Problem 10.70 A thin-walled, hollow spherical shell of mass m and radius r start

Question

Problem 10.70 A thin-walled, hollow spherical shell of mass m and radius r starts from rest and rolls without slipping down the track shown in the figure (Figure 1) . Points A and B are on a circular part of the track having radius R. The diameter of the shell is very small compared to h0 and R, and rolling friction is negligible. Part A What is the minimum height ho for which this shell will make a complete loop-the-loop on the circular part of the track? Express your answer in terms of the variables m, R, and appropriate constants. h0= Part B How hard does the track push on the shell at point B, which is at the same level as the center of the circle? Express your answer in terms of the variables m, R, and appropriate constants. N= Part C Suppose that the track had no friction and the shell was released from the same height h0 you found in part (a). Would it make a complete loop-the-loop? yes no Part D In part (C), how hard does the track push on the shell at point A, the top of the circle? Express your answer in terms of the variables m, R, and appropriate constants. |N| My Answers Give Up Part E How hard did the track push on the shell at point A in part (a)? Express your answer in terms of the variables m, R, and appropriate constants. |N|=

Explanation / Answer

Rolls without slipping ...that means we have translational and rotational kinetic energies Ket and Ker respectively.
Maximum potential energy on the top equal to total energy of the system

Pe(max)=E (max)
Pe(max)=Ket + Ker + Pe(h)
h- in this case any height above the lowest part of the track.
Then lets use H as the maximum height we have

Pe(max)=mgH
Pe= mgh
Ket= 0.5 mV^2
Ker=0.5 I w^2 ;
w=V/R and
I(hollow sphere)= (2/3)mR^2 we have
Ker=0.5 I w^2
Ker=0.5 (2/3)mR^2 (V/R)^2
Ker=(1/3)mV^2
This was just a warm up so lets do 1)

1) To make a complete loop-the-loop on the circular part of the track
mg=m ac or
ac=g or ac > g (the centripetal acceleration must equal or greater than g)
ac= V^2/R then
g<=V^2/R and
V^2>=gR now we are ready to play with energy

Pe(max) = Pe(h) + Ket + Ker

mgH=mgh + 0.5mV^2 + (1/3)mV^2
since h= 2R we dividing by m have
gH = 2gR + 0.5V^2 + (1/3)V^2
substituting V^2 =gR
gH = 2gR + 0.5 gR + (1/3)gR
H= 2R + 0.5 R + (1/3)R
H= (17/6)R

2) F= m V^2/R here h=R so it is not the same V as above.

N=11 (mg)/5

3) Same as above but Ker=0

Yes

4) since we assumed g=ac the total force, N= 2(mg)/3

5) N=0

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