Download 1. Consider the free-body diagram for a person in the “Rotor

r
1. Consider the free-body diagram for a person in the “Rotor-ride”. FN is the normal force of
r
contact between the rider and the wall, and Ffr is the static frictional force between the back
of the rider and the wall. Write Newton’s 2nd law for the vertical forces, noting that there is
no vertical acceleration.
∑F
y
= Ffr − mg = 0 → Ffr = mg
If we assume that the static friction force is a maximum, then
Ffr = µ s FN = mg → FN = mg µ s .
But the normal force must be the force causing the centripetal
motion – it is the only force pointing to the center of rotation. Thus FR = FN = mv 2 r .
Using v = 2πr T , we have FN =
4π 2 mr
T2
.
Equate the two expressions for the normal force and solve for the coefficient of friction.
Note that since there are 0.5 rev per sec, the period is 2.0 sec.
FN =
4π 2 mr mg
=
µs
T2
→
µs =
(
)
9.8 m s 2 (2 s )
gT 2
=
= 0.22 .
4π 2 r
4π 2 (4.6 m )
2
Any larger value of the coefficient of friction would mean that the normal force could be
smaller to achieve the same frictional force, and so the period could be longer or the
cylinder smaller. There is no force pushing outward on the riders. Rather, the wall pushes
against the riders, so by Newton’s 3rd law the riders push against the wall. This gives the
sensation of being pressed into the wall.
2. Each force is oriented so that it is perpendicular to its lever arm. Call counterclockwise
torques positive. The torque due to the three applied forces is given by
τ applied = ( 28 N )( 0.24 m ) − (18 N )( 0.24 m ) − ( 35 N )( 0.12 m ) = − 1.8 m N .
forces
Since this torque is clockwise, we assume the wheel is rotating clockwise, and so the
frictional torque is counterclockwise. Thus the net torque is
τ net = ( 28 N )( 0.24 m ) − (18 N )( 0.24 m ) − ( 35 N )( 0.12 m ) + 0.40 m N = −1.4 m N
= 1.4 m N , clockwise
3.
(a) The angular acceleration can be found from
α=
∆ω ω v r (10.0 m s ) (0.31 m )
= =
=
= 92.17 rad s 2 ≈ 92 rad s 2
∆t
t
t
0.350 s
(b) The force required can be found from the torque, since τ = Fr sin θ . In this situation the
force is perpendicular to the lever arm, and so θ = 90º . The torque is also given by τ = Iα ,
where I is the moment of inertia of the arm-ball combination. Equate the two expressions
for the torque, and solve for the force.
Fr sin θ = I α
F=
=
4.
Iα
m d 2 + 1 m L2
= ball ball 3 arm arm α
r sin θ
r sin 90º
(1.00 kg )(0.31 m )2 + 13 (3.70 kg )(0.31 m )2 92.17
(
(0.025 m )
)
rad s 2 = 7.9 × 10 2 N
The lines of action of the gravitational and normal forces
both pass through the center of the ball (center of mass),
and thus exert no torque about its center. The frictional
force exerted by the table is much smaller than the
collision force of the cue stick, so its effects during the
collision can be ignored.
The distance between the line of action of the force F
and the axis of rotation gives the lever arm. Since is
tangential (see diagram of forces shown), this distance is
given by d.
The torque about the horizontal axis through the center of the ball is written as:
Apply Newton’s 2nd law for a system in linear motion and Newton’s 2nd law for rotational
motion about the center of the ball:
;
Substitute into :
The nonslip condition relates and α:
Substitute into the previous equation:
Using fig. 8-21 from the textbook, we have :
2 5