A solid disk with a radius of #1 m# and mass of #2 kg# is rotating on a frictionless surface. If #36 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #3 Hz#?

1 Answer
Dec 16, 2016

Answer:

#tau≈2Nm#

Explanation:

In linear motion, we might define power, #P#, as the rate of change of energy of a system with respect to time, #P=(dE_(sys))/(dt)#. The average power then is given by #P_(avg)=(ΔE_(sys))/(Δt)#. By the work-energy theorem, #ΔE_(sys)=W#, and therefore, this can be rewritten as #P=W/t#.

Now, because work is defined by a force applied across some distance (provided there is no angle the force is applied at which would alter the calculation, i.e. #W=FΔrcos(theta)#), this can be rewritten again as #P=(FΔr)/t#, where #Δr# is the displacement of the object. We know that change in distance divided by change in time is the average velocity, and thus #P=F*v#, that is, power is equal to the product of the applied force and velocity of the object.

Units for power are given in watts, where a watt is the equivalent of joules per second.

Linear motion and rotational motion are analogous to each other. Where in linear motion we speak in terms of forces, masses, and velocities, in rotational motion we talk about torques, moments of inertia, and angular velocities, respectively. We can rewrite our equation for power (derived above) in terms of torque and angular velocity.

#P=tauomega#

Where #omega# is the angular velocity of the motion, #tau# is the torque applied during the motion, and, of course, #P# is the power required to maintain this motion.

Using the given values of #P=36W# and #f=3Hz#, we can calculate #tau#. Firstly, we'll want to convert that frequency into an angular velocity, which can be done using #omega=2pif#.

#omega=2pi(3s^-1)#

#omega=6pi(rad)/s#

Solving our power equation for #tau#,

#tau=P/(omega)#

And, using our values for #omega# and #P#,

#tau=(36J/s)/(6pi(rad)/s)#

#tau=1.9Nm#

or #tau≈2Nm#