Centripetal Force with a Spin

Classical Mechanics Level 2

A rock of mass $5 \; kg$ is attached to a spring of equilibrium length $1 \; m$ and spring constant $500 \; N/m$ at one end. The spring has its other end fixed to a point. The rock and spring system are laid flat on a table with respect to the horizontal. The rock engages in uniform circular motion around the fixed point with tangential velocity $10 \; m/s$ . The amount by which the spring stretches with respect to its equilibrium length can be expressed as $\frac{\sqrt{a}-b}{c}$ where where $a$ , $b$ , and $c$ are positive integers and $a$ is not divisible by the square of any prime. Find $a + b + c$ .

Details and Assumptions

Neglect air resistance and friction

Neglect the mass of the spring

Assume the rock is a point particle

The answer is 8.

1 solution

Lucas Chaves Meyles
Apr 7, 2018

The restoring force of the spring equals the centripetal force in this scenario. Let $x$ be the amount by which the spring stretches from its equilibrium position. Realize that the distance the rock is from the fixed point is then $l + x$ due to the stretching of the spring during motion. Thus,

$F_s = F_c$ $kx = \frac{mv^{2}}{l + x}$ $kx^{2} + klx -mv^{2} = 0$ $x = \frac{-kl +\sqrt{k^{2}l^{2}-4k(-mv^{2})}}{2k}$ $x = \frac{-(500)(1) +\sqrt{(500)^{2}(1)^{2}+4(500)(5)(10)^{2}}}{2(500)}$ $x = \frac{-500 + 500\sqrt{5}}{2(500)}$ $x = \frac{\sqrt{5}-1}{2}$ $a + b + c = 5 + 1 + 2 = \boxed{8}$

Centripetal Force with a Spin

The answer is 8.

1 solution

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