Problems and Solution Dynamics of Uniform Circular Motion

 Problem #1

A toy car moving at constant speed completes one lap around a circular track (a distance of 200m) in 25.0 s. (a) What is the average speed? (b) If the mass of the car is 1.50 kg, what is the magnitude of the central force that keeps it in a circle?

Answer:
(a) If a lap around the circular track is of length 200m then the (average) speed of the car is

v = d/t = 200 m/25.0 s = 8.00 m/s

(b) The car undergoes uniform circular motion, moving in a circle of radius r with speed v. The net force on the car points toward the center of the circle and has magnitude

F_cent = mv²/r

Actually, we haven’t found r yet. We are given the circumference of the circle, and from C = 2πr we find

r = C/2π = 200 m/2π = 31.8m

So the net force on the car has magnitude

F_cent = mv²/r = (1.50 kg)(8.00 m/s)²/(31.8m) = 3.02N

The net force on the car has magnitude 3.02 N; its direction is always inward, keeping the car on a circular path.

Problem #2
A mass M on a frictionless table is attached to a hanging mass M by a cord through a hole in the table, as shown in Fig. 1. Find the speed with which m must move in order for M to stay at rest.

Fig.1

Answer:
Taking mass M to be at rest, we see that mass m must be moving in a circle of constant radius r. It is moving at (constant) speed v; so mass m undergoes uniform circular motion. So the net force on m points toward the center of the circle and has magnitude F_cent = mv²/r. The free–body diagram for m is shown in Fig. 2(a). The only force on m is the string tension (pointing toward the center of the circle). This gives us:

T = mv²/r

Fig. 2

Next consider the forces acting on M and its motion. The force diagram for M is shown
in Fig. 2(b). Since mass M is at rest, the net force on it is zero, which gives:


T = Mg
Combining these two results, we get:

mv²/r = Mg

Solving for v, we get:

v² = Mgr/m

v = √[Mgr/m]

Problem #3
A stuntman drives a car over the top of a hill, the cross section of which can be approximated by a circle of radius 250 m, as in Fig. 3. What is the greatest speed at which he can drive without the car leaving the road at the top of the hill?

Fig.3

Answer:
We begin by thinking about the forces acting on the car and its acceleration when it is at the top of the hill.

Fig.4

At the top of the hill, the car is moving in a circular path of radius r = 250m with some speed v. Then the car has a centripetal acceleration of magnitude v²/r which is directed downward. (For all we know, it may also have a tangential acceleration as well, but the problem gives no information on it, and it won’t be relevant for the problem.) By Newton’s Second Law, the net (vertical) force on the car must have magnitude mv²/r and must be directed downward.
The forces acting on the car are shown in Fig. 4. Then the force of gravity is mg downward. The road exerts a normal force of magnitude N upward. One may ask how we know the road’s force goes upward. This is because there is no physical way in which a road can pull downward on a car driving over it. But it can push up.
We combine the results from the last two paragraphs. The net downward force must equal mv²/r. This gives us:

mg − N = mv²/r

however without knowing anything more, we can’t solve for v in this equation because we don’t know N (or, for that matter, m). We have not yet used the condition that the car is on the verge of leaving the road at the top of the hill. What does this condition give us? If we use the last equation to find the normal force:

N = mg – mv²/r

we see that if we increase v there comes a point at which N must be negative in order for the car to stay on the road moving on its circular arc. But as discussed above, N can’t be negative. But it can be zero, and it is for this speed that the car is on the verge of leaving the road at the top of the hill. The critical case has N = 0, and this gives us:

0 = mg − mv²/r
mv²/r = mg

Note that the mass m cancels out of this equation so we don’t need to know m.

We get:

v² = rg = (250 m)(9.80 m/s²) = 2.45 m²/s²

and finally

v = 49.5 m/s

The car may be driven as fast as 49.5 m/s and it will stay on the road.

Problem #4
A coin placed 30.0 cm from the center of a rotating, horizontal turntable slips when its speed is 50.0 cm/s . (a) What provides the central force when the coin is stationary relative to the turntable? (b) What is the coefficient of static friction between the coin and turntable?

Fig.5

Answer:
(a) See Fig. 5. for a fine illustration of the problem. As the coin executes uniform circular motion (before it slips) it is accelerating toward the center of the turntable! So there must be a force (or forces) on the coin causing it to do this. This force can only come from its contact interaction with the turntable, i.e. from friction. Here, since we are dealing with the case where the coin is not sliding with respect to the surface, it is the force of static friction. Furthermore, the force of static friction is directed toward the center of the turntable.

Fig.6

(b) A view of the forces acting on the coin is given in Fig. 6. If the mass of the coin is m then gravity exerts a force mg downward, the turntable exerts a normal force N upward and there is a force of static friction, which as we discussed in part (a) must point toward the center of the turntable. The acceleration of the coin points toward the center of the circle and has magnitude v²/r, (r being the distance of the coin from the center). So the vertical forces must cancel, giving us N = mg. The net force points inward and has magnitude mv²/r, so that f_s = mv²/r.
Now for the conditions at which the coin starts to slip, the force of static friction has reached its maximum value, i.e.

f_s = μ_sN

but here we can use our results to substitute for fs and for N. This give us:

mv²/r = μ_smg

which lets us solve for μ:

μ_s = v²/rg = (50.0 cm/s)²/[(30.0 cm)(9.80 m/s²)]
    = (0.500 m/s)²/[(0.300 m)(9.80 m/s²)]
μ_s = 0.085

So the coefficient of static friction for the turntable and coin is μ_s = 0.085. We were never given the mass of the coin, but we did not need it because it cancelled out of our equations just before the final answer.

Problem #5
A Ferris wheel rotates four times each minute; it has a diameter of 18.0 m. (a) What is the centripetal acceleration of a rider? What force does the seat exert on a 40.0 kg rider (b) at the lowest point of the ride and (c) at the highest point of the ride? (d) What force (magnitude and direction) does the seat exert on a rider when the rider is halfway between top and bottom?

Fig.7

Answer:
(a) First, calculate some numbers which we know are important for circular motion! The wheel turns around 4 times in one minute, so the time for one turn must be

T = 1.0 min/4 = 60.0 s/4 = 15 s

Also, since the radius of the wheel is R = D/2 = 18.0 m/2 = 9.0 m, the circumference of the wheel is

C = 2πR = 2π (9.0m) = 57 m

and then the speed of a rider is

v = C/T = 57 m/15 s = 3.8 m/s

The rider moves at constant speed in a circular path of radius R. So the rider’s acceleration is always directed toward the center of the circle and it has magnitude

a_cent = v²/R = (3.8 m/s)²/(9.0m) = 1.6 m/s²

The centripetal acceleration of the rider is 1.6 m/s²

(b) Consider what is happening when rider is at the lowest point of the ride. His acceleration is upward (toward the center of the circle!) and has magnitude 1.6 m/s².

What are the forces acting on the rider (who has mass M, let’s say) at this point? These are shown in Fig. 7(a). Gravity pulls down on the rider with a force of magnitude Mg, and the seat pushes upward on the rider with a force F_seat. (Usually seats can’t pull downward; also, the force of the seat can’t have any sideways component because here the net force must point upward). Since the net force points upward and has magnitude F_cent = Mv²/R, Newton’s Second Law gives us:

F_seat − Mg = Mv²/R

Since M = 40.0 kg, we get:

F_seat = Mg + Mv²/R
F_seat = (40.0 kg)(9.80 m/s²) + (40.0 kg)(3.8 m/s)²/(9.0m)
       = 460N

The seat pushes upward on the rider with a force of magnitude 460 N. We might say that
when the rider at the lowest point, the rider has an apparent weight of 460 N, since that is
the force of the surface on which the rider rests. Here, the apparent weight is greater than
the true weight Mg of the rider.

(c) When the rider is at the highest point of the wheel, his acceleration is downward. The
forces acting on the rider are shown in Fig. 7(b); these are the same forces as in part (a)
but now the net force points downward and has magnitude Fcent = Mv²/R. Adding up the
downward forces, Newton’s Second Law now gives us:

Mg − F_seat = Mv²/R

which now gives us

F_seat = Mg – Mv²/R

F_seat = (40.0 kg)(9.80 m/s²) − (40.0 kg)(3.8 m/s)²/(9.0m) = 330N

The seat pushes upward on the rider with a force of magnitude 330 N. We would say that at
the top of the ride, the apparent weight of the rider is 330 N. This time the apparent weight
is less than the true weight of the rider.

Fig.8

(d) When the rider is halfway between top and bottom, the net force on him still points toward the center of the circle (and has magnitude Mv²/R), but in this case the direction is horizontal, as  indicated in Fig. 8. (In this picture the rider is on the right side of the Ferris wheel, as we look at it face–on.) The forces acting on the rider are also shown in this picture. The force of gravity, Mg can only pull downward. The only other force on the rider, namely that of the seat does not push straight upward in this case. We know that it can’t, because the sum of the two forces must point horizontally (to the right). The force of the seat must also have a horizontal component; it must point as shown in Fig. 8.

Without being overly formal about the mathematics we can see that the vertical component of F_seat must be Mg so as to cancel the force of gravity. The vertical component of F_seat must have magnitude

F_seat, _vert = Mg = (40.0 kg)(9.80 m/s²) = 392 N

The horizontal component of F_seat must equal Mv²/R since as we’ve seen, that is the net force on the rider. So:

F_seat, horiz = Mv²/R = (40.0 kg)(3.8 m/s)²/(9.0m) = 64 N

Now we can find the magnitude of the force of the seat:

F_seat = √[(F_seat, vert)² + (F_seat, horiz)²]
F_seat = √[(392 N)² + (64 N)²] = 397 N

and this force is directed at an angle θ above the horizontal, where θ is given by

θ = tan⁻¹[F_seat, vert/F_seat, horiz] = tan⁻¹[392 N/64N] = 81⁰

The force of the seat has magnitude 397 N and is directed at 81⁰ above the horizontal.   

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