The combination of an applied force and a friction force produces a constant tot
ID: 778070 • Letter: T
Question
The combination of an applied force and a friction force produces a constant total torque of 35.9 N·m on a wheel rotating about a fixed axis. The applied force acts for 5.80 s. During this time, the angular speed of the wheel increases from 0 to 10.5 rad/s. The applied force is then removed, and the wheel comes to rest in 59.2 s. (a) Find the moment of inertia of the wheel kg m (b) Find the magnitude of the torque due to friction. N-m (c) Find the total number of revolutions of the wheel during the entire interval of 65.0 s S. revolutions Need Help?Read ItExplanation / Answer
Data:
Initial angular speed, 1 = 0 rad/s
Final angular speed, 2 = 10.5 rad/s
Total torque, = 35.9 N.m
Time of acceleration, t1 = 5.80 s
Time of deceleration, t2 = 59.2 s
Solution:
(a)
2 = 1 + t1
10.5 = 0 + ( * 5.8 )
= 1.81 rad/s^2
Moment of inertia, I = /
= 35.9 / 1.81
= 19.83 kg.m^2
Ans:
Moment of inertia, I = 19.83 kg.m^2
(b)
When applied force is removed, then the torque is
due to only frictional force.
2 = 1 + ' t2
0 = 10.5 + ( ' * 59.2 )
' = - 0.177 rad/s^2
Magntitude of angular acceleration, ' = 0.177 rad/s^2
Frictional torque, ' = I '
= 19.83 * 0.177
= 3.5 N.m
Ans:
Frictional torque, ' = 3.51 N.m
(c)
During the application of force:
1 = (1/2) t1^2
= 0.5 * 1.81 * 5.8^2
= 30.44 rad
During the removal of force:
2 = t2 + (1/2) ' t2^2
= 10.5 * 59.2 + [ 0.5 * - 0.177 * 59.2^2 ]
= 311.4 rad
Total angular displacement:
= 1 + 2
= 30.44 + 311.4
= 341.88 rad
= ( 341.88 / 2 ) rev
= 54.44 rev
Ans:
= 54.44 rev