Answer:
the acceleration of point A: [tex]a_A = 41 \ i - 1079 \ j \ ft/s^2[/tex]
the acceleration at point D [tex]a_D = -1030.268i +209.04 j)\ \ ft/s^2[/tex]
Explanation:
To determine the angular velocity ; we use the expression;
[tex]v_o = r \omega[/tex]
[tex]\omega = \frac{v_0}{r}[/tex]
where :
[tex]v_o[/tex] = velocity of the wheel
r = radius of the shaft
Replacing 6 ft/s with [tex]v_o[/tex] and 2 in for r; we have:
[tex]\omega = \frac{6 ft /s }{2 \ in * \frac{1 \ ft}{12 \ in}}[/tex]
[tex]\omega = 36 \ rad/s[/tex]
Calculating the angular acceleration by using the following expression.
[tex]a_o = r \alpha[/tex]
[tex]\alpha = \frac{a_o}{r}[/tex]
Replacing 7 ft/s² for [tex]a_o[/tex] & 2 in for r
[tex]\alpha[/tex] = [tex]\frac{7 \ ft/s^2}{2 \ in * \frac{1 \ ft}{12 \ in}}[/tex]
[tex]\alpha[/tex] = 42 rad/s
The acceleration at point A can be calculated by using derived equation:
[tex]a_A = a_o+ \alpha r_{A/O} + \omega (\omega *r_{A/O} )[/tex]
where;
[tex]r_{A/O[/tex] = distance of point A with respect to O
Substituting 7 ft/s² for [tex]a_o[/tex]; 42 rad/s² for [tex]\alpha[/tex]; 36 rad/s for [tex]\omega[/tex] and 10 for [tex]r_{A/O[/tex] ; we have:
[tex]a_A = (-7 \ i )+(42 k *(10 in* \frac{1 \ ft}{12 \ in})i ) + (36 k *36 K * (10 in* \frac{1 \ ft}{12 \ in})j[/tex]
[tex]a_A = -7 \ i - 34\ i - 1079 \ j[/tex]
[tex]a_A = 41 \ i - 1079 \ j \ ft/s^2[/tex]
Hence, the acceleration of point A: [tex]a_A = 41 \ i - 1079 \ j \ ft/s^2[/tex]
To calculate position of D about O ;we have
[tex]r _{D/O}[/tex] = [tex]\frac{10}{12}cos (sin ^{-1}\frac{2}{10})i - \frac{2}{12}j[/tex]
[tex]r _{D/O}[/tex] = 0.816 i - 0.1667 j
To determine acceleration of D ; we use the following equation
[tex]a_D = a_o+ \alpha r_{D/O} + \omega (\omega *r_{D/O} )[/tex]
replacing 7 ft/s² for [tex]a_o[/tex]; 42 rad/s² for [tex]\alpha[/tex]; 36 rad/s for [tex]\omega[/tex] and 0.816 i - 0.1667 j for [tex]r_{D/O[/tex] ; we have:
[tex]a_D = -7i +(42*(0.816i-0.1667j))+(36 * 36*(0.816-0.1667j)[/tex]
[tex]a_D = -7i +(34.272i-7.0014j))-(1296*(0.816-0.1667j)[/tex]
[tex]a_D = -7i +(34.272i-7.0014j))-(1057.54i+216.04j)[/tex]
[tex]a_D = 27.272i-7.0014j-1057.54i+216.04j[/tex]
[tex]a_D = -1030.268i +209.04 j)\ \ ft/s^2[/tex]
Therefore, the acceleration at point D [tex]a_D = -1030.268i +209.04 j)\ \ ft/s^2[/tex]
The acceleration of points A and D are equal to [tex](-42i-1080j)\;ft/s^2[/tex] and [tex](1084.812i -223.0414j)\;ft/s^2[/tex] respectively.
Given the following data:
First of all, we would calculate the angular velocity and acceleration of the shaft of the wheel:
[tex]V=r\omega\\\\\omega=\frac{V}{r} \\\\\omega=\frac{6}{\frac{2}{12} }\\\\\omega=\frac{6\times 12}{2}\\\\\omega=36\;rad/s[/tex]
For the angular acceleration:
[tex]a=r\alpha \\\\\alpha=\frac{a}{r} \\\\\alpha=\frac{7}{\frac{2}{12}}\\\\\alpha=\frac{7\times 12}{2}\\\\\alpha=42 \;rad/s^2[/tex]
In order to calculate the acceleration at point A, we would use this equation:
[tex]a_A=a+\alpha r_{AO}+\omega(\omega r_{AO})[/tex]
Note: [tex]r_{AO}[/tex] is the outer radius i.e the distance from point A to point O.
[tex]a_A=-7i+42(\frac{10}{12}i) +36 \times (36(\frac{10}{12}j))\\\\a_A=-7i-35i-1080j\\\\a_A=-42i-1080j\;ft/s^2[/tex]
Next, we would calculate the position of D wrt O:
[tex]r_{DO}=\frac{10}{12} cos(sin^{-1}\frac{2}{10} )i-\frac{2}{12} j\\\\r_{DO}=0.816 i - 0.1667 j[/tex]
Next, we would determine the acceleration at point D:
[tex]a_D=a+\alpha r_{DO}+\omega(\omega r_{DO})\\\\a_D=-7i+42(0.816 i - 0.1667 j) +36 \times (36(0.816 i - 0.1667 j))\\\\a_D=-7i+(34.272i-7.0014j)+(1057.54i-216.04j)\\\\a_D=27.272i-7.0014j+1057.54i-216.04j\\\\a_D=1084.812i -223.0414j\;ft/s^2\\\\[/tex]
Read more on acceleration here: brainly.com/question/24728358
