HOMEWORK PROBLEMS FOR CHAPTER 7 CIRCULAR MOTION

C LEVEL PROBLEMS

1)A jet plane traveling 1800 km/h (500 m/s) pulls out of a dive by moving in a circular arc of radius 6.00 km.What is the acceleration of the plane in g’s? (that is, in multiples of 9.80 m/s2)

SOLUTION:

The centripetal acceleration is

            aR = v2/r = (500 m/s)2/(6.00 ´ 103 m)(9.80 m/s2/g) = 4.25g up.

2)A child on a merry-go-round is moving with a speed of 1.35 m/s when 1.20 m from the center of the ride.Calculate (a) the centripetal acceleration of the child, and (b) the net horizontal force exerted on the child (m = 25.0 kg).

SOLUTION:

(a)The centripetal acceleration is
 

            aR = v2/r = (1.35 m/s)2/(1.20 m) = 1.52 m/s2 toward the center.

(b)The net horizontal force that produces this acceleration is

            Fnet = maR = (25.0 kg)(1.52 m/s2) = 38.0 N toward the center.

3)Calculate the force of gravity on a spacecraft 12,800 km above the surface of the Earth if it has a mass of 1400 kg.

SOLUTION:

The gravitational force on the spacecraft is

            F= GMm/r2

            = (6.67 ´ 10–11 N · m2/kg2)(5.98 ´ 1024 kg)(1400 kg)/[3(6.38 ´ 106 m)]2 = 1.52 ´ 103 N.

4)Calculate the acceleration due to gravity on the Moon.The moon’s radius is about 1.74 x 106 m and its mass is 7.35 x 1022 kg.

SOLUTION:

The acceleration due to gravity on the surface of a planet is

            g = F/m = GM/R2.

For the Moon we have

            gMoon = (6.67 ´ 10–11 N · m2/kg2)(7.35 ´ 1022 kg)/(1.74 ´ 106 m)2 = 1.62 m/s2.

5)A hypothetical planet has a radius 2.5 times larger than the Earth, but the same mass.What is the acceleration due to gravity near the surface of the planet?If the planet had a mass 2.5 time that of the Earth but the same radius, what will be its acceleration due to gravity?

SOLUTION:

The acceleration due to gravity on the surface of a planet is

            g = F/m = GM/R2 = (6.67 x 10-11)(5.98 x 1024 kg)/[2.5(6.38 x 106 m)]2 = 1.6 m/s2

I

The acceleration due to gravity on the surface of a planet is

            g = F/m = GM/R2 = (6.67 x 10-11)(2.5 x 5.98 x 1024 kg)/[(6.38 x 106 m)]2 = 24.5 m/s2

6)Calculate the velocity of a satellite moving in a stable orbit about the Earth at a height of 3600 km above the Earth’s surface?

SOLUTION:

The gravitational attraction must provide the centripetal acceleration for the circular orbit:

            GMEm/R2 = mv2/R,

            v2= GME/(RE + h)

            = (6.67 ´ 10–11 N · m2/kg2)(5.98 ´ 1024 kg)/(6.38 ´ 106 m + 3.6 ´ 106 m),

which gives v =6.3 ´ 103 m/s.

7)A 0.35-m diameter grinding wheel rotates at 1800 revolutions per minute.Calculate its angular velocity in rad/s.

SOLUTION:

            w = (1800 rev/min)(2p rad/rev)(1 min/60 s) = 188 rad/s.

8)A centrifuge accelerates from rest to 15,000 rpm in 220 s.Through how many revolutions did it turn in this time?

SOLUTION:

For the angular displacement we use

            q= ½(w0 + w)t

            = ½ [0 + (15,000 rpm)(2p rad/rev)/(60 s/min)](220 s) = 1.73 ´ 105 rad.

Thusq= (1.73 ´ 105 rad)/(2p rad/rev) = 2.75 ´ 104 rev.

9)An automobile engine slows down from 4000 rpm to 1200 rpm in 3.5 s.Calculate (a) it’s angular acceleration if we assume that it is uniform, and (b) the total number of revolutions the engine can make in this time.

SOLUTION:

(a)For motion with constant angular acceleration we use

            w =w0 + at;

            (1200 rev/min)(2p rad/rev)(1 min/60s) = (4000 rev/min)(2? rad/rev)(1 min/60s) + a (3.5 s),

which gives a =– 84 rad/s2.

(b)For the angular displacement we use

            q= ½(w 0 + w)t

            =½ [(4000 rev/min + 1200 rev/min)][(2p rad/rev)/(60 s/min)](3.5 s) = 954 rad.

Thusq= (954 rad)/(2p rad/rev) = 1.5 ´ 102 rev.