What is the angular speed of the rotating wheel?

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Homework Help Overview

The problem involves a mass attached to a cord wrapped around a wheel, focusing on the angular speed of the wheel after a certain time. The subject area includes dynamics and rotational motion, with specific attention to forces, torque, and moment of inertia.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of forces acting on the mass and the wheel, with some attempting to derive the force in the rope and the moment of inertia using different approaches.
  • Questions arise regarding the relationship between linear and angular quantities, particularly how to connect the linear acceleration of the mass to the angular acceleration of the wheel.
  • Some participants express uncertainty about the calculations and seek confirmation on their reasoning and results.

Discussion Status

The discussion is ongoing, with various attempts to clarify the relationships between the forces, torque, and angular quantities. Some participants have provided guidance on setting up equations, while others are exploring different interpretations of the problem. There is no explicit consensus yet, as participants continue to question and refine their approaches.

Contextual Notes

Participants note the importance of correctly identifying the mass of the wheel and the role of tension in the system. There is also mention of the need to relate linear and angular motion accurately, highlighting potential gaps in the information provided.

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Homework Statement


A 12 kg mass is attached to a cord that is wrapped around a wheel with a radius of 10.0 cm. The acceleration of the mass down the frictionless incline of 37 degrees is measured to be 2.0 m/s^2. Assuming the axle of the wheel to be frictionless, determine:
a). the force in the rope.
b). the moment of inertia of the wheel.
c). the angular speed of the wheel 2.0 s after it begins rotating, starting from rest.

Homework Equations


I=1/2mr^2



The Attempt at a Solution


For a). I got 14.44 N by using sin37*2*12
for b). I used I=1/2mr^2 and got .06
for c). I did not understand how to get this answer.
Please help. Thank You.
 
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I got 14.44 N by using sin37*2*12
Not quite right. Set up an equation with F=ma, and sum up the forces. Remember, the block is accelerating, it's not at rest.
I used I=1/2mr^2 and got .06
But you don't know the mass of the wheel, 12 kg is the mass of the block. The tension due to the block is causing a torque on the wheel. You must relate the two to find the moment of inertia.
I did not understand how to get this answer.
You were given the linear acceleration so you can get the angular acceleration. From that, using rotational kinematics, you can find the angular speed.
 
does anyone know about this problem?

for a) what are the external forces? I said F=ma = (12.0 kg)(cos 37)(2.0m/s) = 19.17 N

b) I said Torque = radius*force = (0.1 m)(12 kg*cos(37)*2.00m/s) = 1.92 N*m

thus solving for in [tex]\tau[/tex]=I[tex]\alpha[/tex] , I calculated 0.0958

c) I've calculated 14.1 m/s from using a=r[tex]\alpha[/tex]
then I used [tex]\alpha[/tex]=r[tex]\omega[/tex]2 and solved from there. does that sound correct?
 
sorry for part b I got I=0.0958
 
anyone to confirm?
 
Answer:

a)

F= mgsin(theta) - ma

b) Use T=rF and solve for T , then plug T into T=I*alpha and solve for I using a/r to find alpha.

c) alpha = r*angular vel^2 , solve for ang vel and your done
 
Hi nopistons93,

nopistons93 said:
Answer:

a)

F= mgsin(theta) - ma

b) Use T=rF and solve for T , then plug T into T=I*alpha and solve for I using a/r to find alpha.

c) alpha = r*angular vel^2 , solve for ang vel and your done

The formula highlighted in red is not correct.
 

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