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A quantitative yo-yo is constructed with a disk fixed to an axle and it has string wrapped around it. The radius of the disk is 5 mm. When the disk is dropped it revolves about a perpendicular line through the center as axis. In this process string is unwrapped with each rotation. The length of string unwrapped wit one revolution is equal to the circumference of the disk.

Here, the circumference of the disk is equal to `2*(5/1000)*pi` m. As the disk falls down through a vertical distance 30 cm, the number of revolutions of the disk is equal to `(30/100)/(2*(5/1000)*pi)`

= `(30*1000)/(100*2*5*pi)`

`~~ 95.49`

The disk has revolved approximately 96 times in the vertical drop of 30 cm.

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The torque of a force about a point is by definition the vectorial product between the position vector of the point of force application and the value of the force itself. Therefore

T= rx F

where x represent the vectorial product.

We know that the vectorial product of tho vectors is represented by a determinant having on its first line the unit vectors (`hatx,haty,hatz` ), on its second line the first vector from the product (r here) and on the third line the second vector of the product (F here):

`T = r xx F = |[hatx, haty, hatz],[4,6,0],[3,2,0]| = 4*2*hatz -3*6*hatz =-10hatz (N*m)`

Answer: The absolute value (magnitude) of the torque is `T=10 N*m` and its direction is toward negative values of z axis.

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The rotational inertia of a system is a measure of the system's resistance to angular acceleration. It is the sum of mass x distance squared from the axis of rotation over unit areas of the object rotating about the system, ie

`I = sum_i m_i r_i^2 `   where `i ` indexes the constituent unit areas of the rotating body.

1) Assume this body consists of a solid cylinder (the trunk of the person + the platform on which they stand) with constant density, and 2 point masses which are the 2kg weights that the person is holding 5cm from the axis of rotation. Suppose that the person + stand is h = 2.1m in height (2m for the man + 10cm for the platform) and has girth radius r = 20cm and has mass m = 73.5kg (70kg for the man + 3.5kg for the platform say, so that the man and the platform have the same density).

The rotational inertia of the cylinder is given by the equation

`I_c = 1/12 m(3r^2 + h^2) `  which in this case = `73.5/12(3(0.2)^2 + 2.1^2) = (73.5/12) times 4.53 = 27.74625 ` kg m^2

The rotational inertia of the 2 point masses is given by the equation

`I_p = mu x^2 ` where `mu ` is the reduced mass of the two weights that are `x ` meters apart.

The reduced mass here`= m^2/(2m) = m/2 `  where `m =`2kg. Also, the weights are 10cm apart. Therefore here

`I_p = (2/2) times 0.1^2 = 0.01 ` kg m^2

The total rotational inertia of the person, the platform + weights is given by the sum of the separate rotational inertias, ie

`I = I_c + I_p = 27.8 `  kg m^2 to 3 sf.

2) The person is now holding the 2kg weights at arm's length. The point masses of the weights are now extended at arm's length, 70cm each way say. The arms can be assumed to form a rod of length L = 1.4m and mass m = 0.058 x 70kg (if the person is a man) = 4.06kg.

The rotational inertia of the extended arms as a rod is given by the formula

`I_r = (mL^2)/12 = (4.06 times 1.4^2)/12 = 0.663` kg m^2 to 3sf

The rotational inertia of the two point masses (the 2kg weights, extended at arm's length, 70cm) is now given by

`I_p = (m/2)x^2 = (2/2) 1.4^2 = 1.96 `  kg m^2

The rotational inertia of the solid cylinder (trunk of the person +platform) is still given by

`I_c = 27.74625 ` kg m^2

The total rotational inertia of the new body is the sum of these parts,

`I = I_c + I_p + I_r = 27.74625 + 1.96 + 0.663 = 30.4 `  kg m^2

3) The majority of the total rotational inertia is due to the person themselves (assumed to be a solid cylinder). The combined cylinder of the person plus platform `I_c = 27.74625 `  can be split into that for the person and that for the platform. Since the person makes up 95.2% of the cylinder, the rotational inertia due to the person is `I_(man) = 26.4 ` kg m^2 and that due to the platform `I_(plat) = 1.32` kg m^2.

It is then the person themselves who contributes the most to the total rotational inertia, then the weights they carry and then their arms and hands.

This might be surprising as the weights are held out at arm's length and we know that objects further out on the rotating mass contribute more to the inertia. However, the person is more like a cylinder than a stick and has a considerably larger mass than the 2kg weights that they carry. The weights may seem heavy to them, but their contribution to the total inertia when held out at arm's length is very small compared to that contributed by the person themselves.

Answer:  1) I = 27.8 kg m^2 approx, 2)  I = 30.4 kg m^2 approx, 3) the person themselves contributes most to the total rotational inertia. 

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