An electric motor exerts a constant torque of =10Nm on a grindstone mounted on i
ID: 1461196 • Letter: A
Question
An electric motor exerts a constant torque of =10Nm on a grindstone mounted on its shaft; the moment of inertia of the grindstone is I=2.0kgm2. If the system starts from rest, find the work done by the motor in 8.0 s and the kinetic energy at the end of this time. What was the average power delivered by the motor?
SOLUTION
SET UP AND SOLVE We want to use W= to find the total work done and then divide that by the time interval t=8.0s to find the average power (work per unit time). But first we need to find the total angle (in radians) through which the grindstone turns. The angular acceleration is constant, so we can use =I to find and then apply the formulas for constant angular acceleration.
The angular acceleration is
=I=10Nm2.0kgm2=5.0rad/s2
For constant angular acceleration,
=0+0t+12t2=0+0+12(5.0rad/s2)(8s)2=160rad
The total work is
W==(10Nm)(160rad)=1600J
Finally, the average power is the total work divided by the time interval:
Pav=Wt=1600J8.0s=200J/s=200W
REFLECT We can check the energy relationships. The angular velocity of the grindstone after 8.0 s is
=t=(5.0rad/s2)(8.0s)=40rad/s
The grindstone's kinetic energy at this time is
K=12I2=12(2.0kgm2)(40rad/s)2=1600J
This amount equals the total work done, as of course it must. Note that the power is not constant during this process. The torque is constant, but as the angular velocity increases, so does the rate of doing work. The instantaneous power, , increases linearly with time—from zero, when =0, to 400 W, with an average value of 200 W.
Part A - Practice Problem:
Suppose we now turn off the motor and apply a brake to bring the grindstone from its greatest angular speed of 35 rad/s to a stop. If the brake generates a constant torque of 12 Nm , find the total work done (including signs) needed to stop the grindstone.
Express your answer in joules as an integer.
SubmitMy AnswersGive Up
Incorrect; Try Again
Part B - Practice Problem:
Find the average power (including signs) needed to stop the grindstone in Part A.
Express your answer in joules as an integer.
An electric motor exerts a constant torque of =10Nm on a grindstone mounted on its shaft; the moment of inertia of the grindstone is I=2.0kgm2. If the system starts from rest, find the work done by the motor in 8.0 s and the kinetic energy at the end of this time. What was the average power delivered by the motor?
SOLUTION
SET UP AND SOLVE We want to use W= to find the total work done and then divide that by the time interval t=8.0s to find the average power (work per unit time). But first we need to find the total angle (in radians) through which the grindstone turns. The angular acceleration is constant, so we can use =I to find and then apply the formulas for constant angular acceleration.
The angular acceleration is
=I=10Nm2.0kgm2=5.0rad/s2
For constant angular acceleration,
=0+0t+12t2=0+0+12(5.0rad/s2)(8s)2=160rad
The total work is
W==(10Nm)(160rad)=1600J
Finally, the average power is the total work divided by the time interval:
Pav=Wt=1600J8.0s=200J/s=200W
REFLECT We can check the energy relationships. The angular velocity of the grindstone after 8.0 s is
=t=(5.0rad/s2)(8.0s)=40rad/s
The grindstone's kinetic energy at this time is
K=12I2=12(2.0kgm2)(40rad/s)2=1600J
This amount equals the total work done, as of course it must. Note that the power is not constant during this process. The torque is constant, but as the angular velocity increases, so does the rate of doing work. The instantaneous power, , increases linearly with time—from zero, when =0, to 400 W, with an average value of 200 W.
Part A - Practice Problem:
Suppose we now turn off the motor and apply a brake to bring the grindstone from its greatest angular speed of 35 rad/s to a stop. If the brake generates a constant torque of 12 Nm , find the total work done (including signs) needed to stop the grindstone.
Express your answer in joules as an integer.
JSubmitMy AnswersGive Up
Incorrect; Try Again
Part B - Practice Problem:
Find the average power (including signs) needed to stop the grindstone in Part A.
Express your answer in joules as an integer.
Explanation / Answer
Torque = T = 10 Nm
Moment of inertia = I = 2 kg m^2
time = t = 8 s
angular acceleration a = T / I = 10 / 2 = 5 rad/s^2
Angle rotated in t seconds = theta = 0.5 a t^2 = 0.5 * 5 * 8^2 = 160 rad
(a) Work done = T theta = 10 * 160 = 1600 J
(b) Kinetic energy = work done = 1600 J
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w = 35 rad/s
angular acceleration = a = T / I = - 12 / 2 = - 6 rad/s^2
theta = - w^2 / 2 a = - 35^2 / (2*-6) = 102.083 rad
(a) Work done = T * theta = - 12 * 102.083 = - 1225 J
(b) Average power = Work done / ( w / a ) = - 1225 / ( 35 / -6) = 210 W
(c) Average Power = KE / t = 1600 / 8 = 200 W
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