question stringlengths 24 3.38k | response stringlengths 2 3.87k | CoT stringlengths 1.12k 62.7k |
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A particle moves on the x-axis attracted toward the origin O on it with a force proportional to its instantaneous distance from O. If it starts from rest at \( x = 5 \) cm and reaches \( x = 2.5 \) cm for the first time after 2 s, find (a) the position at any time t after it starts, (b) the speed at \( x = 0 \), (c) th... | (a) \( x = 5 \cos (\pi t/6); \, (b) \, 5\pi/6 \, \text{cm/s}; (c) \, 5 \, \text{cm}, \, 12 \, \text{s}, \, 1/12 \, \text{Hz}; (d) \, 5\pi^2/36 \, \text{cm/s}^2; (e) 5\pi/6 \, \text{cm/s} \) | Let's tackle this problem step by step. It seems like we're dealing with a classic harmonic oscillator scenario, where a particle is moving back and forth around an equilibrium position (the origin) under the influence of a restoring force proportional to its displacement from that equilibrium. This is characteristic o... |
Prove that if a particle moves in a central force field, then its path must be a plane curve. | Let \\( \\mathbf{F} = f(r) \\mathbf{r}_1 \\) be the central force field. Then\n\n\\[\n\\mathbf{r} \\times \\mathbf{F} = f(r) \\mathbf{r} \\times \\mathbf{r}_1 = 0 \\tag{1}\n\\]\n\nsince \\( \\mathbf{r}_1 \\) is a unit vector in the direction of the position vector \\( \\mathbf{r} \\). Since \\( \\mathbf{F} = m \\frac{d... | To prove that the path of a particle moving in a central force field lies in a plane, we need to show that the particle's position vector \(\mathbf{r}(t)\) remains in a two-dimensional plane at all times. A central force field is one in which the force \(\mathbf{F}\) acting on the particle is always directed towards or... |
Two initially static disks, connected by an elastic shaft of stiffness c, are suddenly acted on by a constant torque of moment M_0. The moment of inertia of each disk is J. Neglecting the mass of the shaft, determine the subsequent motion of the system. | ϕ_1 = M/4J t^2 + M/4c (1 - cos sqrt(2c/J) t); ϕ_2 = M/4J t^2 - M/4c (1 - cos sqrt(2c/J) t). | So I've got this problem here involving two disks connected by an elastic shaft, and they're subjected to a sudden constant torque. I need to find out how the system moves after that torque is applied. Let's break this down step by step.
First, let's understand the setup:
- There are two disks, each with a moment of ... |
In Fig. 3-28 the mass \( m \) hangs from an inextensible string \( OA \). It is pulled aside by a horizontal string \( AB \) so that \( OA \) makes an angle \( \alpha \) with the vertical. Find the tension in each string. | Tension in \( AB = mg \tan \alpha \); in \( OA = mg \sec \alpha \). | So I've got this problem here, and I'm trying to figure out the tensions in two strings. There's a mass \( m \) hanging from string \( OA \), and then there's another string \( AB \) that's pulling it aside horizontally, making string \( OA \) make an angle \( \alpha \) with the vertical. I need to find the tensions in... |
(a) Prove that \( x = c_1 e^t + c_2 e^{-3t} + \sin t \) is the general solution of \n\n\[\n\frac{d^2 x}{dt^2} + 2 \frac{dx}{dt} - 3x = 2 \cos t - 4 \sin t\n\]\n\n(b) Find the particular solution such that \( x = 2, \frac{dx}{dt} = -3 \) at \( t = 0 \). | 1. From \( x = c_1 e^t + c_2 e^{-3t} + \sin t \) we have\n\n\[\n\frac{dx}{dt} = c_1 e^t - 3c_2 e^{-3t} + \cos t, \quad \frac{d^2 x}{dt^2} = c_1 e^t + 9c_2 e^{-3t} - \sin t\n\]\n\nThen \n\n\[\n\frac{d^2 x}{dt^2} + 2 \frac{dx}{dt} - 3x = (c_1 e^t + 9c_2 e^{-3t} - \sin t) + 2(c_1 e^t - 3c_2 e^{-3t} + \cos t)\n\]\n\n\[\n- ... | Let's tackle this problem step by step. We have a second-order nonhomogeneous differential equation, and we're given a general solution. Our tasks are to prove that the given function is indeed the general solution and then find a particular solution that satisfies specific initial conditions.
### Part (a): Proving th... |
A uniform beam hinged to a wall has a support 160 cm from the wall. The beam is 400 cm long and weighs 320 kgf. Two weights of 160 kgf and 240 kgf are applied to the beam at the distances of 120 cm and 180 cm, respectively, from the wall. Determine the reactions of the supports. | 790 kgf upwards; 70 kgf downwards. | So I've got this problem here about a uniform beam hinged to a wall with a support 160 cm from the wall. The beam is 400 cm long and weighs 320 kgf. There are two weights applied to the beam: one of 160 kgf at 120 cm from the wall and another of 240 kgf at 180 cm from the wall. I need to determine the reactions of the ... |
Determine the vertical reactions of the supports of the horizontal beam with span \( l \), if a weight \( P \) is applied to the beam at a distance \( x \) from the first support. | **Ans.** \( R_1 = P \frac{l-x}{l} \); \( R_2 = P \frac{x}{l} \). | So I've got this problem here about finding the vertical reactions of the supports for a horizontal beam with a span \(l\), and there's a weight \(P\) applied at a distance \(x\) from the first support. Okay, let's see. I need to find the reactions at the supports, which I'll call \(R_1\) and \(R_2\) for the first and ... |
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