Engineering, 22.04.2020 19:52 dawsoncollins4117

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.

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Answer from: nakarelinp0p303

Complete Question

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

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

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

Explanation:

From the question we are told that

Their radius of the drum is $r = 13 in = \frac{13}{12} ft = 1.083ft$ each

The weight is W = 210 lb

The mass is $M = \frac{210 lb}{32.2 ft /s^2} = 6.563\ lb s^2 ft^{-1}$

Their radius of gyration is $z=30 in= \frac{30 }{12} = 2.5 ft$

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

$m = \frac{30lb}{32.2 ft/s^2} = 0.9317 \ lb ft^{-1} s^2$

a is the acceleration of the hanging block

inputting values we have

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

$30 = 37.87\alpha + \alpha$

$\alpha = \frac{30}{38.87 }$

$\alpha = 0.792 rad/s^2$

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

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

$32.49=41.0\alpha$

$\alpha =\frac{41}{32.49}$

$\alpha =1.262$

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

Each of the two drums and connected hubs of 13-in. radius weighs 210 lb and has a radius of gyration

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