Question

a 100 - g mass is attached to a rubber band with an unstretched length of 13.0 cm. the mass is set into circular motion with a constant speed of 65 cm/s. the spring constant of the rubber band is band 6 n/m.
by how much is the rubber band stretched under these conditions?
note: the extra stretch of the rubber band is not the radius of the circular motion.
your answer:

Explanation:

Step1: Identify centripetal - force and spring - force equations

The centripetal force $F_c = \frac{mv^{2}}{r}$, and the spring force $F_s=k\Delta x$. In circular motion, the spring force provides the centripetal force, so $F_c = F_s$. Let the unstretched length of the rubber - band be $L_0 = 13.0\ cm=0.13\ m$, the mass $m = 100\ g = 0.1\ kg$, the speed $v = 65\ cm/s=0.65\ m/s$, and the spring constant $k = 6\ N/m$. The radius of the circular motion $r = L_0+\Delta x$. So, $\frac{mv^{2}}{L_0 + \Delta x}=k\Delta x$.

Step2: Rearrange the equation

Cross - multiply to get $mv^{2}=k\Delta x(L_0+\Delta x)=kL_0\Delta x + k\Delta x^{2}$. Rearrange it to the quadratic form $k\Delta x^{2}+kL_0\Delta x - mv^{2}=0$. Substitute the values: $6\Delta x^{2}+6\times0.13\Delta x-0.1\times(0.65)^{2}=0$, which simplifies to $6\Delta x^{2}+0.78\Delta x - 0.04225 = 0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 6$, $b = 0.78$, $c=-0.04225$), the quadratic formula is $\Delta x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(0.78)^{2}-4\times6\times(-0.04225)=0.6084 + 1.014 = 1.6224$. Then, $\Delta x=\frac{-0.78\pm\sqrt{1.6224}}{12}=\frac{-0.78\pm1.2737}{12}$. We take the positive root since length cannot be negative. $\Delta x=\frac{-0.78 + 1.2737}{12}=\frac{0.4937}{12}\approx0.041\ m = 4.1\ cm$.

Answer:

$4.1\ cm$