Question on Moment of Inertia/Rotational Inertia

  • #1
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Homework Statement


In the figure, a wheel of radius 0.42 m is mounted on a frictionless horizontal axle. A massless cord is wrapped around the wheel and attached to a 2.7 kg box that slides on a frictionless surface inclined at angle θ = 28 owith the horizontal. The box accelerates down the surface at 1.9 m/s2. What is the rotational inertia of the wheel about the axle?

Diagram attached

Homework Equations


Torque=Moment of Inertia*angular acceleration
F=ma
a=angular acceleration*radius

The Attempt at a Solution


what i did was combinr the equations and yielded the following
I=(F*r^2)/a=0.47628
I don't see where I did wrong but the solution does not match
 

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Answers and Replies

  • #2
haruspex
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I=(F*r^2)/a
You don't say how you calculate the tension. We cannot tell where you went wrong if you do not post your working.
 
  • #3
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I just realised what I calculated was Fa instead of tension. But how do i find the tension?? I dont know what to do with the angle ø=28°. pls help
You don't say how you calculate the tension. We cannot tell where you went wrong if you do not post your working.
 
  • #4
haruspex
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I just realised what I calculated was Fa instead of tension. But how do i find the tension?? I dont know what to do with the angle ø=28°. pls help
You should draw free body diagrams for both objects, the wheel and the block. Consider the forces on each, the accelerations of each and the relationships between them. The tension affects both.
 
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  • #5
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You should draw free body diagrams for both objects, the wheel and the block. Consider the forces on each, the accelerations of each and the relationships between them. The tension affects both.
thanks heaps, i see what went wrong now. Guess i shouldve drawn a bid and clear free body diagram at the first place!
 

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