A 0.2-kg spool slides down along a smooth rod. If the rod has a constant angular rate of rotation \dot{ \theta } = 2 rad/s in the vertical plane, show that the equations of motion for the spool are \ddot{ r } - 4r - 9.81 \sin \theta = 0 and 0.8\dot{ r } + { N }_{ s } - 1.962 \cos \theta = 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 \theta are zero when t = 0, evaluate the constants { C }_{ 1 } and { C }_{ 2 } to determine r at the instant \theta = \pi/4 rad.
Question : A 0.2-kg spool slides down along a smooth rod. If the rod ha...
Kinematic: Here, \dot{ \theta }. = 2 rad/s and \dot{u} = 0. We have
a_r = \ddot{r} – r \dot{\theta}^2 = \ddot{r} – r(2^2) = \ddot{r} – 4r \\ a_\theta = r \ddot{\theta} + 2 \dot{r} \dot{\theta} = r(0) + 2\dot{r}(2) = 4\dot{r}Equation of Motion: We have
\begin{aligned} \Sigma F_r = ma_r; \quad\quad & 1.962 \sin \theta = 0.2(\ddot{r} – 4r) \\ & \ddot{r} – 4r – 9.81 \sin \theta = 0 \quad\quad\quad\quad\quad\quad (1) \end{aligned} \\ \begin{aligned} \Sigma F_\theta = ma_\theta; \quad\quad & 1.962 \cos \theta – N_s = 0.2(4\dot{r}) \\ & 0.8 \dot{r} + N_s – 1.962 \cos \theta = 0 \quad\quad\quad\quad (2) \end{aligned}Since \theta. = 2 rad/s, then \int_0^\theta \dot{\theta} = \int_0^1 2dt, \theta = 2t. The solution of the differential equation (Eq. (1)) is given by
r = C_1 e^{-2t} + C_2 e^{2t} – \frac { 9.81 } 8 \sin 2t \quad\quad\quad\quad (3)
Thus,
\dot{r} = -2 C_1 e^{ -2t } + 2C_2 e^{2t} – \frac { 9.81 } 4 \cos 2t \quad\quad\quad\quad (4)
At t = 0, r = 0. From Eq.(3) 0 = C_1 (1) + C_2 (1) – 0 \quad\quad\quad\quad (5)
At t = 0, \dot{r} = 0. From Eq.(4) 0 = -2 C_1 (1) + 2C_2 (1) – \frac { 9.81 } 4 \quad\quad\quad\quad (6)
Solving Eqs. (5) and (6) yields
C_1 = – \frac { 9.81 } {16} \quad\quad C_2 = \frac {9.81} {16}
Thus,
\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 \text{ h } 2t – \sin 2t) \end{aligned}At \theta = 2t = \frac \pi 4, \quad\quad r = \frac { 9.81 } 8 (\sin \text{ h}\frac \pi 4 – \sin \frac \pi 4 3) = 0.198 m
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