In windup clocks, a strong torsional spring is used to store mechanical energy. Suppose that each week a clock requires four full turns of the winding key to keep running. The initial turn requires a torque of $0.30$ N • m and the final turn a torque of $0.45$ N • m. What amount of mechanical energy do you store in the spring when winding the clock?. What is the torsional spring constant? What is the consumption of mechanical power by the clock?
first, let's find the total energy rotational kinetic energy is juts the integral of torque times the angular displacement : $$\Delta KE = \int_0^{8\pi} \tau d\theta$$ but torque is just $\tau = -K\theta$ where K is called kappa a constant, if we apply a torque clockwise there is gonna be a restoring force counterclockwise replacing torque in the equation we get: $$\Delta KE = \int_0^{8\pi} -K\theta d\theta$$ $$\Delta KE = \frac{1}{2}K(8\pi)^2 $$
now lest find Kappa from the SHO, torque is moment of inertia times angular acceleration hence:
$$-K\theta = I\alpha$$ $$\alpha = \frac{K}{I}\theta$$
from SHO we see that $k/I = \omega^2$ that's the angular frequency hence: $$K = I\omega^2$$
I was trying to find $\omega$ this way since in a week there are 604800 seconds the clock will do one revolution in 151200 seconds since the body rotating:
$$\omega = \frac{2\pi}{T} = 4.15\times 10^-5$$
my problem is that I am not given data to find the moment of inertia "I", is there any other way to do this problem, or how would you find I?
I was TRYING to find "I" this way since we are given torque:
$$\tau = I\alpha$$ $$I = \frac{\tau}{\alpha}$$
but I am not given $\alpha$ angular aceleration
1. What amount of mechanical energy do you store in the spring when winding the clock?
Energy stored in the spring after the four full turns can be obtained by work done, that is
$$E=W=\frac12(0.30+0.45)\cdot (8\pi)=3\pi \;J\approx 9.4 \;J$$
2. What is the torsional spring constant?
Since the first turn require $0.30 \,Nm$ (initial value) and the fourth $0.45 \,Nm$ (final value), the spring constant value is
$$K= \frac{\Delta \tau }{\Delta \theta }=\frac {0.15} {4\cdot2\pi} \frac{Nm}{rad}\approx 6\cdot10^{-3}\frac{Nm}{rad} $$
3. What is the consumption of mechanical power by the clock?
The consumption of mechanical power, that is four full turns each week, is given by
$$P=\frac{E}{t} = \frac {3\pi}{7\cdot 24\cdot 60 \cdot 60}\; \frac{J}{s}\approx 1.56\cdot 10^{-5} \; W$$
The energy stored can also be found as follows.
For the total stored energy stored, we need to consider $0.45/0.375 = 12$ full turns that is
$$E_{TOT}=\frac12 K(24\pi)^2=5.4\pi\; J$$
while for the energy stored before winding the clock we need to consider $0.30/0.375 = 8$ full turns that is
$$E_{0}=\frac12 K(16\pi)^2=2.4\pi\; J$$
therefore energy stored in the spring after the four full turns is
$$E=E_{TOT}-E_{0}=5.4\pi\; J-2.4\pi\; J=3\pi \;J\approx 9.4 \;J$$