Calculate Rotational Inertia of Wheel | 15kg Object on Frictionless Incline

AI Thread Summary
A 15.0-kg object on a frictionless incline is connected to a wheel with a radius of 9.0 cm, and its acceleration is measured at 2.00 m/s². The tension in the rope is calculated to be approximately 58.47 N. The discussion revolves around determining the moment of inertia of the wheel, with participants struggling to apply the correct formulas. The correct formula for calculating the moment of inertia is confirmed as I = Fr²/a. This exchange highlights the importance of using the right equations in rotational dynamics.
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A 15.0-kg object is attached to a cord that is wrapped around a wheel of radius r = 9.0 cm (Fig. P8.60). The acceleration of the object down the frictionless incline is measured to be 2.00 m/s2. Assume the axle of the wheel to be frictionless.

p8-60alt.gif


(a) Determine the tension in the rope.
58.46681 N
(b) Determine the moment of inertia of the wheel.
kg·m2
(c) Determine the angular speed of the wheel 2.00 s after it begins rotating, starting from rest.
rad/s
===================================
The tension is correct. I'm stuck on the inertia. I'll use 't' for torque and 'T' for the tension.

t=Iar => I = t/ar = Fr/ar = T/a but this doesn't work. I also tried using the force of gravity and the sigma force in the above equation and none of them are correct.Due tomorrow morning at 8:30 AM EST... :-(
 
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I guess, the formula is t = Ia/r ... ain't it?
 
saket said:
I guess, the formula is t = Ia/r ... ain't it?
... Crap

Yeah that did it. Thanks.

So for anyone searching the formula becomes
I = Fr^2/a
 
Last edited:

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