1. The problem statement, all variables and given/known data
A trolley consists of an enclosed carriage of mass M supported by 4 wheels each of mass m/4. A simple pendulum consists of a bob of mass μ on a light rod of length l supported at a pivot attached to the roof; its motion is confined to a plane. a) Write down expressions for the kinetic and potential energies in terms of an angle θ to the vertical and its time derivative.
b)Suppose now that the carriage is in motion on a horizontal surface. The plane of swing of the pendulum includes the direction of motion of the carriage). Let y(t) be the displacement of the carriage. Show that the total kinetic energy is $$T = \frac{1}{2}\left[\left(M+\frac{3}{2}m\right)\dot{y}^2 + \mu(\dot{y}^2 + \ell^2 \dot{\theta}^2) \pm 2\mu \ell \cos \theta \dot{y}\dot{\theta}\right],$$ where the sign of the last term is dependant on whether ##\theta## is measured in the same or opposite sense to y.
2. Relevant equations
Results all derived from usual definitions of kinetic and potential energy.
3. The attempt at a solution
In another problem with the surface at an angle ##\phi## (unrelated as far as the pendulum is concerned), the acceleration of the carriage (without the pendulum) was found to be ##(M + \frac{3}{2}m)\ddot{x} = (M+m)g\sin\phi##. So the kinetic energy of the carriage alone is ##T_c = \frac{1}{2} (M+\frac{3}{2} m) \dot{y}^2##. In part a), the kinetic energy of the pendulum is ##T_p = \frac{1}{2} \mu \ell^2 \dot{\theta}^2##. So this accounts for two of the terms. I cannot get the last term and my reasoning for the second term is that the bob's added kinetic energy in the x direction is simply ##1/2 \mu \dot{y}^2## since the pendulum is attached to the carriage which has a constant horizontal velocity of ##\dot{y}##. Is that reasonable? The vertical direction is ##\ell \cos \theta##, but I am not sure how to get the coupled term ##\dot{\theta}\dot{y}##.
Many thanks.
A trolley consists of an enclosed carriage of mass M supported by 4 wheels each of mass m/4. A simple pendulum consists of a bob of mass μ on a light rod of length l supported at a pivot attached to the roof; its motion is confined to a plane. a) Write down expressions for the kinetic and potential energies in terms of an angle θ to the vertical and its time derivative.
b)Suppose now that the carriage is in motion on a horizontal surface. The plane of swing of the pendulum includes the direction of motion of the carriage). Let y(t) be the displacement of the carriage. Show that the total kinetic energy is $$T = \frac{1}{2}\left[\left(M+\frac{3}{2}m\right)\dot{y}^2 + \mu(\dot{y}^2 + \ell^2 \dot{\theta}^2) \pm 2\mu \ell \cos \theta \dot{y}\dot{\theta}\right],$$ where the sign of the last term is dependant on whether ##\theta## is measured in the same or opposite sense to y.
2. Relevant equations
Results all derived from usual definitions of kinetic and potential energy.
3. The attempt at a solution
In another problem with the surface at an angle ##\phi## (unrelated as far as the pendulum is concerned), the acceleration of the carriage (without the pendulum) was found to be ##(M + \frac{3}{2}m)\ddot{x} = (M+m)g\sin\phi##. So the kinetic energy of the carriage alone is ##T_c = \frac{1}{2} (M+\frac{3}{2} m) \dot{y}^2##. In part a), the kinetic energy of the pendulum is ##T_p = \frac{1}{2} \mu \ell^2 \dot{\theta}^2##. So this accounts for two of the terms. I cannot get the last term and my reasoning for the second term is that the bob's added kinetic energy in the x direction is simply ##1/2 \mu \dot{y}^2## since the pendulum is attached to the carriage which has a constant horizontal velocity of ##\dot{y}##. Is that reasonable? The vertical direction is ##\ell \cos \theta##, but I am not sure how to get the coupled term ##\dot{\theta}\dot{y}##.
Many thanks.
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