1.20. Use the fact that the acceleration of \(P\) always points toward \(O\): in polar coordinates, this implies that \(a_\theta=0 = 2\dot{r}\dot{\theta}+ r \ddot{\theta}= \frac{1}{r} \frac{d}{dt}(r^2 \dot{\theta})\). This leads to a quantity which remains constant along the motion. To find this constant, use the initial conditions. This will give you \(\dot{\theta}\) vs \(r\). To find \(\dot{r}\), use the equation of the trajectory \(r = r_0 (1+ \cos\theta)\). To find the time \(T\) taken by \(P\) to reach \(O\), use \(\dot{\theta}\) and \(r\) vs \(\theta\). You can then integrate by separation of the variables: \[ f(\theta) d\theta= C dt \] Integrate from \(A\) to \(O\) and you will find \(T\).
Solution:
- a. We must use the fact that the acceleration of \(P\) always points toward \(O\): in polar coordinates \((r,\theta)\), this implies that \[ \boldsymbol{a}_P = a_r \boldsymbol{\hat{e}}_r \longrightarrow a_\theta=0 = 2\dot{r}\dot{\theta}+ r \ddot{\theta}= \frac{1}{r} \frac{d}{dt}(r^2 \dot{\theta}) \] This shows that \[ r^2 \dot{\theta}= C = \text{constant} \qquad {[1]} \] To find constant \(C\), we must use the initial conditions: \(\boldsymbol{v}_P = v_0 \boldsymbol{\hat{e}}_y\) at the position \((r, \theta) = (2r_0, 0)\) \[ \boldsymbol{v}_P = \dot{r}\boldsymbol{\hat{e}}_x + r\dot{\theta}\boldsymbol{\hat{e}}_y = v_0 \boldsymbol{\hat{e}}_y \] So at \(t=0\), we have \(r\dot{\theta}= 2r_0 \dot{\theta}_0 = v_0\) (and \(\dot{r}=0\)). This gives \(C = 4r_0^2 \dot{\theta}_0 = 2 r_0 v_0\). Now [1] becomes \[ \boxed{ \dot{\theta}= \frac{2 r_0 v_0}{r^2} } \qquad{[2]} \] This result is valid at all time, and is useful to find the speed of \(P\) as a function of \(r\) \[ v^2 = \dot{r}^2 + r^2 \dot{\theta}^2 \] We still need \(\dot{r}\): we use the equation of the trajectory \(r = r_0 (1+ \cos\theta)\) \[ \dot{r}= -r_0 \dot{\theta}\sin\theta \] Then \(v^2\) can be found as a function of \(r\): \[ v^2 = r_0^2 \dot{\theta}^2 \sin^2\theta+ r_0^2 \dot{\theta}^2 (1+\cos\theta)^2 = r_0^2 \dot{\theta}^2 (2+2\cos\theta) =2r_0 r \dot{\theta}^2 \] Now we use [2] \[ v^2 = 2r_0 r \dot{\theta}^2 = 2r_0 r \frac{4 r_0^2 v_0^2}{r^4} = \frac{8 r_0^3 v_0^2}{r^3} \] Finally \[ \boxed{ v= v_0 \sqrt{\frac{8r_0^3}{r^3}} } \]
- b. To find the time \(T\) taken by \(P\) to reach \(O\), we use [2] and take into account \(r= r_0 (1+\cos\theta)\): \[ \dot{\theta}= \frac{2 r_0 v_0}{r_0^2 (1+\cos\theta)^2} \] We separate the variables: \[ (1+\cos\theta)^2 d\theta= 2 \frac{v_0}{r_0} dt \] and integrate from \(A\) to \(O\): \[ \int_0^\pi (1+\cos\theta)^2 d\theta= 2 \frac{v_0}{r_0} \int_0^T dt \] \[ \frac{3}{2}\pi= 2 \frac{v_0}{r_0} T \] giving \[ \boxed{ T = \frac{3\pi r_0}{4v_0} } \]
1.21. Use the intrinsic coordinate system: \(\boldsymbol{a}= (v_0^2/\rho)\boldsymbol{\hat{e}}_N\). You would then need \(\boldsymbol{\hat{e}}_N\) and \(\rho\) versus coordinate \(y\). Follow the procedure outlined in Lecture 4.
Solution:
Step 1: We find the velocity \(\boldsymbol{v}_P = \dot{x}\boldsymbol{\hat{e}}_x + \dot{y}\boldsymbol{\hat{e}}_y = \dot{y}(y/p)\boldsymbol{\hat{e}}_x + \dot{y}\boldsymbol{\hat{e}}_y\). This gives the speed \(v_0 = - \dot{y}(1+ y^2/p^2)^{1/2}\) (the sketch shows that \(\dot{y}<0\).)
Step 2: We find \((\boldsymbol{\hat{e}}_T,\boldsymbol{\hat{e}}_N)\): \[ \boldsymbol{\hat{e}}_T = \frac{(y/p)\boldsymbol{\hat{e}}_x + \boldsymbol{\hat{e}}_y}{(1+ y^2/p^2)^{1/2}} \] \[ \boldsymbol{\hat{e}}_N = \frac{-(y/p)\boldsymbol{\hat{e}}_y + \boldsymbol{\hat{e}}_x}{(1+ y^2/p^2)^{1/2}} \] (we made sure that \(\boldsymbol{\hat{e}}_N\) points toward the center of curvature.)
Step 3: We find the radius of curvature \(\rho\). Here we assume that \(\dot{y}\) is constant (only for the derivation of \(\rho\)). Then \(\boldsymbol{a}_P = (\dot{y}^2 /p)\boldsymbol{\hat{e}}_x\). This gives \(a_N= (\dot{y}^2 /p) /(1+ y^2/p^2)^{1/2} = \dot{y}^2 (1+ y^2/p^2) /\rho\), where in the last equality we used \(a_N= v^2/\rho\), with \(v^2 = \dot{y}^2 (1+ y^2/p^2)\). This gives the expression of \(\rho\) vs \(y\): \[ \frac{1}{\rho} = \frac{1}{p} (1+ y^2/p^2)^{-3/2} \]
Step 4: Now we can find the acceleration of \(P\) as \(v_0^2 /\rho\) directed along \(\boldsymbol{\hat{e}}_N\): \[ \boxed{\boldsymbol{a}_P = \frac{v_0^2}{p} (1+ y^2/p^2)^{-3/2} \; \boldsymbol{\hat{e}}_N} \] As a particular case, when \(y=0\), we find \(\boldsymbol{a}_P = (v_0^2/p) \boldsymbol{\hat{e}}_x\).
1.27. Find the cylindrical coordinates \((r,\theta,z)\) at any given time \(t\). This will then give you the position vector of \(P\): \(\boldsymbol{r}_{OP}= r\boldsymbol{\hat{e}}_r + z \boldsymbol{\hat{e}}_z\). Then differentiate twice w.r.t. time to obtain the velocity and acceleration of \(P\) in cylindrical coordinates. From \((r,\theta,z)\) you easily obtain the Cartesian coordinates \((x,y,z)\) of \(P\). Then \(Q\) has Cartesian coordinates \((x,y,0)\).
Solution:
- a. At time \(t\), the support \(\cal C\) has rotated by angle \(\theta= \omega t = t\) about axis \(Oz\). During the same time, \(P\) has moved along \(\cal C\) so that the angle made by line \(OP\) with the plane \(Oxy\) is \(\phi =\omega t =t\). We introduce the polar unit vectors \(\boldsymbol{\hat{e}}_r\) and \(\boldsymbol{\hat{e}}_\theta\).
The position vector of \(P\) can be written on \((\boldsymbol{\hat{e}}_r,\boldsymbol{\hat{e}}_z)\): \[ \boldsymbol{r}_{OP}= R(\cos\phi \boldsymbol{\hat{e}}_r + \sin\phi \boldsymbol{\hat{e}}_z)= R(\cos t \boldsymbol{\hat{e}}_r + \sin t \boldsymbol{\hat{e}}_z) = R\cos t (\cos\theta\boldsymbol{\hat{e}}_x + \sin \theta\boldsymbol{\hat{e}}_y) + R \sin t \boldsymbol{\hat{e}}_z \] With $R=1 $ m, we obtain the Cartesian coordinates of \(P\) \[ \boxed{ x = \cos^2 t , \qquad y = \cos t \sin t, \qquad z= \sin t } \]
b. To find the velocity and acceleration of \(P\) in cylindrical coordinates, we take the time-derivative of \(\boldsymbol{r}_{OP}= \cos t \boldsymbol{\hat{e}}_r + \sin t \boldsymbol{\hat{e}}_z\): \[ \boldsymbol{v}_P = -\sin t \boldsymbol{\hat{e}}_r + \cos t (\dot{\theta}\boldsymbol{\hat{e}}_\theta) + \cos t \boldsymbol{\hat{e}}_z \] using \(d\boldsymbol{\hat{e}}_t /dt = \dot{\theta}\boldsymbol{\hat{e}}_\theta\). With \(\dot{\theta}=\omega=1\) rad/s, this gives \[ \boxed{ \boldsymbol{v}_P = -\sin t \, \boldsymbol{\hat{e}}_r + \cos t \, \boldsymbol{\hat{e}}_\theta+ \cos t \, \boldsymbol{\hat{e}}_z } \qquad {[1]} \] Then we take \(d/dt\) of [1] \[ \boldsymbol{a}_P = -\cos t \boldsymbol{\hat{e}}_r-\sin t (\dot{\theta}\boldsymbol{\hat{e}}_\theta) - \sin t \boldsymbol{\hat{e}}_\theta+ \cos t (-\dot{\theta}\boldsymbol{\hat{e}}_r) - \sin t \boldsymbol{\hat{e}}_z \] giving \[ \boxed{ \boldsymbol{a}_P = -2\cos t \,\boldsymbol{\hat{e}}_r - 2\sin t \,\boldsymbol{\hat{e}}_\theta- \sin t \,\boldsymbol{\hat{e}}_z } \] Note that we cannot check the dimensions of the components of \(\boldsymbol{a}_P\). To do this, you would need to keep \(R\) and \(\omega\) explicit: a better expression would be \[ \boldsymbol{a}_P = R\omega^2 (-2\cos \omega t \boldsymbol{\hat{e}}_r - 2\sin \omega t \boldsymbol{\hat{e}}_\theta- \sin \omega t \boldsymbol{\hat{e}}_z ) \]
c. The coordinates of \(Q\) are given by \[ x_Q = \cos^2 t = \frac{1}{2}(1+\cos 2t), \qquad y_Q = \cos t \sin t = \frac{1}{2}\sin 2t \] We eliminate \(t\) to find \[ (x_Q - \frac{1}{2})^2 + y_Q^2 = \frac{1}{4} \] This is the equation of a circle of center \(C (1/2, 0, 0)\) and radius \(1/2\). Hence the trajectory of P is the intersection of the sphere of center O and radius R=1 and the cylinder of axis Cz and radius 1/2.
1.29.
Solution:
a. To find the velocity of \(P\) in polar coordinates, we determine the time-derivative of the position vector \(\boldsymbol{r}_{OP}= r \boldsymbol{\hat{e}}_r\): \[ \boxed{ \boldsymbol{v}_P = -2R \dot{\theta}\sin\theta\boldsymbol{\hat{e}}_r + 2R \cos\theta(\dot{\theta}\boldsymbol{\hat{e}}_\theta) = 2R \dot{\theta}(-\sin\theta\boldsymbol{\hat{e}}_r + \cos\theta\boldsymbol{\hat{e}}_\theta) } \] (we used \(d\boldsymbol{\hat{e}}_r /dt = \dot{\theta}\boldsymbol{\hat{e}}_\theta\)). Next we impose that \(\boldsymbol{v}_P\) has magnitude \(v_0\): \[ v_0 = 2R \dot{\theta}\Longrightarrow \boxed{\dot{\theta}= \frac{v_0}{2R}} \] Note that we used the fact that \(\dot{\theta}>0\). Also note that \(\dot{\theta}\) is constant.
b. From the general expression \(\boldsymbol{v}_P = v_0 \boldsymbol{\hat{e}}_T\) and with the expression of \(\boldsymbol{v}_P\) we find \[ \boxed{ \boldsymbol{\hat{e}}_T = -\sin\theta\boldsymbol{\hat{e}}_r + \cos\theta\boldsymbol{\hat{e}}_\theta} \] Then the normal unit vector is given by \[ \boldsymbol{\hat{e}}_N = \pm (-\cos\theta\boldsymbol{\hat{e}}_r - \sin\theta\boldsymbol{\hat{e}}_\theta) \] If we sketch the path of \(P\), we find that only the positive sign is acceptable along the entire path (which appears to be a closed loop since \(r\) is periodic in angle \(\theta\)). \[ \boxed{ \boldsymbol{\hat{e}}_N = (-\cos\theta\boldsymbol{\hat{e}}_r - \sin\theta\boldsymbol{\hat{e}}_\theta) } \]
c. To find the acceleration \(\boldsymbol{a}_P\), we start by taking \(d/dt\) of \(\boldsymbol{v}_P\): \[ \boldsymbol{a}_P = \frac{d}{dt} [v_0 (-\sin\theta\boldsymbol{\hat{e}}_r + \cos\theta\boldsymbol{\hat{e}}_\theta)] = v_0 \dot{\theta}(-\cos\theta\boldsymbol{\hat{e}}_r -\sin\theta\boldsymbol{\hat{e}}_\theta)+ v_0 \big(-\sin\theta(\dot{\theta}\boldsymbol{\hat{e}}_\theta) + \cos\theta(-\dot{\theta}\boldsymbol{\hat{e}}_r) \big) \] Finally, we obtain \[ \boldsymbol{a}_P =2v_0 \dot{\theta}(-\cos\theta\boldsymbol{\hat{e}}_r -\sin\theta\boldsymbol{\hat{e}}_\theta) \] We then use \(\dot{\theta}= v_0 /2R\): \[ \boldsymbol{a}_P = \frac{v_0^2}{R} \underbrace{(-\cos\theta\boldsymbol{\hat{e}}_\theta-\sin\theta\boldsymbol{\hat{e}}_r)}_{\boldsymbol{\hat{e}}_N} \] Since \(\boldsymbol{a}_P = \frac{v_0^2}{\rho} \boldsymbol{\hat{e}}_N\) (the speed is constant), we see that \[ \boxed{ \boldsymbol{a}_P = \frac{v_0^2}{R} \boldsymbol{\hat{e}}_N } \] and \[ \boxed{ \rho = R } \] It appears that the path of \(P\) is a circle of radius \(\rho= R\). It is easy to verify that the center is located on axis \(Ox\) at a distance \(R\) from \(O\).
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