Physics, 08.04.2020 01:07 ethanyayger

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.

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Answer from: jadaroyval

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

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

$v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s$

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

$\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}$

$\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}$

with this value we can compute the angular velocity

$\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}$

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

$v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}$

hope this helps!!

Answer from: uuuuyyu

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 = $-218.6 \ i \ ft/s^2$

Explanation:

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

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

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

$v_{D/A} = \frac{L}{2} (sin \theta \ i' - cos \theta \ j')$

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

$d_{HD} = \frac{L}{2}$

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

$\omega__{AB}} = \frac{V_D}{d_{HD}}$

where:

$d_{HD} = \frac{L}{2}$

and

V_D = velocity of point D = 18.5 ft/s

L = length of the bar = 8 ft

Then;

$\omega__{AB}} = \frac{18.5}{4}$

$\omega__{AB}} = 4.625 \ rad/s$

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:

$a_D = ( \alpha_{AB} Lcos \theta - \omega^2 _{AB}L sin \theta ) i' + (a_A + \alpha_{AB} \frac{L}{2}sin \theta + \omega ^2 _{AB} \frac{L}{2}cos \theta )j'$

The y - component of a_D ⇒ $a__{Dy}} = a_A + \alpha_{AB} \frac{L}{2}sin \theta + \omega ^2 _{AB} \frac{L}{2}cos \theta$

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

$- 23 ft/s^2 \ = a_A + \alpha_{AB} \frac{L}{2}sin \theta + \omega ^2 _{AB} \frac{L}{2}cos \theta$

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

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