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

| {\vec{F}}_{1} | = 20 N,

| {\vec{F}}_{2} | = 30 N

| {\vec{F}}_{3} | = 30 N

, and

r = 0.5 m

. Find the net torque on the flywheel about an axis through the center.

Figure shows a flywheel with three forces acting on it at different locations and angles. Force F3 is applied at the center and is perpendicular to the axis of rotation. Force F2 is applied at the left edge and is perpendicular to the axis of rotation. Force F1 is applied at the center and forms 30 degree angle with the axis of rotation. — Three forces acting on a flywheel.

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

{\vec{F}}_{1}

. If we look at [link] , we see that

{\vec{F}}_{1}

makes an angle of

90 ° + 60 °

with the radius vector

\vec{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:

| {\vec{τ}}_{1} | = r F_{1} sin 150 ° = 0.5 m (20 N) (0.5) = 5.0 N \cdot m .

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

| {\vec{τ}}_{2} | = − r F_{2} sin 90 ° = −0.5 m (30 N) = −15.0 N \cdot m .

When we evaluate the torque due to ${\vec{F}}_{3}$ , we see that the angle it makes with $\vec{r}$ is zero so $\vec{r} \times {\vec{F}}_{3} = 0 .$ Therefore, ${\vec{F}}_{3}$ does not produce any torque on the flywheel.

We evaluate the sum of the torques:

τ_{net} = \sum_{i} | τ_{i} | = 5 - 15 = −10 N \cdot 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,

{\vec{F}}_{3}

would cause the flywheel to translate, as well as

{\vec{F}}_{1}

. Its motion would be a combination of translation and rotation.

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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 $5.0 \times 10^{5} N$ 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] )?

Figure shows a ship that lies at an angle on the seashore. A force of 50000 N is applied at 10 degree angle to the normal at a point that 100 meters above the point of contact between the ship and the seashore. — A ship runs aground and tilts, requiring torque to be applied to return the vessel to an upright position.

The angle between the lever arm and the force vector is $80 °;$ therefore, $r_{⊥} = 100 m(sin80 °) = 98.5 m$ .

The cross product $\vec{τ} = \vec{r} \times \vec{F}$ gives a negative or clockwise torque.

The torque is then $τ = − r_{⊥} F = −98.5 m (5.0 \times 10^{5} N) = −4.9 \times 10^{7} N \cdot m$ .

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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 $| \vec{τ} | = r_{⊥} F$ , where $r_{⊥}$ 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 $\vec{r}$ and $\vec{F}$ , then $\vec{r} \times \vec{F}$ 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

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

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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.

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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