A 0.2-kg spool slides down along a smooth rod. If the rod has a constant angular rate of rotation θ = 2 rad/s in the vertical plane, show that the equations of motion for the spool are r – 4r – 9.81 sinθ = 0 and 0.8 r + Ns – 1.962 cosθ = 0, where Ns is the magnitude of the normal force of the rod

Question

A 0.2-kg spool slides down along a smooth rod. If the rod has a constant angular rate of rotation \dot{ heta } = 2 rad/s in the vertical plane, show that the equations of motion for the spool are \ddot{ r } - 4r - 9.81 \sin heta = 0 and 0.8\dot{ r } + { N }_{ s } - 1.962 \cos heta = 0, where { N }_{ s } is the magnitude of the normal force of the rod on the spool. Using the methods of differential equations, it can be shown that the solution of the first of these equations is r = { C }_{ 1 } { e }^{ -2t } + { C }_{ 2 } { e }^{ 2t } - (9.81/8) \sin 2t. If r, \dot{ r }, and heta are zero when t = 0, evaluate the constants { C }_{ 1 } and { C }_{ 2 } to determine r at the instant heta = \pi/4 rad.

Accepted Answer

Kinematic: Here, [latex]\dot{ heta }. = 2 rad/s[/latex] and [latex]\dot{u} = 0[/latex]. We have

[latex]a_r = \ddot{r} - r \dot{ heta}^2 = \ddot{r} - r(2^2) = \ddot{r} - 4r \ a_ heta = r \ddot{ heta} + 2 \dot{r} \dot{ heta} = r(0) + 2\dot{r}(2) = 4\dot{r}[/latex]

Equation of Motion: We have

[latex]\begin{aligned} \Sigma F_r = ma_r; \quad\quad & 1.962 \sin heta = 0.2(\ddot{r} - 4r) \ & \ddot{r} - 4r - 9.81 \sin heta = 0 \quad\quad\quad\quad\quad\quad (1) \end{aligned} \ \begin{aligned} \Sigma F_ heta = ma_ heta; \quad\quad & 1.962 \cos heta - N_s = 0.2(4\dot{r}) \ & 0.8 \dot{r} + N_s - 1.962 \cos heta = 0 \quad\quad\quad\quad (2) \end{aligned}[/latex]

Since [latex] heta. = 2 rad/s[/latex], then [latex]\int_0^ heta \dot{ heta} = \int_0^1 2dt, heta = 2t[/latex]. The solution of the differential equation (Eq. (1)) is given by

[latex]r = C_1 e^{-2t} + C_2 e^{2t} - \frac { 9.81 } 8 \sin 2t \quad\quad\quad\quad (3)[/latex]

Thus,

[latex]\dot{r} = -2 C_1 e^{ -2t } + 2C_2 e^{2t} - \frac { 9.81 } 4 \cos 2t \quad\quad\quad\quad (4)[/latex]

At [latex]t = 0, r = 0[/latex]. From Eq.(3) [latex]0 = C_1 (1) + C_2 (1) - 0 \quad\quad\quad\quad (5)[/latex]

At [latex]t = 0, \dot{r} = 0[/latex]. From Eq.(4) [latex]0 = -2 C_1 (1) + 2C_2 (1) - \frac { 9.81 } 4 \quad\quad\quad\quad (6)[/latex]

Solving Eqs. (5) and (6) yields

[latex]C_1 = - \frac { 9.81 } {16} \quad\quad C_2 = \frac {9.81} {16}[/latex]

Thus,

[latex]\begin{aligned} r &= - \frac { 9.81 } {16} e^{ -2t } + \frac {9.81} {16} e^{2t} - \frac {9.81} 8 \sin 2t \ &= \frac {9.81} 8 (\frac { -e^{-2t} + e^{2t} } 2 - \sin 2t) \ &= \frac {9.81} 8 (\sin ext{ h } 2t - \sin 2t) \end{aligned}[/latex]

At [latex] heta = 2t = \frac \pi 4, \quad\quad r = \frac { 9.81 } 8 (\sin ext{ h}\frac \pi 4 - \sin \frac \pi 4 3) = 0.198 m[/latex]

Question : A 0.2-kg spool slides down along a smooth rod. If the rod ha...

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