# 10.6 Torque  (Page 3/6)

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## Calculating torque on a rigid body

[link] shows several forces acting at different locations and angles on a flywheel. We have $|{\stackrel{\to }{F}}_{1}|=20\phantom{\rule{0.2em}{0ex}}\text{N},$ $|{\stackrel{\to }{F}}_{2}|=30\phantom{\rule{0.2em}{0ex}}\text{N}$ , $|{\stackrel{\to }{F}}_{3}|=30\phantom{\rule{0.2em}{0ex}}\text{N}$ , and $r=0.5\phantom{\rule{0.2em}{0ex}}\text{m}$ . Find the net torque on the flywheel about an axis through the center.

## Strategy

We calculate each torque individually, using the cross product, and determine the sign of the torque. Then we sum the torques to find the net torque.

## Solution

We start with ${\stackrel{\to }{F}}_{1}$ . If we look at [link] , we see that ${\stackrel{\to }{F}}_{1}$ makes an angle of $90\text{°}+60\text{°}$ with the radius vector $\stackrel{\to }{r}$ . Taking the cross product, we see that it is out of the page and so is positive. We also see this from calculating its magnitude:

$|{\stackrel{\to }{\tau }}_{1}|=r{F}_{1}\text{sin}\phantom{\rule{0.2em}{0ex}}150\text{°}=0.5\phantom{\rule{0.2em}{0ex}}\text{m}\left(20\phantom{\rule{0.2em}{0ex}}\text{N}\right)\left(0.5\right)=5.0\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}.$

Next we look at ${\stackrel{\to }{F}}_{2}$ . The angle between ${\stackrel{\to }{F}}_{2}$ and $\stackrel{\to }{r}$ is $90\text{°}$ and the cross product is into the page so the torque is negative. Its value is

$|{\stackrel{\to }{\tau }}_{2}|=\text{−}r{F}_{2}\text{sin}\phantom{\rule{0.2em}{0ex}}90\text{°}=-0.5\phantom{\rule{0.2em}{0ex}}\text{m}\left(30\phantom{\rule{0.2em}{0ex}}\text{N}\right)=-15.0\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}.$

When we evaluate the torque due to ${\stackrel{\to }{F}}_{3}$ , we see that the angle it makes with $\stackrel{\to }{r}$ is zero so $\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\stackrel{\to }{F}}_{3}=0.$ Therefore, ${\stackrel{\to }{F}}_{3}$ does not produce any torque on the flywheel.

We evaluate the sum of the torques:

${\tau }_{\text{net}}=\sum _{i}|{\tau }_{i}|=5-15=-10\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}.$

## Significance

The axis of rotation is at the center of mass of the flywheel. Since the flywheel is on a fixed axis, it is not free to translate. If it were on a frictionless surface and not fixed in place, ${\stackrel{\to }{F}}_{3}$ would cause the flywheel to translate, as well as ${\stackrel{\to }{F}}_{1}$ . Its motion would be a combination of translation and rotation.

Check Your Understanding A large ocean-going ship runs aground near the coastline, similar to the fate of the Costa Concordia , and lies at an angle as shown below. Salvage crews must apply a torque to right the ship in order to float the vessel for transport. A force of $5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}$ acting at point A must be applied to right the ship. What is the torque about the point of contact of the ship with the ground ( [link] )?

The angle between the lever arm and the force vector is $80\text{°};$ therefore, ${r}_{\perp }=100\text{m(sin80}\text{°}\right)=98.5\phantom{\rule{0.2em}{0ex}}\text{m}$ .

The cross product $\stackrel{\to }{\tau }=\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}$ gives a negative or clockwise torque.

The torque is then $\tau =\text{−}{r}_{\perp }F=-98.5\phantom{\rule{0.2em}{0ex}}\text{m}\left(5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\text{N}\right)=-4.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\text{N}·\text{m}$ .

## Summary

• The magnitude of a torque about a fixed axis is calculated by finding the lever arm to the point where the force is applied and using the relation $|\stackrel{\to }{\tau }|={r}_{\perp }F$ , where ${r}_{\perp }$ is the perpendicular distance from the axis to the line upon which the force vector lies.
• The sign of the torque is found using the right hand rule. If the page is the plane containing $\stackrel{\to }{r}$ and $\stackrel{\to }{F}$ , then $\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}$ is out of the page for positive torques and into the page for negative torques.
• The net torque can be found from summing the individual torques about a given axis.

## Conceptual questions

What three factors affect the torque created by a force relative to a specific pivot point?

magnitude of the force, length of the lever arm, and angle of the lever arm and force vector

Give an example in which a small force exerts a large torque. Give another example in which a large force exerts a small torque.

When reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame?

The moment of inertia of the wheels is reduced, so a smaller torque is needed to accelerate them.

two particles rotate in a rigid body then acceleration will be ?
same acceleration for all particles because all prticles will be moving with same angular velocity.so at any time interval u find same acceleration of all the prticles
Zaheer
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units as in how
praise
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no, only the frequency and the material of the spring
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means?
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AMIT
meaning
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briefly discuss rocket in physics
ok let's discuss
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physics is the study of natural phenomena with concern with matter and energy and relationships between them
Ibrahim
a potential difference of 10.0v is connected across a 1.0AuF in an LC circuit. calculate the inductance of the inductor that should be connected to the capacitor for the circuit to oscillate at 1125Hza potential difference of 10.0v is connected across a 1.0AuF in an LC circuit. calculate the inducta
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can you explain it
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this greatly depend on the kind of energy. for gravitational energy, it is result of the shattering effect violent collision of two black holes on the space-time which caused space time to be disturbed. this is according to recent study on gravitons and gravitational ripple. and many other studies
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and not every thing have to pop into existence. and it could have always been there . and some scientists think that energy might have been the only entity in the euclidean(imaginary time T=it) which is time undergone wick rotation.
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2 dimensional motion under constant acceleration due to gravity
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Not always 2D Awais
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bullet travel in x and y comment same as rock which is 2 dimensional
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components
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no all pf you are wrong. projectile is any object propelled through space by excretion of a force which cease after launch
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