Answer:
Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.
Explanation:
Considering the complete question attached in figure below.
Time period for balance wheel is:
![T=2\pi\sqrt{\frac{I}{K}}](https://tex.z-dn.net/?f=T%3D2%5Cpi%5Csqrt%7B%5Cfrac%7BI%7D%7BK%7D%7D)
![I=mR^{2}](https://tex.z-dn.net/?f=I%3DmR%5E%7B2%7D)
m = mass of balance wheel
R = radius of balance wheel.
Angular frequency is related to Time period as:
![\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D%5C%5C%5Comega%3D%5Csqrt%7B%5Cfrac%7BK%7D%7BI%7D%7D%20%5C%5C%5Comega%3D%5Csqrt%7B%5Cfrac%7BK%7D%7BmR%5E%7B2%7D%7D)
As dimensions of new balance wheel are one-third of their original values
![R_{new}=\frac{R}{3}](https://tex.z-dn.net/?f=R_%7Bnew%7D%3D%5Cfrac%7BR%7D%7B3%7D)
![\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega](https://tex.z-dn.net/?f=%5Comega_%7Bnew%7D%3D%5Csqrt%7B%5Cfrac%7BK%7D%7BmR_%7Bnew%7D%5E%7B2%7D%7D%7D%5C%5C%5C%5C%5Comega_%7Bnew%7D%3D%5Csqrt%7B%5Cfrac%7BK%7D%7Bm%28%5Cfrac%7BR%7D%7B3%7D%29%5E%7B2%7D%7D%7D%5C%5C%5C%5C%5Comega_%7Bnew%7D%3D%7B3%7D%5Csqrt%7B%5Cfrac%7BK%7D%7BmR%5E%7B2%7D%7D%7D%5C%5C%5C%5C%5Comega_%7Bnew%7D%3D%7B3%7D%5Comega)
Johannes Kepler was a main stargazer of the Scientific Revolution known for detailing the Laws of Planetary Motion. A stargazer, obviously, is a man who contemplates the sun, stars, planets and different parts of room. Kepler was German and lived in the vicinity of 1571 and 1630.
Despite the fact that Kepler is best known for characterizing laws in regards to planetary movement, he made a few other striking commitments to science. He was the first to discover that refraction drives vision in the eye and that utilizing two eyes empowers profundity recognition.
The force applied would be 1.05*9.8 = 10.3 N
the pressure is equal to F/a
area will be πr^2 = 0.002826
thus pressure will be = 10.3/0.002826= 3644.72 N/m^2
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
![9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}](https://tex.z-dn.net/?f=9.8%5Ctimes%2012.5%3DW%5Ctimes%20%2850-12.5%29%5C%5C%5C%5CW%3D%5Cdfrac%7B9.8%5Ctimes%2012.5%7D%7B37.5%7D)
W = 3.266 N
The mass of the meters stick is :
![m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7BW%7D%7Bg%7D%5C%5C%5C%5Cm%3D%5Cdfrac%7B3.266%7D%7B9.81%7D%5C%5C%5C%5Cm%3D0.333%5C%20kg)
So, the mass of the meter stick is 0.333 kg.