t = 0 समय पर,2, kg के एक कण का स्थिति सदिश मू | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) कण पर कार्य करने वाला टॉर्क $\vec{	au} = \vec{r} 	imes \vec{F} = \vec{r} 	imes \frac{d\vec{p}}{dt} = \frac{d}{dt}(\vec{r} 	imes \vec{p}) = \fr

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(A) कण पर कार्य करने वाला टॉर्क $\vec{	au} = \vec{r} 	imes \vec{F} = \vec{r} 	imes \frac{d\vec{p}}{dt} = \frac{d}{dt}(\vec{r} 	imes \vec{p}) = \frac{d\vec{L}}{dt}$ द्वारा दिया जाता है।द्रव्यमान $m = 2\, kg$,स्थिति $\vec{r}(t) = \vec{r}_0 + \int_0^t \vec{v} dt = (4\hat{i} - 2\hat{j}) + \int_0^t (2t^2\hat{i}) dt = (4 + \frac{2}{3}t^3)\hat{i} - 2\hat{j}$.वेग $\vec{v}(t) = 2t^2\hat{i}$.कोणीय संवेग $\vec{L} = \vec{r} 	imes m\vec{v} = m [((4 + \frac{2}{3}t^3)\hat{i} - 2\hat{j}) 	imes (2t^2\hat{i})] = 2 [0 - 2\hat{j} 	imes 2t^2\hat{i}] = 2 [4t^2\hat{k}] = 8t^2\hat{k}$.टॉर्क $\vec{	au} = \frac{d\vec{L}}{dt} = \frac{d}{dt}(8t^2\hat{k}) = 16t\hat{k}$.$t = 2\, s$ पर,$\vec{	au} = 16(2)\hat{k} = 32\hat{k}\, N\cdot m$.

Similar Questions

Vedclass Test Series

Exam Paper Generator

Online Exam Module