p08_020.pdf

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Chapter 8 - 8.20
20. (a) At Q the block (which is in circular motion at that point) experiences a centripetal acceleration
v 2 /R leftward. We find v 2
from energy conservation:
K P + U P = K Q + U Q
0+ mgh =
1
2 mv 2 + mgR
Using the fact (mentioned in problem 6) that h =5 R , we find mv 2 =8 mgR . Thus, the horizontal
component of the net force on the block at Q is mv 2 /R =8 mg and points left (in the same direction
as a ).
(b) The downward component of the net force on the block at Q is the force of gravity mg downward.
(c) To barely make the top of the loop, the centripetal force there must equal the force of gravity:
mv t
R
= mg =
⇒
mv t = mgR
This requires a different value of h than was used above.
K P + U P = K t + U t
0+ mgh =
1
2 mv t + mgh t
1
2 ( mgR )+ mg (2 R )
mgh =
Consequently, h =2 . 5 R .
(d) The normal force N , for speeds v t greater than √ gR (which are the only possibilities for non-zero
N – see the solution in the previous part), obeys
N = mv t
R −
mg
from Newton’s second law. Since v t is related to h by energy conservation
K P + U P = K t + U t =
⇒
gh =
1
2 v t
+2 gR
then the normal force, as a function for h (solongas h
≥
2 . 5 R – see solution in previous part),
becomes
2 mg
R
N =
h
−
5 mg .
2 . 5 R consists of a straight line of positive slope 2 mg/R (which can be set
to some convenient values for graphing purposes). For h
≥
≤
2 . 5 R , the normal force is zero. In the
interest of saving space, we do not show the graph here.
Thus, the graph for h
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