fisica1-youn-e-freedman-exercicios-resolvidos

10.69: a) The distance the marble has fallen is y = h − ( 2R
− r)
= h + r − 2R.
The
radius of the path of the center of mass of the marble is R − r,
so the condition that the
2
ball stay on the track is v = g(
R − r).
The speed is determined from the work-energy
2
2
theorem, mgy = (1 2) mv + (1 2) Iω . At this point, it is crucial to know that even for the
curved track, ω = v r;
this may be seen by considering the time T to move around the
circle of radius R − r at constant speed V is obtained from 2π ( R − r)
= Vt,
during which
R
time the marble rotates by an angle 2π ( − 1) = ω T,
from which ω = V r.
The work-
2
energy theorem then states mgy = (7 10) mv , and combining, canceling the factors of m
and g leads to ( 7 10)( R − r)
= h + r − 2R,
and solving for h gives
2
h = ( 27 10) R − (17 10) r.
b) In the absence of friction, mgy = (1 2) mv , and substitution
2
of the expressions for y and v in terms of the other parameters gives
( 1 2)( R − r)
= h − r − 2R,
which is solved for h = ( 5 2) R − (3 2) r.
10.70: In the first case, F → and the friction force act in opposite directions, and the
friction force causes a larger torque to tend to rotate the yo-yo to the right. The net force
to the right is the difference F − f , so the net force is to the right while the net torque
causes a clockwise rotation. For the second case, both the torque and the friction force
tend to turn the yo-yo clockwise, and the yo-yo moves to the right. In the third case,
friction tends to move the yo-yo to the right, and since the applied force is vertical, the
yo-yo moves to the right.
10.71: a) Because there is no vertical motion, the tension is just the weight of the hoop:
T = Mg = 0 .180 kg 9.8 N kg = 1.76 b) Use τ = Iα to find α.
The torque is
( )( ) N
2
RT , so α = RT / I = RT MR = T / MR = Mg MR,
2
2
( 9.8 m s ) ( 0.08 m) 122.5 rad/s
so α = g R =
=
2
c) a = Rα
= 9.8 m s
d) T would be unchanged because the mass M is the same, α and a would be twice as
great because I is now 1 2
MR .
2