Question

Each of the two drums and connected hubs of 13-in. radius weighs 210 lb and has a radius of gyration about its center of 30 in. Calculate the magnitude of the angular acceleration of each drum. Friction in each bearing is negligible.

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The angular acceleration for first drum
`\alpha = 0.792 rad/s^2`

The angular acceleration for the second drum is
`\alpha =1.262`

Step-by-step explanation:

From the question we are told that

Their radius of the drum is
`r = 13 in = (13)/(12) ft = 1.083ft` each

The weight is
`W = 210 lb`

The mass is
`M = (210 lb)/(32.2 ft /s^2) = 6.563\ lb s^2 ft^(-1)`

Their radius of gyration is
`z=30 in= (30 )/(12) = 2.5 ft`

The free body diagram of a drum and its hub and 30lb and in the case the weight is connect to the hub separately is shown on the second uploaded image

The T in the diagram is the tension of the string

Now taking moment about the center of the the drum P we have

`\sum M_p = I_p \alpha`

=>
`T * r = Mz^2 * \alpha`

Where r is the radius ,z is the radius of gyration about the center O , M is the mass of the drum including the hub, and
`\alpha` is the angular acceleration

Inputting

`T * 1.083 = 6.563 * 2.5^2 \alpha`

=>
`T = 37.87\alpha`

Considering the force equilibrium in the vertical direction (Looking at the second free body diagram now )

The first on is

`\sum F_y = ma`

=>
`30lb - T = m(r \alpha )`

Where m is the mass of the hanging block which has a value of

`m = (30lb)/(32.2 ft/s^2) = 0.9317 \ lb ft^(-1) s^2`

a is the acceleration of the hanging block

inputting values we have

`30- 37.87 \alpha = 0.9317* 1.083 \alpha`

`30 = 37.87\alpha + \alpha`

`\alpha = (30)/(38.87 )`

`\alpha = 0.792 rad/s^2`

So the angular acceleration for first drum
`\alpha = 0.792 rad/s^2`

The free body diagram of a drum and its hub when the only on the string is 30lb is shown on the third uploaded image

So here we would take the moment about O

`\sum M_o = I_O \alpha`

So
`\sum M_o = 30* 1.083`

and
`I = M z^2`

Therefore we will have

`30 * 1.083 = (Mz^2 )\alpha`

inputting values

`30 * 1.083 = 6.563 * 2.5^2 \alpha`

`32.49=41.0\alpha`

`\alpha =(41)/(32.49)`

`\alpha =1.262`

So the angular acceleration for the second drum is
`\alpha =1.262`