Imagine two disks on a shared (frictionless) axle. In the initial state the lowe

Question

Imagine two disks on a shared (frictionless) axle. In the initial state the lower disk, of mass 430 g and radius 3.3 cm , is rotating at 180 rpm, while the upper disk, of mass 260 g and radius 2.5 cm , is held at rest. Now, the upper disk is pushed down onto the lower disk, and they rub against each other until they come to a common final rotational speed. Assume that no external torques act on the system (the only torques involved are due to the friction between the disks while the upper disk is speeding up and the lower disk is slowing down, before they become stuck together and rotate as one unit).

Explanation / Answer

given

m1 = 430 g = 0.43 kg

r1 = 3.3 cm = 0.033 m

I1 = 0.5*m1*r1^2 = 0.5*0.43*0.033^2 = 2.34*10^-4 kg.m^2

m2 = 260 g = 0.26 kg

r2 = 2.5 cm = 0.025 m

I2 = 0.5*m1*r1^2 = 0.5*0.26*0.025^2 = 0.8125*10^-4 kg.m^2

Ii = I1 = 2.34*10^-4 kg.m^2

wi = 180 rpm

If = (I1 + I2) = (2.34 + 0.8124)*10^-4 = 3.1524*10^-4 kg.m^2

wf = ?

Apply conservation of angular momentum

If*Wf = Ii*wi

==> wf = Ii*wi/If

= 2.34*180/3.1524

= 133.6 rpm <<<<<<<<<<<-----------------Answer

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

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