fisica1-youn-e-freedman-exercicios-resolvidos

2 2θ
2θ
2θ
I
10.57: t = = = .
α ( τ I)
τ
The angle in radiants is π 2,
the moment of inertia is
2
3
( 1 3)((750 N) (9.80 m s )(1.25 m)) = 39.9 kg ⋅
and the torque is ( 220 N)(1.25 m) = 275 N ⋅ m. Using these in the above expression gives
t 0.455 s
2 , so t = 0.675 s.
2 =
m
2
10.58: a) From geometric consideration, the lever arm and the sine of the angle
r
between F and are both maximum if the string is attached at the end of the rod. b) In
terms of the distance x where the string is attached, the magnitude of the torque is
Fxh
2 2
x + h . This function attains its maximum at the boundary, where x = h,
so the
string should be attached at the right end of the rod. c) As a function of x, l and h, the
torque has magnitude
τ = F
xh
.
2 2
( x − l 2) + h
This form shows that there are two aspects to increasing the torque; maximizing the lever
arm l and maximizing sin φ . Differentiating τ with respect to x and setting equal to zero
2
gives x
max
= ( l 2)(1 + (2 h l)
). This will be the point at which to attach the string unless
2 h > l, in which case the string should be attached at the furthest point to the right, x = l.
10.59: a) A distance L 4 from the end with the clay.
2
b) In this case I = ( 4 3) ML and the gravitational torque is
( 3L 4)(2Mg) sinθ
= (3Mg
L 2) sinθ
, so α = ( 9 g 8L) sin θ .
2
c) In this case I = ( 1 3) ML and the gravitational torque is
( L 4)(2Mg) sin θ = ( Mg L 2) sin θ , soα
= (3 g 2L) sin θ . This is greater than in part (b).
d) The greater the angular acceleration of the upper end of the cue, the faster you would
have to react to overcome deviations from the vertical.