Question

An electric motor exerts a constant torque of τ=10N⋅m on a grindstone mounted on its shaft; the moment of inertia of the grindstone is I=2.0kg⋅m2. 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? Part A:Suppose we now turn off the motor and apply a brake to bring the grindstone from its greatest angular speed of 36 rad/s to a stop. If the brake generates a constant torque of 13 N⋅m , find the total work done (including signs) needed to stop the grindstone. Part A:Suppose we now turn off the motor and apply a brake to bring the grindstone from its greatest angular speed of 36 rad/s to a stop. If the brake generates a constant torque of 13 N⋅m , find the total work done (including signs) needed to stop the grindstone. Part B:Find the average power (including signs) needed to stop the grindstone in Part A.

Answer 1

= constant torque applied by motor = 10 Nm

I = moment of inertia of grindstone = 2 kgm²

angular acceleration of the grindstone is given as

= /I

= 10/2 = 5 rad/s²

w_o = initial angular velocity = 0 rad/s

w_f = final angular velocity = ?

t = time = 8 sec

final angular velocity is given as

w_f = w_o + t

w_f = 0 + (5) (8)

w_f = 40 rad/s

work done is given as the change in rotational kinetic energy

W = (0.5) I (w_f² - w_o²)

W = (0.5) (2) ((40)² - (0)²)

W = 1600 J

average power delivered is given as

P = W/t

P = 1600/8

P = 200 Watt

a)

w_o = initial angular velocity = 36 rad/s

w_f = final angular velocity = 0 rad/s

work done is the change in rotational kinetic energy , hence

W = (0.5) I (w_f² - w_o²)

W = (0.5) (2) ((0)² - (36)²)

W = - 1296 J

b)

= constant torque applied by brakes = - 13 Nm

I = moment of inertia of grindstone = 2 kgm²

angular acceleration of the grindstone is given as

= /I

= - 13/2 = - 6.5 rad/s²

w_o = initial angular velocity = 36 rad/s

w_f = final angular velocity = 0 rad/s

t = time = ?

final angular velocity is given as

w_f = w_o + t

0 = 36 + (- 6.5) t

t = 5.5 sec

average power is given as

P = W/t

P = - 1296/5.5

P = - 235.6 Watt