Question

A solid uniform disc of mass M and Radius R is oscillating about an axis through P. The axis is perpendicular to the palne of the disc. Friction at P is negligibly small and can be ignored. The distance from P to the centre, C, of the disc is b. The gravitation acceleration is g.
Answer the following:
a) When the displacement angle is theta, what then is the torque relative to point P?
b) What is the moment of inertia for rotation about the axis through P?
c) The torque causes an angular acceleration about the add through P. Write down the equation of motion in terms of angle theta and the angular acceleration.
d) As the disc oscillates, the maximum ange of displacement, theta(max), is very small, and the motion is near perfect SHM. Then,
i) What is the period of oscillation?
ii) As the disc oscillates, is there any force that the axis at P exerts on the disc? Explain your answer. ​

Answer 1

Answer:

$\sf{\green{\underline{\underline{\orange{Answer :-}}}}}$

It is well-known that there is a 90 deg phase shift between the current and voltage in the capacitor (when supplied by a sinusoidal signal) and it varies from 0 to 90 degrees in the RC integrating circuit when the frequency changes from zero to infinity. But there are not good "physical" explanations why and how the phase shift appears and why it varies with the frequency.

I have tried to explain intuitively the phenomenon on the Wikipedia talk page about the RC circuit (see the attachment below). Let's use it as an initial point for this discuss.

Answer 2

a) ε is 0 ahead i

Because I and R is P in phase and i.e, Z and ε are in phase as ε = ε_0 sin ωt from diagram

$i = \frac{e0}{z} \sin(wt - 0) \\ \\ where. \: 0 = { \tan }^{ - 1} \frac{ xl}{r}$

b) I = 1.0A(given)

$\sf{ V_R=160=iR=> R=}$$\sf\frac{160}{1}=160$

$\sf{ V_L= 160V=iXL=>X_L=}$$\sf\frac{120}{1}=120$

$\sf{V_net = }$

$\sqrt{ {v}^{2}r + {v}^{2}l } \\ \\ \sqrt{ {(160)}^{2} + {(120)}^{2} } \\ \\ \sqrt{25600 + 14400} \\ \\ 200...effective \: value \: of \: apphead \\ voltage \\ \\ z = \sqrt{ {r}^{2} + x \times \frac{2}{l} } = \sqrt{ {(160)}^{2} + {(120)}^{2} } \\ \\ 200...impendence$

c) when direct current is passes through the circuit then R will be zero 0.

at t=w, inductor will act as plane wire and potential drop will only act occur at R ( i.e, i=E/R)

For immediate time

E=iR+

$\frac{ldi}{dt} \\ \\ where \: i \: will \: vary \: accrodingly$

hope this will be helpful to you!! don't spam