fisica1-youn-e-freedman-exercicios-resolvidos

10.23:
n = mg cos α
mg sin θ − µ mg cos θ = ma
g(sin
θ − µ
s
cos θ)
= a (eq.1)
n and mg act at the center of the ball and provide no torque.
∑ τ = τ
f
= µ mg cosθR;
I =
s
∑ τ = Iα gives µ mg cos θ =
s
2
5
mR
2
2
mR α
2
5
2
, so µ
sg
cosθ
5
No slipping means α = a R
=
s
a
(eq.2)
We have two equations in the two unknowns a and
s.
= 5 g sinθ
and µ = tanθ
=
7 2 tan65.0°
= 0.613
a
7
s
7 2
2
b) Repeat the calculation of part (a), but now I = 2
mR .
2
a = g sinθ
and µ
2
s
= tanθ
= tan65.0°
5
5
5
3 =
The value of
hollow ball.
0.858
µ Solving gives
µ
s
calculated in part (a) is not large enough to prevent slipping for the
c) There is no slipping at the point of contact.
3