Knowing that inner gear A is stationary and outer gear C starts from rest and has a constant angular acceleration of 4 rad/s² clockwise, determine at t = 5 s (a) the angular velocity of arm AB, (b) the angular velocity of gear B, (c) the acceleration of the point on gear B that is in contact with

Question

Knowing that inner gear A is stationary and outer gear C starts from rest and has a constant angular acceleration of 4 rad/s² clockwise, determine at t = 5 s (a) the angular velocity of arm AB, (b) the angular velocity of gear B, (c) the acceleration of the point on gear B that is in contact with gear A.

Accepted Answer

Angular velocity of gear C at t = 5 s.

[latex]\omega_{c}=\alpha_{c}t=(4\;\mathrm{rad/s}^{2})(5\;\mathrm{s})=20\;\mathrm{rad/s}\quad \boldsymbol{\omega}_{c}=20\;\mathrm{rad/s}\circlearrowright [/latex]

Let Point 1 be the contact point between gears A and B. Let Point 2 be the contact point between gears B and C. Points A, B, and C are the centers, respectively, of gears A, B, and C.

Positions: Take x axis along the straight line A1 B2.

[latex]\begin{aligned}& x_{A}=x_{C}=0\& x_{1}=r_{A}=80\,\mathrm{mm}\& x_{B}=(80\,\mathrm{mm}+40\,\mathrm{mm})=120\,\mathrm{mm}\& x_{2}=120\,\mathrm{mm}+80\,\mathrm{mm}=200\,\mathrm{mm}\end{aligned}[/latex]

Velocity at 1. Since gear A is stationary, [latex] u_{1}=0[/latex] .

Velocity at 2. [latex]\qquad u_{2}=x_{2}\omega_{C}=(200)(20)[/latex]

[latex]\displaystyle\mathbf{v}_{2}=4000\,\mathrm{mm}/\mathrm{{s}}\downarrow[/latex]

Point 1 is the instantaneous center of gear B.

[latex]\begin{aligned}& u_{2}=(x_{2}-x_{1})\omega_{B}=(120\,\mathrm{mm})\omega_{B}\& \omega_{B}={\frac{ u_{2}}{120}}={\frac{4000}{120}}=33.333\,\,\mathrm{rad/s}\& u_{B}=(x_{B}-x_{1})\omega_{B}=(40\;\mathrm{mm})(33.333\;\mathrm{rad/s})=1333.33\;\mathrm{mm}/\mathrm{s}\& \omega_{A B}=\frac{ u_{B}}{x_{B}}=\frac{1333.33}{120}=11.111\,\mathrm{rad/s}\end{aligned}[/latex]

(a) Angular velocity of arm AB: [latex]\qquad\qquad \boldsymbol{\omega}_{A B}=11.11\,\mathrm{rad/s}\circlearrowright \blacktriangleleft[/latex]

(b) Angular velocity of gear B:[latex]\qquad\qquad \boldsymbol{\omega}_{ B}=33.3\,\mathrm{rad/s}\circlearrowright \blacktriangleleft[/latex]

Calculate tangential accelerations:

[latex]\begin{aligned}& (a_{1})_{t}=0\& {(a_{2})_{t}}=(a_{C})_{t}=r_{C}\alpha_{2}=(200\;\mathrm{mm})(4\;\mathrm{rad/s}^{2})\& \qquad\qquad (a_{2})_{t}=800\,\mathrm{mm/s^{2}}\downarrow\& (a_{2})_{t}=(x_{2}-x_{1})\alpha_{B}=(120\operatorname{mm})\alpha_{B}\& \quad \alpha_{B}={\frac{(a_{2})_{t}}{120}}={\frac{800}{120}}=6.667\,\,\mathrm{rad/s}^{2}\& (a_{B})_{t}=(x_{B}-x_{1})\alpha_{B}=(40\,\mathrm{mm})(6.667\,\,\mathrm{rad/s^{2}})\& \qquad\qquad\qquad =266.67{\mathrm{~mm/s}}^{2}\& \alpha_{A B}=\frac{(a_{B})_{t}}{r_{B}}=\frac{2666.67}{120}=2.222\,\mathrm{rad/s}^{2}\end{aligned}[/latex]

Angular accelerations:

[latex]\boldsymbol{\alpha}_{A B}=2.222\,\mathrm{rad/s^{2}}\circlearrowright \qquad \boldsymbol{\alpha}_{B}=6.667\,\mathrm{rad/s}\circlearrowright[/latex]

Acceleration of Point B.

[latex]\begin{aligned}& {\bf a}_{B}=[\alpha_{A B}\ x_{B}\downarrow]+[\omega_{A B}^{2}\ x_{B} \longleftarrow]\& \quad\quad =[(2.222)(120)\downarrow]+[(11.11)^{2}(120)\longleftarrow]\& =(266.67\mathrm{\ mm}/\mathrm{s}^{2})\downarrow_{+}(1481{5\mathrm{\ mm}}/\mathrm{s}^{2})\longleftarrow \end{aligned}[/latex]

(c) Acceleration of Point 1 on gear B.

[latex]\begin{aligned}& {\bf a}_{1}={\bf a}_{B}+[\alpha_{B}(x_{B}-x_{1})\uparrow]+[\omega_{B}^{2}(x_{B}-x_{1})\longrightarrow]\& \quad\quad =[266.67\downarrow]+[14815\longleftarrow]+[(6.667)(40)\uparrow]+[(33.333)^{2}(40)\longrightarrow]\& \quad\quad =[266.67\downarrow]+[14815\longleftarrow]+[266.67\uparrow]+[444444\longrightarrow]\& \quad\quad =29629\ \mathrm{mm/s}\longrightarrow & \mathbf{a}_{1}=29.6\operatorname{m}\!/\mathrm{s}^{2}\longrightarrow \blacktriangleleft\end{aligned}[/latex]

Note that the tangential component of acceleration is zero as expected.

Question 15.251: Knowing that inner gear A is stationary and outer gear C sta...

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