A hanger of mass m is attached to a light string that is wrapped around the rim

Question

A hanger of mass m is attached to a light string that is wrapped around the rim of a wheel of radius r. The wheel is released, allowing the weight to fall and cause the wheel to accelerate rotationally. In the experiment, the rotational acceleration %u03B1 is measured. The acceleration of gravity is g (defined as a positive quantity), and the moment of inertia is I. Work out a formula for I by following the steps given below.

All answers are formulas, and should contain only symbols from the following set: hanger mass m (type "m"), acceleration of gravity g (type "g"), moment of inertia I (type "I"), radius r (type "r"), linear acceleration of the falling weight a (type "a"), string tension T (type "T"), and rotational acceleration %u03B1 (type "alpha").

Important: Regard m, g, T, r and I as positive quantities. The accelerations a and %u03B1 may be positive or negative, depending on direction. In the coordinate system assumed "positive" means up for linear quantities and counterclockwise for angular quantities.

Write both sides of Newton's 2nd law for the falling weight: the net force, given in terms of the tension T and the gravity force mg should equal the weight's mass times it acceleration.

Sum of Forces Mass x Acceleration

? = ?

Write both sides of Newton's 2nd law in rotational form for the wheel. Express the torque in terms of the tension T and wheel radius r. Express the angular acceleration %u03B1 in terms of %u2013a/r, since a counterclockwise angular acceleration is positive, but this corresponds to a negative (downward) linear acceleration of the hanger. A counterclockwise torque is positive.

Sum of torques Moment of Inertia x Angular Acceleration

? = ?

Combine the two previous equations in order to eliminate the tension T and come up with a formula for the acceleration of the falling weight a in terms of the remaining variables. To make sure you have the correct equation check that it acts the way you would expect it to if I were to be infinitely large or zero. For example, if I = 0 the weight would drop with the acceleration of gravity, or a = %u2013g, while if I=(infinity), then the acceleration would go to 0.

a=?

Finally, substitute a = %u2013r%u03B1 into your formula and solve it for I to obtain the required formula in terms of the variables m, g, r and %u03B1.

I = ?

Please show all work necessary to complete the problem most importantly the last equation. Thanks in advance!!!

Explanation / Answer

mg-T = ma

torque = I*alpha

T*r = I*alpha

substituting T

m(g-a)*r = I*alpha

or, g-a = I*alpha/m*r

or, a = g - (I*alpha)/(m*r)

a = r*alpha

alpha(r + I/m*r) = g

or, I = ((g/alpha) -r)m*r

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