A uniform solid disk has a radius of R = 0.45 m and mass of M = 4.35 kg and is f

Question

A uniform solid disk has a radius of R = 0.45 m and mass of M = 4.35 kg and is free to rotate about a fixed axis through its center of mass as shown below. It is initially at rest, and at t = 0 a constant torque is applied such that the disk starts to turn with angular acceleration a = 8*pi rad/s2.

If the magnitude of the total torque acting on the disk were suddenly halved at t = 10 sec, at what time (relative to t = 0) would the kinetic energy of the flywheel be four times as big as it was at t = 10 sec?

Explanation / Answer

= 8 * t .

So at t = 10 s, = 80 rad/s. And = I*

Since, is proportional to alpha. So when it is halved, the will also get halved to 4 rad /s2

Since, K.E. is proportional to ^2. So, K.E. will be four times when will become double.

So, we need to find time when = 2* 80 = 160 rad/s

=> 160 = 80 + _new * t

=> t = 80 / (4) = 20 .

So, relative to t = 0, total time at which K.E. is four time will be t = 20 + 10 = 30 second

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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