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  • Please help! The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin . If the force acting on it is 1 N , and the distance of the particle from the origin is 5 m, the angle between the force and the position vector is (in radians) :

The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin . If the force acting on it is 1 N , and the distance of the particle from the origin is 5 m, the angle between the force and the position vector is (in radians) :

  • Option 1)

    \frac{\pi}{6}

  • Option 2)

    \frac{\pi}{4}

  • Option 3)

    \frac{\pi}{8}

  • Option 4)

    \frac{\pi}{3}

Answers (1)

best_answer

 

Torque -

\underset{\tau }{\rightarrow}= \underset{r}{\rightarrow}\times \underset{F}{\rightarrow}   

 

- wherein

This can be calculated by using either  \tau=r_{1}F\; or\; \tau=r\cdot F_{1}

r_{1} = perpendicular distance from origin to the line of force.

F_{1} = component of force perpendicular to line joining force.

 

 

T=\vec{r}\times \vec{F}

T=|\vec{r}|\cdot | \vec{F}|\cdot \sin \theta -----(1)

T=2.5Nm

|\vec{r}|=5m

|\vec{F}|=1N --------------------put in (1)

T=2.5=1\times 5\times \sin \theta

\sin \theta =0.5=\frac{1}{2}

\theta =\frac{\pi }{6}

 


Option 1)

\frac{\pi}{6}

Option 2)

\frac{\pi}{4}

Option 3)

\frac{\pi}{8}

Option 4)

\frac{\pi}{3}

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