1.1 You are given \(\boldsymbol{v}_p\) in polar coordinates: this gives you the components \(v_r\) and \(v_\theta\). Since \(v_r = \dot{r}\) and \(v_\theta = r \dot{\theta}\), you can easily obtain the trajectory by integration. Conversely, you obtain the acceleration by time-differentiating \(v_\theta \,\boldsymbol{\hat{e}}_\theta\): this will give you the terms: \(\dot{v}_\theta\, \boldsymbol{\hat{e}}_\theta\) and \(v_\theta\, d \boldsymbol{\hat{e}}_\theta/dt\).
Solution:
a. Scalar \(a\) must have the dimension of a velocity \([a] = LT^{-1}\). Since \(\lambda t\) must be dimensionless, \([\lambda] = T^{-1}\). (\(L\) stands for length, \(T\) for time).
b. In polar coordinates, we have \(\boldsymbol{v}_p = \dot{r}\boldsymbol{\hat{e}}_r + r \dot{\theta}\boldsymbol{\hat{e}}_\theta\): This gives two equations \[ \dot{r} = 0 , \qquad r\dot{\theta} = a \exp(-\lambda t) \] which gives by integration \[ \boxed{r = r_0} \qquad \boxed{\theta= \frac{a}{r_0\lambda}( 1- \exp (-\lambda t) ) } \] after imposing the BC \(r=r_0\) and \(\theta= 0\) at time \(t=0\). We can verify that the ratio \(a/r_0\lambda\) is dimensionless.
c. Since \(r=r_0\) and \(0 \leq \theta\leq 1\), the path is an arc of circle (if \(\lambda >0\)) of radius \(r_0\) centered at \(O\). If \(\lambda >0\), the entire circle is travelled by \(P\) in the clockwise direction.
d. We obtain \(\boldsymbol{a}_P\) as \(d/dt\) of \(v_\theta\boldsymbol{\hat{e}}_\theta\): this gives \[ \boldsymbol{a}_P = -\lambda a \exp(-\lambda t) \boldsymbol{\hat{e}}_\theta+a \exp(-\lambda t) (-\dot{\theta}\boldsymbol{\hat{e}}_r) \] with \(\dot{\theta}= (a/r_0) \exp(-\lambda t)\) \[ \boxed{ \boldsymbol{a}_P = -\lambda a \exp(-\lambda t) \boldsymbol{\hat{e}}_\theta-\frac{a^2}{r_0} \exp(-2\lambda t) \boldsymbol{\hat{e}}_r } \]
1.2 You start with the position vector \(\boldsymbol{r}_{OP}= r(\theta) \boldsymbol{\hat{e}}_r\). Differentiate twice w.r.t. time. You should arrive at \(\boldsymbol{a}_P = a_r \boldsymbol{\hat{e}}_r\). This will hinge upon exploiting the constraint \(r^2\dot{\theta} = Cst\).
Solution: First draw a sketch which would show the quantities \(r, \theta, \boldsymbol{\hat{e}}_r, \boldsymbol{\hat{e}}_\theta\). We want to show that the acceleration is radial \(\boldsymbol{a}= a_r \boldsymbol{\hat{e}}_r\), knowing that \(r= p/(1+e \cos\theta)\) and \(r^2 \dot{\theta}= p C\) (\(p, e, C\) are constants).
a. \(p\) has the dimension of a length: so \([C] = L T^{-1}\). Constant \(C\) has the dimension of a speed.
b. First we note that \(a_\theta= 2\dot{r}\dot{\theta}+ r\ddot{\theta}\) from the general expression of \(\boldsymbol{a}_P\) in polar coordinates. We can take \(d/dt\) of \(r^2\dot{\theta} = Cst\) to find \[ 2 r \dot{r}\dot{\theta}+ r^2 \ddot{\theta}= r (2\dot{r}\dot{\theta}+ r\ddot{\theta}) = 0 \] This implies that \(a_\theta=0\).
The next step is to prove that \(a_r = \ddot{r}- r\dot{\theta}^2= -C \dot{\theta}\). From \(1/r = (1+ e \cos\theta)/p\), we find by time-differentiation \[ - \dot{r}/r^2 = (-e/p) \dot{\theta}\sin\theta \] With \(\dot{\theta}= pC/r^2\), this simplifies to \[ \dot{r}= eC \sin\theta \] One more differentiation gives \[ \ddot{r}= e C \dot{\theta}\cos\theta \] The second second term can be expressed as \[ r \dot{\theta}^2 = \frac{1}{r} (r^2 \dot{\theta}) \dot{\theta}= C \dot{\theta}(1+e\cos\theta) \] Now we find \(a_r\) \[ a_r = e C \dot{\theta}\cos\theta- C\dot{\theta}(1+e\cos\theta) = -C \dot{\theta} \] Note that \(a_r\) is also given by \(-C \dot{\theta}= -p C^2 /r^2\). We recover the law of gravitation: the force \(m \boldsymbol{a}_P\) points toward \(O\) and its magnitude is proportional to $1/r^2 $ (which varies along the motion!)
1.3 First find \(r\) vs \(\theta\). Use triangle \(OAC\) (\(C\) is the center of circle). To find the velocity of \(A\), take the time-derivative of \(\boldsymbol{r}_{OA}= r(\theta)\boldsymbol{\hat{e}}_r\). Verify that \(\boldsymbol{v}_A\) is tangent to the circular path at \(A\). Check that the motion is uniform. Then take \(d/dt\) of \(\boldsymbol{v}_A\) keeping in mind that \(\dot{\theta}\) is constant. Does \(\boldsymbol{a}_A\) point toward \(O\) or \(C\) ? Yes/No? Why?
Solution:
a. To find the trajectory of A in polar coordinates we need to express \(r = |OA|\) versus angle \(\theta\). In isosceles triangle \(OAC\) (\(C\) is the center of the circle), we find that \(r/2 = R \cos\theta\). This gives \[ \boxed{ r = 2R \sin\theta } \] We define the corresponding unit vectors \((\boldsymbol{\hat{e}}_r, \boldsymbol{\hat{e}}_\theta)\) on the sketch shown below.
b. To find the velocity of \(A\) we take the time-derivative of \(\boldsymbol{r}_{OA}= 2R \sin\theta \,\boldsymbol{\hat{e}}_r\): \[ \boldsymbol{v}_A = 2R (\dot{\theta}\cos\theta\boldsymbol{\hat{e}}_r + \sin \theta(\dot{\theta}\boldsymbol{\hat{e}}_\theta)) \] which gives \[ \boxed{ \boldsymbol{v}_A = 2R \dot{\theta}(\cos\theta\boldsymbol{\hat{e}}_r + \sin \theta\boldsymbol{\hat{e}}_\theta) } \qquad [1] \] We verify that \(\boldsymbol{v}_A\) is tangent to the circular path at \(A\) (see Figure below). The speed is the constant \(v_A = 2R \dot{\theta}\): the motion is uniform. To find the acceleration we take the time-derivative of [1] keeping in mind that \(\dot{\theta}\) is constant \[ \boldsymbol{a}_A = 2R \dot{\theta}(-\dot{\theta}\sin\theta\boldsymbol{\hat{e}}_r +\dot{\theta}\cos\theta\boldsymbol{\hat{e}}_\theta+ \dot{\theta}\cos \theta\boldsymbol{\hat{e}}_\theta- \sin \theta\boldsymbol{\hat{e}}_r) \] which gives \[ \boxed{ \boldsymbol{a}_A = 4R \dot{\theta}^2 (-\sin\theta\boldsymbol{\hat{e}}_r + \cos\theta\boldsymbol{\hat{e}}_\theta) } \qquad {[2]} \] The magnitude of \(\boldsymbol{a}_A\) is \(|\boldsymbol{a}_A | = 4R\dot{\theta}^2\) is also constant. We can check that \(\boldsymbol{a}_A\) points toward \(C\) as is expected for a uniform circular motion.
We find \(v_A = 0.4\) m/s and \(|\boldsymbol{a}_A | = 0.32\) m/s\(^2\).
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Note: both results [1] and [2] must match the expressions of \(\boldsymbol{v}_P\) and \(\boldsymbol{a}_P\) in the polar coordinate system whose pole is the center \(C\). For instance \(\boldsymbol{v}_P = R\dot{\phi} \boldsymbol{\hat{e}}_\phi\) and \(\boldsymbol{a}_P = -R\dot{\phi}^2\boldsymbol{\hat{e}}_R\). With \(\phi = 2\theta\), we have \(\dot{\phi} = 2\dot{\theta}\).