g, what is the torque?


Homework Equations


not sure, this is my problem. i think its torque=angular acceleration*inertia. but i don't know how to turn "500 rpm in 3.0 revolutions" into angular acceleeration


The Attempt at a Solution


nothing, i don't how to make the equations work
 
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You have your analogy from linear kinematics equations to rotational motion equations.

a = delta v / delta t

So the same is true for rotational acceleration, and change in rotational velocity
 
but how does torque fit into this equation?
 
You can use: \omega^{2}=2\alpha s , where s=2\pi r * 3 (3 times the circumference)

and 500 rev/min=4770 rad/sec
 

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