Kinetic Energy Calculation

Classical Mechanics Level 3

One end of a rod of length L = 1 L = 1 is located at coordinates ( x 0 , y 0 ) (x_0, y_0) . The other end is attached to a particle of mass m = 1 m = 1 . The rod makes an angle θ \theta (in radians) with the horizontal.

At a particular instant, the state of the system is as follows:

x ˙ 0 = 1 y ˙ 0 = 1 θ = π / 3 θ ˙ = − 1 \dot{x}_0 = 1 \\ \dot{y}_0 = 1 \\ \theta = \pi / 3 \\ \dot{\theta} = -1

What is the kinetic energy of the particle?


The answer is 1.866.

Steven Chase by Steven Chase , 37, USA

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1 solution

Krishna Karthik
Jan 22, 2021

First, we know that x = L cos ⁡ θ x = L\cos{\theta} and y = L sin ⁡ θ y = L\sin{\theta}

With some differentiation with respect to time, x ˙ = − L sin ⁡ θ θ ˙ \dot{x} = -L\sin{\theta} \dot{\theta} and y ˙ = L cos ⁡ θ θ ˙ \dot{y} = L \cos{\theta} \dot{\theta}

But let's not forget the initial conditions. So:

x ˙ = 1 − L sin ⁡ θ θ ˙ \dot{x} = 1 - L\sin{\theta} \dot{\theta}

y ˙ = 1 + L cos ⁡ θ θ ˙ \dot{y} = 1 + L\cos{\theta} \dot{\theta}

Where θ ˙ = − 1 \dot{\theta} = -1 and θ = π 3 \displaystyle \theta = \frac{\pi}{3}

The square of the speed v 2 = 2 + 2 l ( sin ⁡ θ − cos ⁡ θ ) + L 2 v^2 = 2 + 2l(\sin{\theta} - \cos{\theta}) + L^2 ; this is just algebra.

Substituting values, we get E = 2 + 3 2 \displaystyle \boxed{E = \frac{2+\sqrt{3}}{2}}

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