# Torque and rotational inertia

##### Intros

##### Examples

###### Lessons

- A person exerts a force of 45N on the handle of a door 65cm wide.
- Calculate the net torque about the axle of the of the wheel shown below. Assume that a friction torque of 0.60 $N.m$ opposes the motion.
- A torque of 66 $N.m$ is required to tighten the bolts on the cylinder head of a car engine. The wrench used to tighten the bolt is 26cm long.
- Two small weights of mass 4.0kg and 6.0kg are mounted 6.0m apart on a light rod as shown below. Calculate the moment of inertia of the system;
- A 12.0N force is applied to a cord wrapped around a pulley of mass $M$ = 6.00kg and radius $R$= 22.0cm. The pulley accelerates uniformly from rest to an angular speed of 40.0 rad/s in 2.00s. If there is a frictional torques $\tau _{fr}$ = 1.20$m.N$ at the axle, determine the moment of inertia of the pulley as it rotates around its center.
- Determine the moment of inertia for the following uniform objects:
- An Atwood's machine consists of two masses, $m_{1}$ = 4.0kg and $m_{2}$ = 7.0kg, which are connected by a massless inelastic cord that passes over a pulley. The pulley has radius $R$ = 0.4m and moment of inertia $I$ =4.0 kg. $m_{2}$
about its axle.

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###### Topic Notes

In this lesson, we will learn:

- Definition of Torque
- Translational equilibrium Vs. Rotational equilibrium
- Rotational Inertia
- Moment of inertia of uniform objects

__Notes:__- To have better understanding of the meaning of “torque”, let’s try the following activity and compare the motion of object as different forces are exerted.
- A ruler is placed on a horizontal flat surface.

**(a)**The force is applied at the center of mass

$\quad$The whole object (ruler) will accelerate in the direction of the force exerted.

**(b)**The force is applied away from the center of mass

As the result of exerting force, the object will rotate about the “Pivot” or “Axis of Rotation”.

Therefore, we can define Torque as the “Turning Effect of a Force”.

Torque is represented by the Greek letter $\tau$ and the standard unit is $N.m.$

Conclusion:

*Force*causes acceleration; $F=ma$*Torque*causes angular acceleration: $\tau = r F \sin \theta \;$ (vector quantity)

$\theta$ : angle between $r$ and $F$

$r$: distance between the pivot and point at which the force is exerted on.

$\tau \, \propto \, r$: the further the force, the bigger the torque

$\tau \, \propto \, \theta$: the larger the angle, the bigger the torque

The object could be in Translational Equilibrium and Rotational Equilibrium;

$\sum F = 0 \qquad \qquad$ “Translational Equilibrium” $\; \Rightarrow \,$ acceleration is zero

$\sum \tau = 0 \qquad \qquad$ “Rotational Equilibrium” $\; \Rightarrow \,$ angular acceleration is zero

**Rotational Inertia**

Let’s consider a mass m rotating in a circle of radius r about a fixed point. The object is going to experience tangential and angular acceleration.

^{2}

*Large Rotational Inertia:*the mass of the object is distributed far from the axis of rotation.

Objects with larger Rotational Inertia are harder to get rotating and harder to stop rotating.-
*Small Rotational Inertia:*the mass of the object is distributed close to the axis of rotation.

$I = mr^{2} \qquad$ (single mass rotating of a single radius)

$I = \sum mr^{2} \enspace \;$ (multiple individual masses rotating in circles of different radius)

- Moment of inertia for objects of uniform composition is constant;

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