Question

A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of point D is 23 ft/s2 downward. At this instant, compute the angular acceleration of the bar and the acceleration of point B.

Answer 1

a) the angular acceleration of the bar AB is
`\alpha_(AB) = - 29.31 \ k' \ rad/s^2`

b) the acceleration
`a_B` of point B =
`-218.6 \ i \ ft/s^2`

Step-by-step explanation:

The sketched diagram below shows an illustration of what the question comprises of:

Now, from the diagram ; we can deduce the following relations;

`v_(B/A) = L (sin \theta \ i' - cos \theta \ j')`

`v_(D/A) = (L)/(2) (sin \theta \ i' - cos \theta \ j')`

Taking point H as the instantaneous center of rotation of line HD. The distance between H and D is represented as:

`d_(HD) = (L)/(2)`

The angular velocity of the bar AB from the diagram can be determined by using the relation:

`\omega__(AB)} = (V_D)/(d_(HD))`

where:

`d_(HD) = (L)/(2)`

and

`V_D` = velocity of point D = 18.5 ft/s

L = length of the bar = 8 ft

Then;

`\omega__(AB)} = (18.5)/(4)`

`\omega__(AB)} = 4.625 \ rad/s`

Using vector approach to acceleration analysis;

Acceleration about point B can be expressed as;

`a_B = ( \alpha_(AB) Lcos \theta - \omega^2 _(AB)L sin \theta ) i + (a_A + \alpha_(AB) Lsin \theta + \omega ^2 _(AB) Lcos \theta )j` ---equation(1)

The y - component of
`a_B` ⇒
`a__(By)}` =
`a_A + \alpha_(AB) Lsin \theta + \omega ^2 _(AB) Lcos \theta`

where;
`a__(By)}` = 0

Then

0 =
`a_A + \alpha_(AB) Lsin \theta + \omega ^2 _(AB) Lcos \theta`

Making
`a_A` the subject of the formula; we have:

`a_A = - ( \alpha_(AB) Lsin \theta + \omega ^2 _(AB) Lcos \theta)` ------- equation (2)

Acceleration about point D is expressed as follows:

`a_D = ( \alpha_(AB) Lcos \theta - \omega^2 _(AB)L sin \theta ) i' + (a_A + \alpha_(AB) (L)/(2)sin \theta + \omega ^2 _(AB) (L)/(2)cos \theta )j'`

The y - component of
`a_D` ⇒
`a__(Dy)} = a_A + \alpha_(AB) (L)/(2)sin \theta + \omega ^2 _(AB) (L)/(2)cos \theta`

Replacing - 23 ft/s² for
`a_y`; we have:

`- 23 ft/s^2 \ = a_A + \alpha_(AB) (L)/(2)sin \theta + \omega ^2 _(AB) (L)/(2)cos \theta`

Also; replacing equation (2) for
`a_A` in the above expression; we have

`- 23 ft/s^2 \ = - ( \alpha_(AB) Lsin \theta + \omega ^2 _(AB) Lcos \theta) + \alpha_(AB) (L)/(2)sin \theta + \omega ^2 _(AB) (L)/(2)cos \theta`

`- 23 ft/s^2 \ = - \alpha_(AB) (L)/(2)sin \theta - \omega ^2 _(AB) (L)/(2)cos \theta`

Making
`\alpha _(AB)` the subject of the formula ; we have:

`\alpha_(AB) = (46 \ ft/s^2)/(Lsin\theta ) - \omega^2 _(AB) cot \theta`

Replacing 8 ft for L; 27° for θ; 4.625 rad/s for
`\omega __(AB)`

Then;

`\alpha_(AB) = (46 \ ft/s^2)/(8 sin27^0 ) - (4.625^2)( cot 27)`

`\alpha_(AB) = 12.67 - 41.98`

`\alpha_(AB) = - 29.31 rad/s^2`

`\alpha_(AB) = - 29.31 \ k' \ rad/s^2`

Thus, the angular acceleration of the bar AB is
`\alpha_(AB) = - 29.31 \ k' \ rad/s^2`

b)

To calculate the acceleration of point B using equation (1); we have:

`a_B = ( \alpha_(AB) Lcos \theta - \omega^2 _(AB)L sin \theta ) i + (a_A + \alpha_(AB) Lsin \theta + \omega ^2 _(AB) Lcos \theta )j`

Replacing

`\alpha_(AB) = - 29.31 rad/s^2`

L = 8 ft

θ = 27°

`\omega __(AB)` = 4.625 rad/s

`a_A + \alpha_(AB) Lsin \theta + \omega ^2 _(AB) Lcos \theta` = y = 0

Then;

`a_B = [-29.31(8)cos 27^0- (4.625^2)(8sin27^0)] i +0j`

`a_B = -218.6 \ i \ ft/s^2`

Thus, the acceleration
`a_B` of point B =
`-218.6 \ i \ ft/s^2`

Answer 2

alpha=53.56rad/s

a=5784rad/s^2

Step-by-step explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

`v=v_0+at\\\\t=(v)/(a)=((23(ft)/(s)))/(32.17(ft)/(s^2))=0.71s`

Now, we can calculate the angular acceleration (w0=0rad/s)

`\theta=\omega_0t +(1)/(2)\alpha t^2\\\alpha=(2\theta)/(t^2)`

`\alpha=(27)/((0.71s)^2)=53.56(rad)/(s^2)`

with this value we can compute the angular velocity

`\omega=\omega_0+\alpha t\\\omega = (53.56(rad)/(s^2))(0.71s)=38.02(rad)/(s)`

and the tangential velocity of point B, and then the acceleration of point B:

`v_t=\omega r=(38.02(rad)/(s))(4)=152.11(ft)/(s)\\a_t=(v_t^2)/(r)=((152.11(ft)/(s))^2)/(4ft)=5784(rad)/(s^2)`

hope this helps!!