Physics 221

Name:

Homework 3

Spring 2013

03.1 A wheel with radius 3.00 m is turning about a horizontal axis through its center. The linear

speed of a point A on the rim is constant and equal to 7.00 m/s. What are the magnitudes and direction of As acceleration when it is a) at the lowest point of its trajectory?

b) at the highest point?

c) How much time does it take for the wheel to make one revolution?

Solution:

IDENTIFY: Uniform circular motion.

SET UP: Since the magnitude of !v is constant. vtan =d !vdt

= 0 and the resultant acceleration is perpendicular to the

velocity. For circular motion, the acceleration is directed in toward the center of the circular path and its magnitude is given by 2 / .v R

EXECUTE: (a) arad =

v2

R=

(7.00 m/s)2

3.00 m= 16.3 m/s2 , upward.

(b) The radial acceleration has the same magnitude as in part (a), but now the direction is downward. (c) SET UP: The time to make one rotation is the period T, and the speed v is the distance for one revolution divided by T.

EXECUTE: 2 RvT

= so T = 2!R

v=

2! (3.00 m)7.00 m/s

= 2.69 s

EVALUATE: The radial acceleration is constant in magnitude since v is constant and is at every point in the motion directed toward the center of the circular path. The acceleration is perpendicular to !v and is nonzero because the direction of !v changes.

Physics 221

Name:

Homework 3

Spring 2013

water rel.shorevr

boat rel.shorevr

boat rel.watervr

110 m

260 m

o45

03.2 A boat must cross a 260-m-wide river and arrive at a point 110 m upstream from where it starts (see figure). To do so, the pilot must head the boat at a 45 upstream angle. The current has speed 2.3 m/s.

a) What is the speed of the boat in still water?

b) How much time does the journey take?

Solution:

a) Call the direction of the flow of the river the x direction, the direction straight across the river the y direction, the speed of the boat relative to still water vbw, and the speed of the current (water relative to the shore) vws. Then the equations for distance traveled in the x and y directions are

!x = vxt = (vws " vbw sin 45o )t = "110m!y = vyt = (vbw cos45o )t = 260m.

Note that the y components of x and vbw are negative

since they are upstream, and that by taking the ratio the time cancels:

!x!y =

(vws " vbw sin 45o )(vbw cos45o )

="110m260m = "0.423.

Then solving for vbw

vbw =vws

sin45o ! 0.423cos45o =2.3 m/s

0.707! 0.423(0.707) = 5.64m/s.

or 5.6 m/s to two significant figures.

b) From the y equation

t = !y / (vbw cos45o ) = (260 m) / (5.64 m/s cos45) = 65 s.

Physics 221

Name:

Homework 3

Spring 2013

03.3 A boat is heading upstream when a float is dropped overboard and carried downstream.

After 1 hour, when the boat has traveled to a point 2.0 km farther upstream, it turned around and caught up with the float 8.0 km downstream from the turnaround point. If the boat is traveling with constant speed throughout the journey,

a) How fast is the river flowing? b) What is the boats speed in still water?

Solution:

IDENTIFY: Apply the relative velocity relation. SET UP: Let B/Wv be the speed of the boat relative to water and W/Gv be the speed of the water relative to the ground. EXECUTE: (a) Taking all units to be in km and h, we have three equations. We know that heading upstream B/W W/G 2v v = . We know that heading downstream for a time t, (vB/W + vW/G )t = 8. We also know that for the float vW/G (t +1) = 6. Solving these three equations for W/G B/W, 2v x v x,= = +

therefore 8 or (2 2 ) 5.x t+ = Also t = 6 / x !1, so (2+ 2x) 6x!1

"

#$

%

&' = 8 or 2x2 ! 2x !12 = 0. The positive

solution is x = vW/G = 3.0 km/h. (b) vB/W = 2 km/h + vW/G = 5.0 km/h. EVALUATE: When the boat headed upstream, its speed relative to the ground was 5.0 km/h !3.0 km/h = 2.0 km/h . When it headed downstream, its speed relative to the ground was 5.0 km/h +3.0 km/h = 8.0 km/h . The float was moving downstream at 3.0 km/s relative to the earth, so the boat was able to overtake it. As a check we can find the time, t = 8 / (vB/W + vW/G ) = 8 / (5+3) =1.0 h, so the boat spent equal times going up and down the stream. Therefore vavg = [8.0+(-2.0)]/2 = 3.0 km/h, which is the same as for the float, as it must be if they start together and end together.

Physics 221

Name:

Homework 3

Spring 2013

03.4 Three forces are applied to a ring that lies on a frictionless surface in the x-y plane as shown and has a mass of 100 kg. The angle between

!FA and

!FB is 135

o, the angle between

!FA and

!FC is 105

o, and the magnitude of FC is 240 N.

(a) If the system remains stationary what is the value of FB? (b) If the system has an acceleration of 0.50 m/s2 in the positive x direction, what is the value of FB? Solution:

a) The ring does not move, so the net force has to be zero in both the x and y directions: From the information given, the angle between

!FC and the x-axis must be 30

o , so

in the x-direction: ! FA cos45+ 240N cos30o = 0

in the y-direction: ! FB + FA sin 45+ 240N sin30o = 0

From the first eq, we have FA = (240 N cos30o ) / cos45o! FA = 294 N.

From the second eq. we get FB = FA sin 45+ 240N sin30o , or

= (294 N sin45o + 240 N sin30o ) = 328 N.

b) In this part, there is a net force in the positive x direction.

in the x-direction: ! FA cos45+ 240N cos30o =max = (100 kg)(0.5 m/s

2 ) = 50 N

in the y-direction: ! FB + FA sin 45+ 240N sin30o = 0

From the first eq, we have FA = (240 N cos30o !50 N) / cos45o! FA = 223 N.

From the second eq. we get FB = FA sin 45+ 240N sin30o , or

FB = (223 N sin45o + 240 N sin30o ) = 278 N.