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Umnica [9.8K]
3 years ago
10

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 p

oint D is 23 ft/s2 downward. At this instant, compute the angular acceleration of the bar and the acceleration of point B.

Physics
2 answers:
user100 [1]3 years ago
7 0

Answer:

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

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

Making \alpha _{AB} the subject of the formula ; we have:

\alpha_{AB} = \frac{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} = \frac{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

777dan777 [17]3 years ago
6 0

Answer:

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!!

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