Recent content by Bling Fizikst

si There seems to be too many things going on at the same time , including center of mass shifts . I am trying to find the accelerations as functions of time . I tried to write the position vector of B ...

Is this correct? Seems fairly complicated . How do i handle all these tension forces , moreover since it's an ideal pully , the net force acting on it would be zero .

Consider the wedge as '1' and block as '2' : i need to find ##a_{2/F}=a_{2/1}+a_{1/F}## . The FBD of '1' : $$Mg\sin\theta - f=Ma_{1/F}$$ $$Mg\cos\theta=N$$ The FBD of '2' : $$mg - T =ma_{2/1}$$ Assuming kinetic friction : ##f=\mu Mg\cos\theta##. But how do i find ##T##?

My bad , the answer matches if I took the mass as given in the diagram , i.e, ##3m## instead of ##2m## . The answer turns out to be : ##\frac{F}{3m}## .

I assumed the length of the rod to be ##6l## for simpler calculations . Here , $$a_{cm}=\frac{F}{m+2m}=\frac{F}{3m}$$ Here , CM is located at a distance of ##2l## from the ##2m## mass . Writing the moment equation : $$\alpha \times \left[ m(4l)^2+2m(2l)^2\right]=Fl\implies...

I just used work energy theorem which includes ##W_{\text{ext}}+W_{mg}=\triangle KE## , all i need is ##W_{\text{ext}}## . Morever , i don't even think you can find tension in that way explicitly . So , yeah just ignore that work done by tension thing , it's useless imo .

You are right , but here ##mg## is a constant force .

Based on the above discussions , i wrote a general relation for ##v,r,\alpha## : $$T_x=T\sin\alpha=\frac{mv^2}{r}$$ $$T_y=T\cos\alpha =mg$$ From here , $$\tan\alpha=\frac{r}{y} =\frac{v^2}{rg}$$ $$\implies y\equiv y(r,v)=\frac{r^2g}{v^2}$$ Now , if i were to compute $$\int \vec{T}\cdot \vec{dy}...

From the FBD , i assumed the angle between the thread and the vertical is ##\theta## . Now , writing the force equations : $$\frac{mv_{\circ}^2}{R}=T\sin\theta $$ $$mg=T\cos\theta$$ From here we can find : $$T=\frac{m\sqrt{v_{\circ}^4+R^2g^2}}{R}$$ and $$\tan\theta=\frac{v_{\circ}^2}{Rg}$$ . But...

Say some external agent does the above work ##W_{\text{ext}}## , say the inital depth was ##h## , then final depth ##=\frac{h}{2}## . Now , we don't know the radius of the new circular path , as well as the new velocity of the particle . Employing the work energy theorem ...

Tried to find patterns on the coefficients of the polynomials . But could only go as far as : $$x: \frac{i+1}{2}$$ $$x^2: \frac{i(i+2)}{8}$$ for the ##i##th fibonacci polynomial . Quite stuck on this one for a while , so , not sure if i have to change routes . But seems like if i find the...

Not a rigorous post . But how i interpret ##g(x)=f(x+a)## is just setting the argument of ##f## to zero ,i.e, ##x+a=0## this means : ##x=-a## . So , whatever graph you have for ##f(x)## , you get the same value of ##f(0)## at ##x=-a##,i.e, ##g(-a)=f(-a+a)=f(0)## (##a>0##) . So , it is as good...

thank you

edited it , yet it is not rendering

I asked chatgpt to correct it yet it is not rendering here