Physics 11 HW #5 Solutions Chapter 4:

Problems: 65, 78, 81

Chapter 5:

Focus On Concepts: 3, 7, 10, 15 Problems: 8, 11, 23, 24, 41, 45

Focus On Concepts 5-3 (b) The magnitude of the centripetal acceleration is equal to v2/r, where v is the speed of the object and r is the radius of the circular path. Since the radius of the track is smaller at A compared to B, the centripetal acceleration of the car at A has a greater magnitude. Focus On Concepts 5-7 (d) The radius of path 1 is twice that of path 2. The tension in the cord is the centripetal force. Since the centripetal force is inversely proportional to the radius r of the path, T1 must be one-half of T2.

Focus On Concepts 5-10 2 (a) The magnitude of the centripetal force is given by Fc = mv /r. The two cars have the same speed v and the radius r of the turn is the same. The cars also have the same mass m, even though they have different weights due to the different accelerations due to gravity. Therefore, the centripetal accelerations are the same. Focus On Concepts 5-15 (b) According to Newton’s second law, the net force, FN − mg , must equal the mass m times the centripetal acceleration v2/r.

Problem 4-65 REASONING The toboggan has a constant velocity, so it has no acceleration and is in equilibrium. Therefore, the forces acting on the toboggan must balance, that is, the net force acting on the toboggan must be zero. There are three forces present, the kinetic frictional force, the normal force from the inclined surface, and the weight mg of the toboggan. Using Newton’s second law with the acceleration equal to zero, we will obtain the kinetic friction coefficient.

SOLUTION In drawing the free-body diagram for the toboggan we choose the +x axis to be parallel to the hill surface and downward, the +y direction being perpendicular to the hill surface. We also use fk to symbolize the frictional force. Since the toboggan is in equilibrium, the zero net force components in the x and y directions are

+y FN

fk

8.00º +x mg 8.00º

= ΣFx mg sin 8.00= ° − mk FN 0 ΣF= FN − mg cos8.00= ° 0 y

In the first of these expressions we have used Equation 4.8 for fk to express the kinetic frictional force. Solving the second equation for the normal force FN and substituting into the first equation gives = ° 0 mg sin 8.00° − mk mg cos8.00    F

mk or=

sin 8.00° = tan= 8.00° 0.141 cos8.00°

N

Problem 4-81 REASONING AND SOLUTION If the +x axis is taken to be parallel to and up the ramp, then ΣFx = max gives T – fk – mg sin 30.0° = max where fk = µkFN . Hence, T = max + µkFN + mg sin 30.0° Also, ΣFy = may gives

(1)

FN – mg cos 30.0° = 0

since no acceleration occurs in this direction. Then FN = mg cos 30.0° Substitution of Equation (2) into Equation (1) yields T = max + µkmg cos 30.0° + mg sin 30.0° T = (205 kg)(0.800 m/s2) + (0.900)(205 kg)(9.80 m/s2)cos 30.0° + (205 kg)(9.80 m/s2)sin 30.0° = 2730 N

(2)

Problem 4-81 REASONING The forces acting on the motorcycle are the normal force FN, the 3150-N propulsion force, the 250-N force of air resistance, and the weight, which is mg = (292 kg)(9.80 m/s2 = 2860 N. All of these forces must be considered when determining the net force for use with Newton’s second law to determine the acceleration. In particular, we note that the motion occurs along the ramp and that both the propulsion force and air resistance are directed parallel to the ramp surface. In contrast, the normal force and the weight do not act parallel to the ramp. The normal force is perpendicular to the ramp surface, while the weight acts vertically downward. However, the weight does have a component along the ramp. SOLUTION In drawing the free-body diagram for the motorcycle we choose the +x axis to be parallel to the ramp surface and upward, the +y direction being perpendicular to the ramp surface. The diagram is shown at the right. Since the motorcycle accelerates along the ramp and we seek only that acceleration, we can ignore the forces that point along the y axis (the normal force FN and the y component of the weight). The x component of Newton’s second law is

+y FN

+x 3150 N

250 N 30.0º 30.0º

2860 N

(2860 N) sin 30.0º

= ΣFx 3150 N − ( 2860 N ) sin 30.0° − 250 = N max

Solving for the acceleration ax gives

= ax

3150 N − ( 2860 N ) sin 30.0° − 250 N = 5.03 m/s 2 292 kg

Problem 5-8 2 REASONING The centripetal acceleration is given by Equation 5.2 as ac = v /r. The value of the radius r is given, so to determine ac we need information about the speed v. But the

speed is related to the period T by v = (2π r)/T, according to Equation 5.1. We can substitute this expression for the speed into Equation 5.2 and see that

b

2π r / T v2 = ac = r r

g

2

=

4π 2 r T2

SOLUTION To use the expression obtained in the reasoning, we need a value for the period T. The period is the time for one revolution. Since the container is turning at 2.0 revolutions per second, the period is T = (1 s)/(2.0 revolutions) = 0.50 s. Thus, we find that the centripetal acceleration is ac =

b

g

2 4 π 2 r 4 π 0.12 m = = 19 m / s 2 2 2 T 0.50 s

b

g

Problem 5-11 REASONING AND SOLUTION T = 2π r/v. The speed is

The sample makes one revolution in time T as given by

v2 = rac = (5.00 × 10–2 m)(6.25 × 103)(9.80 m/s2)

so that

v = 55.3 m/s

The period is T = 2π (5.00 × 10–2 m)/(55.3 m/s) = 5.68 × 10–3 s = 9.47 × 10–5 min The number of revolutions per minute = 1/T = 10 600 rev/min .

Problem 5-23 REASONING a. The free body diagram shows the swing ride and the two forces that act on a chair: the tension T in the cable, and the weight mg of the chair and its occupant. We note that the chair does not accelerate vertically, so the net force ∑ Fy in the vertical direction must be zero, ∑ Fy = 0 . The net force consists of the upward vertical component of the tension and the downward weight of the chair. The fact that the net force is zero will allow us to determine the magnitude of the tension.

60.0°

T +y

15.0 m +x

r mg

b. According to Newton’s second law, the net force ∑ Fx in the horizontal direction is equal to the mass m of the chair and its occupant times the centripetal acceleration Fx ma = mv 2 / r . There is only one force in the horizontal ( ac = v 2 / r ) , so that ∑= c direction, the horizontal component of the tension, so it is the net force. We will use Newton’s second law to find the speed v of the chair. SOLUTION a. The vertical component of the tension is +T cos 60.0°, and the weight is −mg, where we have chosen “up” as the + direction. Since the chair and its occupant have no vertical acceleration, we have that ∑ Fy = 0 , so +T cos 60.0° − mg = 0    ∑ Fy

Solving for the magnitude T of the tension gives

(1)

mg = T = cos 60.0°

(179 kg ) ( 9.80 m/s 2 )

= 3510 N cos 60.0°

b. The horizontal component of the tension is +T sin 60.0°, where we have chosen the direction to the left in the diagram as the + direction. Since the chair and its occupant have a centripetal acceleration in this direction, we have

 v2  T sin 60.0 = = m   ° ma c  r  ∑ Fx

(2)

From the drawing we see that the radius r of the circular path is r = (15.0 m) sin 60.0° = 13.0 m. Solving Equation (2) for the speed v gives

= v

N ) sin 60.0° (13.0 m )( 3510=

r T sin 60.0° = m

179 kg

14.9 m/s

Problem 5-24 REASONING The angle θ at which a friction-free curve is banked depends on the radius r of the curve and the speed v with which the curve is to be negotiated, according to Equation 5.4: tan θ = v2 /(rg). For known values of θ and r, the safe speed is

v = rg tan θ Before we can use this result, we must determine tan θ for the banking of the track. SOLUTION The drawing at the right shows a cross-section of the track. From the drawing we have tan θ =

18 m = 0.34 53 m

a. Therefore, the smallest speed at which cars can move on this track without relying on friction is = vmin

(112 m ) ( 9.80 m/s 2 ) ( 0.34 ) =

19 m/s

b. Similarly, the largest speed is = vmax

(165 m ) ( 9.80 m/s 2 ) ( 0.34 ) =

23 m/s

Problem 5-41 REASONING As the motorcycle passes over the top of the hill, it will experience a centripetal force, the magnitude of which is given by Equation 5.3: FC = mv 2 / r . The centripetal force is provided by the net force on the cycle + driver system. At that instant, the net force on the system is composed of the normal force, which points upward, and the weight, which points downward. Taking the direction toward the center of the circle (downward) as the positive direction, we have FC = mg − FN . This expression can be solved for FN , the normal force. SOLUTION a. The magnitude of the centripetal force is FC =

mv 2 (342 kg)(25.0 m / s) 2 = 1.70 × 10 3 N = 126 m r

b. The magnitude of the normal force is FN = mg − FC = (342 kg)(9.80 m/s 2 ) − 1.70 ×103 N = 1.66 ×103 N

Problem 5-45 REASONING The magnitude Fc of the centripetal force is Fc = mv 2 / r (Equation 5.3). Since the speed v, the mass m, and the radius r are fixed, the magnitude of the centripetal force is the same at each point on the circle. When the ball is at the three o’clock position, the force of gravity, acting downward, is perpendicular to the stick and cannot contribute to the centripetal force. (See Figure 5.21 in the text, point 2, for a similar situation.) At this point, only the tension of T = 16 N contributes to the centripetal force. Considering that the centripetal force is the same everywhere, we can conclude that it has a magnitude of 16 N everywhere. At the twelve o’clock position the tension T and the force of gravity mg both act downward (the negative direction) toward the center of the circle, with the result that the centripetal force at this point is –T – mg. (See Figure 5.21, point 3.) At the six o’clock position the tension points upward toward the center of the circle, while the force of gravity points downward, with the result that the centripetal force at this point is T – mg. (See Figure 5.21, point 1.) SOLUTION Assuming that upward is the positive direction, we find at the twelve and six o’clock positions that