A small block of mass m=0100kg can slide along the frictionless loop-the-loop, w
ID: 2260851 • Letter: A
Question
A small block of mass m=0100kg can slide along the frictionless loop-the-loop, with loop radius R= 0.150m , as shown in the figure. The block is given an initial velocity along the track of Vi=2.00 m/s at the point P, at height h=2.00R above the bottom of the loop.
(a)Determine the normal force at point Q, and free body diagram
(b)determine the normal force at the top of the loop.
(c)S is a point higher than P. The ball is released from rest at S, and still makes it around the track without falling off the loop at the top.Determine the minimum height h, above the base of the loop, of point S.
A small block of mass m=0100kg can slide along the frictionless loop-the-loop, with loop radius R= 0.150m , as shown in the figure. The block is given an initial velocity along the track of Vi=2.00 m/s at the point P, at height h=2.00R above the bottom of the loop.Explanation / Answer
a) free body diaggram at Q
there are 3 forces
1) weight acting the downward direction
2) centrifugal force=mv^2/R actingg in the rightward direction
3) Normal force (N) acting in the leftward direction
now applying energy conservation between point P and point Q
m*g*2R +0.5*m*2^2=mgR +0.5*m*v^2
==> 0.5*v^2=9.8*0.15 + 0.5*4=3.47
==> v=2.634 m/s
now
applying the newtons laws
N=mv^2/R=0.1*2.634^2/0.15= 4.6253 N
b) free body diaggram at top
there are 3 forces
1) weight acting the downward direction
2) centrifugal force=mv^2/R actingg in the upward direction
3) Normal force (N) acting in the downward direction
now applying energy conservation between point P and Top
m*g*2R +0.5*m*2^2=mg*2R +0.5*m*v^2
==> v=2 m/s
now
applying the newtons laws
N=mv^2/R-mg=0.100*2^2/0.15-0.100*9.8= 1.68667 N
c) for not falling
N=0
==> mg=mv^2/R at the top
==> v=sqrt(Rg)=sqrt(0.15*9.8)=1.212 m/s
now applying energy conservation between point P and Top
m*g*h=mg*2R +0.5*m*1.212^2
==> H=2R+0.0756
==> H=0.3+0.0756
==> H=0.3756