Posted by **COFFEE** on Tuesday, March 27, 2007 at 8:41pm.

In Figure 11-32 (which shows a ball at the top of an incline, at the bottom of the incline a loop begins with radius R and Q a point on the loop lined up with the center of the loop), a solid brass ball of mass m and radius r will roll without slipping along the loop-the-loop track when released from rest along the straight section.

(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.

For Further Reading

* Physics - KE - bobpursley, Tuesday, March 27, 2007 at 8:21pm

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

*******************

I am thinking that maybe I solved for the velocity incorrectly. Do I use V=sqrt(gR) or do I have to take into account the center of mass and do Vcom and do an additional step under the sqrt???

## Answer This Question

## Related Questions

- Physics - KE/rotation - In Figure 11-32 (which shows a ball at the top of an ...
- Physics - KE - In Figure 11-32 (which shows a ball at the top of an incline, at...
- College Physics - A student of mass M = 82 kg takes a ride in a frictionless ...
- Physics (Rolling) - In Fig. 11-35, a solid brass ball of mass m will roll ...
- Physics - A solid brass ball of mass 9.5 g will roll smoothly along a loop-the-...
- Physics - drwls? - A small block of mass m = 4.0 kg can slide along the ...
- Physics - A pilot, whose mass is 70.0 kg, makes a loop-the-loop in a fast jet. ...
- Physics - A pilot, whose mass is 91.0 kg, makes a loop-the-loop in a fast jet. ...
- physics - A mass m = 76.0 kg slides on a frictionless track that has a drop, ...
- physics - A small block of mass m can slide along the frictionless loop-the loop...

More Related Questions