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August 28, 2014

August 28, 2014

Posted by **COFFEE** on Tuesday, March 27, 2007 at 6:18pm.

(a) From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? (The radius of the loop-the-loop is R = 2.16 m. Assume R>>r, and the mass of the ball is 3.2 kg.)

(b) If the brass ball is released from height 6R above the bottom of the track, what is the magnitude of the horizontal component of the force acting on it at point Q?

*My work so far*

for finding I for a sphere I believe that I need to use I = (2/5)MR^2 which is I = 5.97 kg*m^2. Can I use V = sqrt(gR) to find the V...if so, V = 4.6 m/s. I am not sure what to do now. Can I use -Fs = m(-V^2/R)???

For Further Reading

* Physics - KE/rotation - drwls, Monday, March 26, 2007 at 5:06am

To stay on track at the top of the loop, MV^2/R must equal or exceed Mg there. Use conservation of enewrgy to relate the velocity there to the distance of the release point above the bottom of the loop.

The total kinetic energy when the velocity is V is

KE = (1/2) M V^2 + (1/2)(2/5)Mr^2*(V/r)^2 = (3/5) M V^2

If H is the initial height of the ball above the bottom of the loop, then at the top of the loop, 2R from the bottom,

MgH = (3/5) M V^2 + 2 MgR

V^2 = (5/3) gH -(10/3)gR > g

Solve for the minimum required H

My work:

V^2 = (5/3)*gH - (10/3)gR...I am not sure what he meant by > g in the equation above???

using V^2 = (5/3)*gH - (10/3)gR I found H to be 5.6155 meters which is the wrong answer. Any thoughts??? Thanks.

Mv^2/R >= Mg to stay on the top of the loop.

But mv^2 = 5/3 MgH -10/3 gR from

MgH = (3/5) M V^2 + 2 MgR

Therefore,

5/3 MgH -10/3 MgR >= MgR

divide thru by Mg

5/3 H >= R+10R/3

or H >= 5.616M or to three places 5.62m

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