Answer :
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)
[tex]v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s[/tex]
Now, we can calculate the angular acceleration (w0=0rad/s)
[tex]\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}[/tex]
[tex]\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}[/tex]
with this value we can compute the angular velocity
[tex]\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}[/tex]
and the tangential velocity of point B, and then the acceleration of point B:
[tex]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}[/tex]
hope this helps!!
Answer:
a) the angular acceleration of the bar AB is [tex]\alpha_{AB} = - 29.31 \ k' \ rad/s^2[/tex]
b) the acceleration [tex]a_B[/tex] of point B = [tex]-218.6 \ i \ ft/s^2[/tex]
Explanation:
The sketched diagram below shows an illustration of what the question comprises of:
Now, from the diagram ; we can deduce the following relations;
[tex]v_{B/A} = L (sin \theta \ i' - cos \theta \ j')[/tex]
[tex]v_{D/A} = \frac{L}{2} (sin \theta \ i' - cos \theta \ j')[/tex]
Taking point H as the instantaneous center of rotation of line HD. The distance between H and D is represented as:
[tex]d_{HD} = \frac{L}{2}[/tex]
The angular velocity of the bar AB from the diagram can be determined by using the relation:
[tex]\omega__{AB}} = \frac{V_D}{d_{HD}}[/tex]
where:
[tex]d_{HD} = \frac{L}{2}[/tex]
and
[tex]V_D[/tex] = velocity of point D = 18.5 ft/s
L = length of the bar = 8 ft
Then;
[tex]\omega__{AB}} = \frac{18.5}{4}[/tex]
[tex]\omega__{AB}} = 4.625 \ rad/s[/tex]
Using vector approach to acceleration analysis;
Acceleration about point B can be expressed as;
[tex]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[/tex] ---equation(1)
The y - component of [tex]a_B[/tex] ⇒ [tex]a__{By}}[/tex] = [tex]a_A + \alpha_{AB} Lsin \theta + \omega ^2 _{AB} Lcos \theta[/tex]
where; [tex]a__{By}}[/tex] = 0
Then
0 = [tex]a_A + \alpha_{AB} Lsin \theta + \omega ^2 _{AB} Lcos \theta[/tex]
Making [tex]a_A[/tex] the subject of the formula; we have:
[tex]a_A = - ( \alpha_{AB} Lsin \theta + \omega ^2 _{AB} Lcos \theta)[/tex] ------- equation (2)
Acceleration about point D is expressed as follows:
[tex]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'[/tex]
The y - component of [tex]a_D[/tex] ⇒ [tex]a__{Dy}} = a_A + \alpha_{AB} \frac{L}{2}sin \theta + \omega ^2 _{AB} \frac{L}{2}cos \theta[/tex]
Replacing - 23 ft/s² for [tex]a_y[/tex]; we have:
[tex]- 23 ft/s^2 \ = a_A + \alpha_{AB} \frac{L}{2}sin \theta + \omega ^2 _{AB} \frac{L}{2}cos \theta[/tex]
Also; replacing equation (2) for [tex]a_A[/tex] in the above expression; we have
[tex]- 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[/tex]
[tex]- 23 ft/s^2 \ = - \alpha_{AB} \frac{L}{2}sin \theta - \omega ^2 _{AB} \frac{L}{2}cos \theta[/tex]
Making [tex]\alpha _{AB}[/tex] the subject of the formula ; we have:
[tex]\alpha_{AB} = \frac{46 \ ft/s^2}{Lsin\theta } - \omega^2 _{AB} cot \theta[/tex]
Replacing 8 ft for L; 27° for θ; 4.625 rad/s for [tex]\omega __{AB}[/tex]
Then;
[tex]\alpha_{AB} = \frac{46 \ ft/s^2}{8 sin27^0 } - (4.625^2)( cot 27)[/tex]
[tex]\alpha_{AB} = 12.67 - 41.98[/tex]
[tex]\alpha_{AB} = - 29.31 rad/s^2[/tex]
[tex]\alpha_{AB} = - 29.31 \ k' \ rad/s^2[/tex]
Thus, the angular acceleration of the bar AB is [tex]\alpha_{AB} = - 29.31 \ k' \ rad/s^2[/tex]
b)
To calculate the acceleration of point B using equation (1); we have:
[tex]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[/tex]
Replacing
[tex]\alpha_{AB} = - 29.31 rad/s^2[/tex]
L = 8 ft
θ = 27°
[tex]\omega __{AB}[/tex] = 4.625 rad/s
[tex]a_A + \alpha_{AB} Lsin \theta + \omega ^2 _{AB} Lcos \theta[/tex] = y = 0
Then;
[tex]a_B = [-29.31(8)cos 27^0- (4.625^2)(8sin27^0)] i +0j[/tex]
[tex]a_B = -218.6 \ i \ ft/s^2[/tex]
Thus, the acceleration [tex]a_B[/tex] of point B = [tex]-218.6 \ i \ ft/s^2[/tex]
