Question

5. The entrance of a science museum features a funnel into which marbles are rolled one at a time. The marbles circle around the wall of the funnel, eventually spiraling down into the neck of the funnel. The internal radius of the funnel at the top is 0.54 m. At the bottom, the funnel's neck nar- rows to an internal radius of 0.040 m. A 2.5 x 10-2 kg marble begins rolling in a large circular orbit around the funnel's rim at 0.35 rev/s. If it continues moving in a roughly circular path, what will the marble's angular speed be as it passes throught the neck of the funnel? (Consider only the effects of the conservation of angular momentum.)​

Answer 1

Answer:

400.7886829 rad/s

Explanation:

First we have to turn our 0.35 rev/s into rad/s using the equation

(Xrev/s)*2pi=Xrad/s we can plug in .35*2pi=.7pi rad/s

Now we can us the equation m_1*v_1*r_1^2=m_1*v_2*r_2^2 we can plug in the given. Because the mass remains the same we can cross it off of both sides giving us just: v_1*r_1^2=v_2*r_2^2

(.7pi)*(.54)^2=(v_2)*(.04)^2

(.20412pi)=(v_2)*(.0016)     [.20412pi=.6412618925]

then using division on both sides we get

(.6412618925/.0016)=v_2=400.79rad/s(This answer is rounded to the nearest hundreth)

See you in Mr.K's class tomorrow! -Ruben