6.2 Torque and Work

Learning objectives

6.2A—Describe the work done on a rigid system by a given torque or collection of torquesDescribe the work done on a rigid system by a given torque or collection of torques.• A torque can transfer energy into or out of an object or rigid system if the torque is exerted over an angular displacement.• The amount of work done on a rigid system by a torque is related to the magnitude of that torque and the angular displacement through which the rigid system rotates during the interval in which that torque is exerted. Relevant equation:• Work done on a rigid system by a given torque can be found from the area under the curve of a graph of torque as a function of angular position. TOPIC 6.3 Angular Momentum and Angular Impulse

Connect torque to energy transfer

A torque transfers energy into or out of a rigid system only while the system undergoes an angular displacement. Positive rotational work adds energy; negative rotational work removes energy.

Use the constant-torque relationship

$W=\tau\,\Delta\theta$

Quantity or condition	Meaning
$\tau$	Constant signed torque during the interval; use $\tau_{\mathrm{net}}$ for net work
$\Delta\theta$	Signed angular displacement in radians
Same rotational sense	$W>0$ : energy transferred into the system
Opposite rotational senses	$W<0$ : energy transferred out

Calculate constant-torque work

Constant torque: a net torque of $+4.0\,\text{N}\,\text{m}$ acts while a wheel turns through $+3.0\,\text{rad}$ .

$W=\tau_{\mathrm{net}}\Delta\theta=(+4.0)(+3.0)=+12\,\text{J}$ .

The positive sign means $12\,\text{J}$ of energy is transferred into the wheel.

Read work as signed graph area

Torque–angular-position graph feature	Rotational meaning
Horizontal axis	Angular position $\theta$ in radians
Vertical axis	Signed torque $\tau$ in $\text{N}\,\text{m}$
Area above the axis	Positive work
Area below the axis	Negative work
Total signed area	Net work over the interval

Calculate varying-torque work

Varying torque: suppose torque increases linearly from $0$ to $6.0\,\text{N}\,\text{m}$ while angular position changes by $4.0\,\text{rad}$ . The graph region is a triangle, so

$W=\tfrac12(4.0\,\text{rad})(6.0\,\text{N}\,\text{m})=12\,\text{J}$ .

Choose product or area correctly

Do not multiply one instantaneous torque value by the entire angular displacement when torque varies. Use signed graph area instead. In $W=\tau\Delta\theta$ , the angle is in radians; degrees must be converted first.

Next Topic

—6.3 Angular Momentum and Angular ImpulseRead