Since torque is a rotational force, this distance is also a radius. For this reason, you’ll sometimes see it represented with an “r” in the basic torque equation.
In torque problems, you’ll typically be given the magnitude force. However, if you have to work it out yourself, you’ll need to know the mass of the object and the acceleration of the object in m/s2. According to Newton’s Second Law, force is equal to mass times acceleration (F=m×a{\displaystyle F=m\times a}).
For example, suppose you have a force perpendicular to your object exerting 20 Newtons of force on the object 10 meters from the axis. The magnitude of the torque is 200 N∙m: τ=20×10=200{\displaystyle \tau =20\times 10=200}
For example, if the object is moving clockwise and the magnitude of the torque is 200 N∙m, you would express this as -200 N∙m of torque. No sign is necessary if the magnitude of the torque is positive. The value given for the magnitude of the torque remains the same. If a negative sign appears before the value, it simply means that the object in question is rotating clockwise.
For example, suppose you’re told that the net torque is zero. The magnitude of the torque on one side of the axis is 200 N∙m. On the other side of the axis, force is being exerted from the axis in the opposite direction 5 meters from the axis. Since you know that net torque is 0, you know that the 2 forces must add up to 0, so you can construct your equation to find the missing force:200+(F×5)=0{\displaystyle 200+(F\times 5)=0}F×5=−200{\displaystyle F\times 5=-200}F=−2005{\displaystyle F=-{\frac {200}{5}}}F=−40{\displaystyle F=-40}
For most physics problems, this distance is measured in meters. In the torque equation, this distance is represented by “r” for radius or radial vector.
If you’re not provided with the amount of force, you would multiply mass times acceleration to find the force, which means you would need to be given those values. You might also be given the torque and told to solve for the force. In the torque equation, force is represented by “F. "
In the torque equation, this angle is represented by the Greek letter theta, “θ. " You’ll typically see it referred to as “angle θ” or “angle theta. "
If you were determining the sine of the angle by hand, you would need the measurements for the opposite side and the hypotenuse side of a right triangle. Since most torque problems don’t involve making exact measurements, however, you shouldn’t have to worry about this.
For example, suppose you have a radial vector 10 meters long. You’re told that 20 Newtons of force is being applied to that radial vector at a 70° angle. You would find that the torque is 188 N∙m: τ=10×20×sin70∘=10×20×0. 94=188{\displaystyle \tau =10\times 20\times sin70^{\circ }=10\times 20\times 0. 94=188}
For example, suppose you’re trying to figure out the magnitude of torque on a solid disc. The moment of inertia for a solid disc is 12MR2{\displaystyle {\frac {1}{2}}MR^{2}}. The “M” in this equation stands for the mass of the disc, while the “R” stands for the radius. If you know that the mass of the disc is 5 kg and the radius 2 meters, you can determine that the moment of inertia is 10 kg∙m2: 12(5×22)=12(5×4)=12(20)=10{\displaystyle {\frac {1}{2}}(5\times 2^{2})={\frac {1}{2}}(5\times 4)={\frac {1}{2}}(20)=10}
Remember that the angular acceleration can be zero if the object is moving at a constant speed and is neither speeding up nor slowing down.
For example, suppose you know that the moment of inertia for an object is 10 kg∙m2. You’re also told that the torque is 20 N∙m, but you need to find out the angular acceleration. Since you know that τ=Iα{\displaystyle \tau =\mathrm {I} \alpha }, you also know that α=τI{\displaystyle \alpha ={\frac {\tau }{\mathrm {I} }}}. When you put in the variables you know, you’ll find that the angular acceleration for the object is 2 radians/s2: α=2010=2{\displaystyle \alpha ={\frac {20}{10}}=2}
