LEVEL B PROBLEMS
1.A flat puck(mass M) is rotated in a circle on a frictionless air hockey tabletop, and is held in its orbit by a light cord which is connected to a dangling mass (mass m) through the central hole as shown in the figure. Show that the speed of the puck is given by:

SOLUTION:

We write SF = ma from the force diagram for the stationary hanging mass, with down positive: 

            mg – FT= ma = 0;which gives 

            FT= mg.

For the rotating puck, the tension provides the centripetal acceleration,

            SFR = MaR:

            FT= Mv2/R.

When we combine the two equations, we have

            Mv2/R = mg, which gives v = (mgR/M)1/2.

2.What is the minimum speed with which a 1050-kg car can round a turn of radius 70 m on a flat road if the coefficient of friction between the tires and road is 0.80?Is the result independent of the mass of the car? [vmax = 23 m/s] 
SOLUTION:
If the car does not skid, the friction is static, with Ffr =msFN.

This friction force provides the centripetal acceleration.We take a coordinate system with the x-axis in the direction of the centripetal acceleration.

We write SF = ma from the force diagram for the auto:

x-component:Ffr = ma = mv2/R;

y-component:FN– mg = 0.

The speed is maximum when Ffr = Ffr,max = msFN.

When we combine the equations, the mass cancels, and we get

            msg = vmax2/R;

            (0.80)(9.80 m/s2) = vmax2/(70 m), which gives vmax = 23 m/s.

The mass canceled, so the result isindependent of the mass.

3.Calculate the effective value of g, the acceleration due to gravity, at (a) 3200 m, and (b) 3200 km, above the Earth’s surface. [9.8 m/s2, 4.3 m/s2.] 
SOLUTION:
(a) g = GME/r2 = GME/(RE + h)2 = (6.67 x 10-11)(5.98 x 1024 kg)/(3200 m + 6.38 x 106 m)2 = 9.80 m/s2

(b) g = GME/r2 = GME/(RE + h)2 = (6.67 x 10-11)(5.98 x 1024 kg)/(3.2 x 106 m + 6.38 x 106 m)2 = 4.30m/s2

4.We know that the acceleration due to gravity on Mars is 0.38 that of the Earth’s. If mars has a radius of 3400 km, what is the mass of Mars? [6.4 ´1023 kg]
SOLUTION:
The acceleration due to gravity on the surface of a planet is

            g = GMM/r2 = GMM/(RM )2 = (6.67 x 10-11)MM/(3.4  x 106 m)2
            (0.38)(9.80 m/s2) = 6.67 x 10-11MM/(3.4  x 106 m)2
            MM = 6.4 ´1023 kg
5.During an Apollo lunar landing mission, the command module continued to orbit the Moon at a height of 100 km. How long did it take to go around the Moon once? [2.0 h] 
SOLUTION:
We relate the speed to the period of revolution from

            v = 2pR/T.

The required centripetal acceleration is provided by the gravitational attraction:

            GMMm/R2 = mv2/R = m(2pR/T)2/R = m4p2R/T2, so we have

            GMM = 4p2(RM + h)3/T2;

            (6.67 ´ 10–11 N · m2/kg2)(7.4 ´ 1022 kg) = 4p2(1.74 ´ 106 m + 1.00 ´ 105 m)3/T2, 

which gives T = 7.06 ´ 103 s = 2.0 h.

6.Calculate the angular velocity of the Earth (a) as it orbits the Sun, and (b) as it rotates in its own axis. [1.99 ´ 10–7 rad/s; 7.27´ 10–5 rad/s]
SOLUTION:
(a)The Earth moves one revolution around the Sun in one year, so we have

            worbit= Dq/Dt 

            = (2p rad)/(1 yr)(3.16 ´ 107 s/yr) = 1.99 ´ 10–7 rad/s.

(b)The Earth rotates one revolution in one day, so we have

            wrotation= Dq/Dt 

            = (2p rad)/(1 day)(24 h/day)(3600 s/h) = 7.27 ´ 10–5 rad/s.

7.A 40-cm diameter wheel accelerates uniformly from 240 rpm to 360 rpm in 6.5 s. How far will a point on the edge of the wheel have traveled in this time? [38 m] 
SOLUTION:
We find the total angle the wheel turns from 
            q=½ (wo + w))t

            =½ [(210 rev/min + 350 rev/min)][(2p rad/rev)/(60 s/min)](6.5 s) = 191 rad = 30.3 rev.

For each revolution the point on the edge will travel one circumference, so the total distance traveled is

            d = qD = (30.3 rev)(0.40 m) =38 m.