E. Circular Motion Homework Part I



Things you might need to know for some of these problems:     \

1 mile = 1609 meters

Mearth = 5.98 x 1024 kg        Rearth = 6.378 x 106 m (equator)    

Msun = 2.0 x 1030 kg            Rsun = 7 x 108 m (equator)

Mmoon = 7.4 x 1022 kg         Rmoon = 1.75 x 106 m  (equator)

Rearth-sun = 1.5 x 1011 m (average distance) Known as 1 A.U. (astronomical unit)

Rearth-moon = 3.8 x 108 m (average distance)


7.1 Angular Measures

1, 2, 5, 7, 8, 10, 15


10:  How many degrees will the Earth turn in those 3 months?  It takes the Earth 12 months to make a complete (360o) circle.  Look up the distance from Earth to Sun in your textbook (this distance is called 1 astronomical unit (1 AU))


7.2  Angular Speed and Velocity

19, 20, 21, 22, 24, 30, 31, 34


21:  It is so important that you know the difference between these two ideas.

22:  Again, you need to know this.

30:  Vtangent/r = w.  And then, Dt = Dq/w where Dq = 2p rad since it maintains the angular velocity (w) over one complete revolution (2prad)

34:  Aw come on…. You wanted to know this anyway.  Weren’t you curious?


7.3 Uniform Circular Motion and Centripetal Acceleration


36, 37, 40, 46, 48, 49, 50, 51, 53, 55 (you have got to at least try this one!), 56


40:  Think about it!  What is the direction of acceleration for any rotating object?  The bobs will always move in the direction of acceleration.  Just like a helium balloon on a string in your car will lurch forward when you step on the gas (accelerating in the forward direction).  But be careful, because acceleration in this case is NOT in the forward direction!

46:  ac = rw2

48:  a)  v = (d/t)=(2pr)/t             b)  Fc = mac = m(v2/r)   c)  No, there has to be some kind of upward component tension force to balance the downward force of gravity.

49:  q = tan-1(mg/Fc)

50:  To keep the water from coming out of the bucket, you have to have a velocity which will match the gravitational force to the centripetal force.  Mg = Fc = m(v2/r).   Therefore, v =

51.  fs = msN = msmg = Fc = m(v2/r)      such that v =   (that “s” should be subscripted.  And that “u” should be a “m”.  But the darn computer wouldn’t let me do that!)

53:  a)  A normal force by the loop must act on the block to provide extra centripetal force.  Minimum speed corresponds to a minimum normal force of zero so the only force on the block is gravity.  Gravity is the only source of centripetal force in this case.

Fc = mg = m(v2/r) so that v =

        b)  From the energy conservation equation:  U + KE = Uo + KEo such that:

     mgh + 1/2 m(0)2 = +mg(2r) + 1/2 mv2

  Therefore, h = 2r + 1/2 (rg/g) = (5/2) r

55:  The key to this problem is the positioning of the x and y axis.  Typically, we have been referring to the ramp as being the x axis and the normal force to be the y axis (since the normal force always acts ^ to the ramp).  But now, we are interested in the force acting towards the center of the curve (which is N sin q).  We will say, in this case, that the x axis is along the line of N sin q. And therefore, the y axis will be along N cos q. (Look at that figure 7.28 on p. 246 – can you see how N sin q is ^ to N cos q?)

     As the car travels faster and faster, it produces a greater force against the curve and therefore increases the N sin q force acting on it towards center (remember, the magnitude of the normal force is a direct response to the pushing on the ground by the object – mantra:  “you can’t hit the ground any harder than it hits you back”) . 

     Think about it:  If its Vtangential increases, it is “trying” to go into the track, but since the track is rigid, it cannot expand the radius of its turn.  (What if this were the space shuttle, not on a track, but in orbit around the Earth.  If the space shuttle were to speed up in the Vtangential direction its orbit would grow in radius right?)

     Okay, now how do you solve this confounded problem?

       Along the y axis, Ncosq = mg (look at the drawing!).  Along the x axis, N sin q is the force which acts centripetally (Fc).  N sin q is the only force keeping the car in its circular acceleration.

     We can re-arrange N cos q = mg to say:          N = mg/cosq

      We can say also (along the x axis) that Fc = N sin q

       Since N = mg/cosq, we can substitute this in the Fc equation above to say:

       Fc = (mg/cosq)  sin q     and by trig identity (sin q /cosq = tan q)

              Therefore……..        Fc = mg tan q

       We also know that Fc = mv2/r 

        So, let’s set the two equations we have for Fc equal to each other

                              mg tan q = mv2/r      (how about that, the m’s cancel!!!!!!)

                 So, tan q = v2/gr

56:  The static friction between you and the wall provides the upward force to balance your weight.  The normal force on you by the wall provides the centripetal force.

     N = Fc = mrw2.    So…..   fs = msN = msmrw2 = mg

          Therefore, w =

Even Answers:

2:  c

8:  1.4 x 102 m

10:  2.4 x 108 km

20:  d
22:  Remember the earth turns counterclockwise in north and clock in south

24:  3.1 rad/s

30:  0.42 sec

34:  0.120 sec

36:  b

40:  In the direction of the acceleration:  inward

46:  2.31 x 10-3 m/s2

48:  a)  7.85 m/s      b)  10.3 N     c)   no

50:  3.1 m/s       56:  3.6 rad/sec