Fix my mistakes The combination of an applied force and a constant frictional fo

Question

Fix my mistakes

The combination of an applied force and a constant frictional force produces a constant total torque of 36.4 N·m on a wheel rotating about a fixed axis. The applied force acts for 6.08 s. During this time the angular speed of the wheel increases from 0 to 9.6 rad/s. The applied force is then removed, and the wheel comes to rest in 60.8 s.

(a) Find the moment of inertia of the wheel.


(b) Find the magnitude of the frictional torque.
3.978 J is what I got.
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.

(c) Find the total number of revolutions of the wheel.
52.94 rad is what I got
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.

Explanation / Answer

a)
Angular acceleration = (2- 1)/t = 9.6/ 6.08 = 1.5789 rad/s^2

I = I *1.5789 = 36.4

I = 23.05 kg m/s^2
===========================
b)
The wheel comes to rest only due to the frictional torque (friction)

(friction) = I ' = 23.05 '
' = - 9.6/60.8 = - 0.15789 minus to show that the speed reduces.
(friction) = - 23.05*0.15789 = 3.639 N.m or 3.63 Nm
==============================
c)
in time 6.08s
= 0.5 t^2 = 0.5*1.5789*6.08^2 = 29.18 radians.

Or from average angular velocity *time = (9.6/2)*6.08 =29.18 radians

In time 60.8 s
(9.6/2)*60.8 = 291.84 radians

Total angle traversed
291.84 +29.18 = 321.02 radians or 321 rad.
321 / (2) revolutions = 51.08 revolutions

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.