Simple concept help: Pulley and Net Torque

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SUMMARY

The discussion centers on calculating the angular acceleration, linear acceleration, and tension in a cord wrapped around a hollow sphere. The problem involves a hollow sphere with a radius of 0.11m and mass of 0.444kg, supporting a 0.02kg object. The correct approach requires careful attention to the direction of forces and proper sign conventions. The final linear acceleration calculated is 0.72 m/s², indicating the downward acceleration of the object.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with torque and rotational dynamics
  • Knowledge of moment of inertia for a hollow sphere
  • Ability to apply free body diagrams for analyzing forces
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  • Review the concept of moment of inertia for different shapes, particularly hollow spheres
  • Learn about the relationship between linear and angular acceleration
  • Study the application of free body diagrams in rotational motion problems
  • Explore the effects of frictionless systems on torque and acceleration calculations
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john562
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Homework Statement


A ball (hollow sphere) of radius .11m and mass .444kg is mounted on a frictionless axle. A massless cord is wrapped around the ball supporting an object of mass .02kg. Find the angular acceleration of the wheel, linear acceleration of the object and tension in the cord.


Homework Equations





The Attempt at a Solution


\SigmaF=Fg+T=m*a
-.2+T=.02a

\Sigma\tau=F*r=I*\alpha
-T*.11=(.0036*a)/.11
T = -.2975*a

-.2 + .2975*a = .02*a
-.2 = -.2775a
.72 = a

That's what I tried, but it's wrong.
Can anyone help?
 

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john562 said:

Homework Statement


A ball (hollow sphere) of radius .11m and mass .444kg is mounted on a frictionless axle. A massless cord is wrapped around the ball supporting an object of mass .02kg. Find the angular acceleration of the wheel, linear acceleration of the object and tension in the cord.


Homework Equations





The Attempt at a Solution


\SigmaF=Fg+T=m*a
-.2+T=.02a
0.2 -T = .02a
\Sigma\tau=F*r=I*\alpha
-T*.11=(.0036*a)/.11
T*.11 = .0036a/.11
T = -.2975*a
T = .2975a
-.2 + .2975*a = .02*a
-.2 = -.2775a
.72 = a

That's what I tried, but it's wrong.
Can anyone help?
You must watch your signage. The acceleration is downwards. Choose the downward and clockwise directions as positive, for compatability and simplicity.
 

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