CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Tangential acceleration of a particle moving in a circle of radius $$1\ m$$ varies with time $$t$$ as (initial velocity of particle is zero). Time after which total acceleration of particle makes and angle of $$30^{o}$$ with radial acceleration is

1094761_f2c3fd795035476ea6353acfdaced29e.png


A
4 sec
loader
B
4/3 sec
loader
C
22/3 sec
loader
D
2 sec
loader

Solution

The correct option is C $$2^{2/3}\ sec$$
using equation for a straight line $$(y=mx+c)$$ where,
$$m=\tan 60^o=\sqrt{3} $$ (slop) $$C=0$$ from graph
relation b/w $$a_T$$ and $$t$$ is obtained as:-
$$a_T=\tan 60^ot+0$$
this tangential acceleration increases velocity (v) of the particless as.
$$a_T =\dfrac{dv}{dt} \Rightarrow \dfrac{dv}{dt}=\sqrt{3}t$$
$$\Rightarrow \int dv =\int \sqrt{3}+dt \Rightarrow v=\dfrac{\sqrt{3}t^2}{2}$$
so, $$a_T$$ a particular time, centripetal accin  $$(a_c)$$ is given by :-
$$a_c=\dfrac{v^2}{r}=\dfrac{3}{4}t^4 (r=1) \, \vec{a_T}$$
let a time $$\vec{a}$$ masses angle $$30^o$$ with $$\vec{a_c}$$. so it makes angles $$60^o$$ with $$\vec{a_T}$$ ($$\vec{a_c} K \vec{a_T}$$ are perpendicular)
$$\Rightarrow \vec{a_T}=\sqrt{3t^2+\dfrac{9}{16}+8}\dfrac{1}{2}$$ that given,
$$\Rightarrow 3t^2=\dfrac{1}{11}(3t^2+\dfrac{9}{16}+8)$$
$$\Rightarrow t=0$$ or $$t=2^{2/3}$$.

1094248_1094761_ans_08e967412686489eb53154e62d4a3e37.png

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image