When solving for the normal force at C, why is there no tangential acceleration?

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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When solving for the normal force at C, why is there no tangential acceleration? 

Solutions
Conservation of total energy
1. The spring of constant k is unstretched when the slider of
mass m passes position B. If the slider is released from rest
in position A, determine its speed as it passes point B and
C. What is the normal force exerted by the guide on the
slider at position C? Neglect friction between the mass and
the circular guide, which lies in a vertical plane.
(Meriam 3/148)
|A
k
В
R
Solution:
C
The system is a conservative system, for position A, B and C
we have
E1 = E2= E3
T1 + Vị = T2 + V2 = T3 + V3
HA
R
T1 + Vị = 0 + mgR +
k
1
T2 + V2
mv, + 0
2
R
1
mv} – mgR
T3 + V3
-R
3
C
kR?
(ans)
→ v2 = 2gR + 0.1716-
т
N
kR?
|4gR+0.1716-
m
(ans)
→ Vz =
m
mg
At C: N- mg
= m
R
→ N= 5mg + 0.1716kR (ans)
Transcribed Image Text:Solutions Conservation of total energy 1. The spring of constant k is unstretched when the slider of mass m passes position B. If the slider is released from rest in position A, determine its speed as it passes point B and C. What is the normal force exerted by the guide on the slider at position C? Neglect friction between the mass and the circular guide, which lies in a vertical plane. (Meriam 3/148) |A k В R Solution: C The system is a conservative system, for position A, B and C we have E1 = E2= E3 T1 + Vị = T2 + V2 = T3 + V3 HA R T1 + Vị = 0 + mgR + k 1 T2 + V2 mv, + 0 2 R 1 mv} – mgR T3 + V3 -R 3 C kR? (ans) → v2 = 2gR + 0.1716- т N kR? |4gR+0.1716- m (ans) → Vz = m mg At C: N- mg = m R → N= 5mg + 0.1716kR (ans)
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