Bank Soal Circular Motion

For Questions #2-#5, identify the type of force which causes the following bold-faced objects to travel along a circular path.

2. An eraser is tied to a string swung in a horizontal circle.

a. gravity b. normal c. tension d. appliede. friction f. spring g. electrical h. magnetic

3. The moon orbits the earth.


4. A car makes a sharp right-hand turn along a level roadway.


5. A roller coaster car passes through a loop. Consider the car at the bottom of the loop.


Answers:

2. C - Tension (A string is attached to the eraser and pulls it towards the center point of the circle.)

3. A - Gravity (All masses attract with a force of gravity. In the case of the moon and the earth, gravity pulls on the moon in a direction which is roughly perpendicular to its path.)

4. E - Friction (Once the wheels are turned, friction can push perpendicular to the wheels' direction, pushing the car towards the center of the circle.)

5. B - Normal (There are two forces at the bottom of the loop; gravity pulls outwards from the center; but the norma

6. A physics teacher ties an eraser to the end of a string and then whirls it in a counter-clockwise circle. If the teacher lets go of the string, then the eraser hits a student (or several students) in the classroom. If the string is let go when the eraser is at point X on the diagram at the right, then which student(s) in the class will the eraser hit? Write the initials in this space: ________________

Answer: JG and AF

15. In the diagram at the right, draw vector arrows (straight lines with arrowheads) which indicate the following for an object which is moving in a clockwise circle.

the net force at point A. the acceleration at point B. the velocity at point C.

Answer:

See diagram for answers. The force and acceleration vectors are directed inwards towards the center of the circle and the velocity vector is directed tangent to the circle.

18. Construct a free-body diagram showing the direction and types of forces acting upon the car at the top and the bottom of a loop. Be sure to label the forces according to type.

Anna experiences a downward acceleration of 24.0 m/s2 at the top of the loop and an upwards acceleration of 12.0 m/s2 at the bottom of the loop. Use Newton's second law to fill in all the blanks and to ultimately determine the normal force acting upon Anna's 800. kg roller coaster car. PSYW

Top BottomFnet = 19200 N, down Fnet = 9600 N, upFgrav = 7840 N, down Fgrav = 7840 N, down

Fnorm = 11360 N, down Fnorm = 17440 N, up

Answer: See diagram and table above.

This problem is very similar to question #17 above. The net force is found by calculating m•a; its direction is in the same direction as the a vector. The gravitational force is found from m•g; the value of 9.8 m/s/s is used for g. Now the free-body diagram can be used to generate the following equations for the top and the bottom of the loop:

Top: Fnet = Fgrav + Fnorm which can be rearranged to Fnorm = Fnet - Fgrav

Bottom: Fnet = Fnorm - Fgrav which can be rearranged to Fnorm = Fnet + Fgrav

The difference in these two equations is attributed to the fact that at the top of the loop, the Fnorm and the Fgrav point in the same direction as the net force vector; at the bottom of the loop, they point in opposite directions. Using the calculated values of Fnet and Fgrav, the normal force can now be calculated.

19. A fullback is running a sweep around the left side of the line. As he rounds the turn, he is momentarily moving in circular motion, sweeping out a quarter-circle with a radius of 4.17 meters. If the 83.5-kg fullback makes the turn with a speed of 5.21 m/s, then what is his acceleration, the net force, the angle of lean (measured to the vertical), and the total contact force with the ground? PSYW

a = 6.51 m/s2

Fnet = 544 N

Fgrav = 818 N

Fnorm = 818 N

Ffrict = 544 N

Fcontact = 982 N

Angle of Lean = 33.6 degrees

Answer: See table above,

The acceleration is found using the equation: a = v2/R. For the top of the loop, a = (5.21)2/(4.17) = 6.51 m/s2. The net force is always found by Fnet = m•a. For the fullback, Fnet = (83.5 kg)•(6.51 m/s2) = 544 N. This net force is directed inwards and as such, it is equal to the force of friction supplied by the ground. Subsequently, Ffrict = 544 N. The force of gravity is always found by m•g where g = 9.8 m/s2. Using the fullback's mass, this value is 818 N. Since the normal force must support the fullback's weight, it is equal in magnitude to the force of gravity; so Fnorm = 818 N. The contact force is the resultant force of the Fnorm and Ffrict. Its value can be determined using the Pythagorean theorem: Fcontact = SQRT[(544 N)2 + (818 N)2] = 982 N. The angle of lean (theta in the diagram above) can be found using either the sine or the tangent function. From the diagram, it is seen that the sine(theta) = Ffrict /Fcontact. Thus, theta is equal to sin-1(Ffrict /Fcontact) = sin-1 (544 N/982 N) = 33.6 degrees.

20. Pete Zaria is riding the Cliff Hanger at Great America. Pete enters a cylindrical barrel which makes 18.5 revolutions every minute. The diameter of the barrel is 6.92 m. Pete's mass is 64.7 kg. Construct a free-body diagram showing the forces acting upon Pete. Calculate the

acceleration and net force acting upon Pete Zaria. Finally, determine the coefficient of friction between Pete and the walls that would be required to support Pete's weight. PSYW

Answer: 0.755

As usual, Fgrav = m•g = (64.7 kg)•(9.8 m/s/s) = 634 N. This downward force must be balanced by the friction force since there is no vertical acceleration. Therefore, Ffrict = 634 N. The net inward force is supplied by the normal force; so a calculation of the net force will yield the value of the normal force. Acceleration is found from speed and radius. The speed is:

v = (18.5 rev/60 sec)• (2 • pi • 3.46 m/rev) = 6.703 m/s.

Now the acceleration is:

a = v2/R = (6.703 m/s)2/(3.46 m) = 12.986 m/s2

Now the net force (and the normal force) can be calculated:

Fnet = Fnorm = m•a = (64.7 kg)•(12.986 m/s/s) = 840.20 N

Now that the normal force and the friction force have been found, the coefficient of friction can be determined:

µ = (Ffrict)/(Fnorm) = (634 N)/(840.20 N) = 0.755

29. A roller coaster car loaded with passengers, has a mass of 500 kg; the radius of curvature of the track at the bottom point of a dip is 12 m. The vehicle has a speed of 18 m/s at this point.

a. In the space below, draw a free-body diagram for the car (label forces according to type).

b. Calculate the acceleration and the net force acting upon the car. PSYWc. Calculate the force exerted on the vehicle by the track? PSYW

Answers:

The free-body diagram (part a) is shown at the right.

The acceleration of the car can be computed as follows:

a = v2/R = (18.0 m/s)2/(12.0 m) = 27.0 m/s2 (part b)

The net force can be found in the usual manner:

Fnet = m•a = (500. kg)•(27.0 m/s2) = 13500 N (part b)

Since the center of the circle (see diagram) is above the riders, then both the net force and the acceleration vectors have an upward direction. The force of gravity is downwards, so the net force is equal to the upward force minus the downward force:

Fnet = Fnorm - Fgrav

where Fgrav = m • g = (500. kg) • (9.8 m/s/s) = 4900 N

Thus,

Fnorm = Fnet + Fgrav = 13500 N + 4900 N = 18400 N (part c)

30. What is the acceleration of a piece of dust on an old-fashioned record album, 15 cm from the center, if the record is spinning at 33.3 rpm?

Answer: 1.82 m/s/s

To find acceleration, the speed and the radius must be known. The radius is given; the speed can be computed as distance per time. The dust moves a distance equivalent to 33.3 circumferences in 60 seconds. So

v = 33.3 • 2 • pi • (0.15 m) / (60 s) = 0.523 m/s.

The acceleration can now be computed using the centripetal acceleration equation:

a = v2/R = (0.523 m/s)2 / (0.15 m) = 1.82 m/s/s

31. What is the tension in a 0.500 meter rope which carries a 2.50 kg bucket of water in a vertical circle with a velocity of 3.00 m/s when the bucket is at the bottom of its swing?

Answer: 69.5 N

Begin with a free-body diagram for the bucket at the bottom of its string. Since the bucket is at the bottom of the circle, the net force is upwards (inwards). Notice that the tension force must be greater than the force of gravity in order to have a net upwards (inwards) force. Also note that while Ftens is directed in the direction of the net force, Fgrav is in the opposite direction. Thus, the equation can be written:

Fnet = Ftens - Fgrav

This equation can be rearranged to solve for tension; the expressions for Fgrav (m•g) and Fnet (m•v2/R) can be substituted into it:

Ftens = Fnet + Fgrav = m•v2/R + m•g = (2.50 kg)•(3.00 m/s)2/(0.500 m) + (2.50 kg)•(9.8 m/s2)

Ftens = 69.5 N

32. A 53.5-kg rider on a roller coaster car is moving 10.3 m/s at the top of a loop which has a radius of curvature of 7.29 m. Determine the normal force experienced by the rider.

Answer: 254 N

At the top of the loop, both the gravitational and the normal force are directed inwards (as shown in the free-body diagram at the right). Thus, the Fnet equation can be written as:

Fnet = Fgrav + Fnorm

This equation can be rearranged and the expressions for Fgrav (m•g) and Fnet (m•v2/R) can be substituted into it:

Fnorm = Fnet - Fgrav = m • v2 / R - m • g

Fnorm = (53.5 kg) • (10.3 m/s)2 / (7.29 m) - (53.5 kg) • (9.8 m/s2)

Fnorm = 254 N

33. Determine the minimum coefficient of friction required to keep a 920.-kg car on a turn with a radius of 26.8 m. The car is traveling 29.9 m/s and the roadway is level.

Answer: 3.40

As usual, begin with a free-body diagram. The gravity force is balanced by (and equal to) the normal force and the force of friction is the net force. The solution then begins by equating m • a to Ffrict and carrying out the customary substitutions and algebra steps (using the fact that a = v2 / R and Ffrict = µ • Fnorm and Fgrav = m • g).

m•a = Ffrict

m • v2/R = µ • Fnorm

m • v2/R = µ • m • g

The mass now cancels and the equation can be rearranged to solve for mu:

µ = v2/(g• R) = (29.9 m/s)2 / [ (9.8 m/s2) • (26.8 m)] = 3.40

(This is not a likely mu value and as such the car does not have much of a chance negotiating the turn at this speed.)

34. With what speed (in mi/hr) must you drive your car across the crest of a hill of radius 37.1-m in order to feel weightless (1.00 m/s = 2.24 mi/hr)?

Answer: 42.8 mi/hr (19.1 m/s)

The feeling of weightlessness is experienced when a person is under the sole influence of gravity. So the free-body diagram for such a person would reveal only one force - gravity, pointing straight down. In such a case, the seat does not push upward upon the person. The car has passed over the crest of the hill at such a high speed that it is momentarily raised off the roadway and is unable to support the weight of its passengers. The force of gravity is the net force and the following equation can be written:

m • a = m • g

The mass can be canceled from both sides of the equation (sigh!) and the expression for acceleration can be substituted into the equation, yielding:

v2/R = g

Rearrangement and substitution yields the solution:

v = SQRT(g • R) = [(9.8 m/s2) • (37.1 m)] = SQRT[364 m2/s2] = 19.1 m/s

This value for speed can be converted to mi/hr using the equivalency: 2.24 mi/hr = 1m/s; the car is traveling with a speed of 42.8 mi/hr.

36. An airplane pilot makes a vertical loop with a 56-2-m radius. Determine the normal force acting upon the pilot's 62.6-kg body at the top of the loop if his air speed is 64.1 m/s.

Answer: 3.96 x 103 N

As usual, begin with a free-body diagram (shown at the right). From the diagram, the equation relating net force to the individual forces can be written:

Fnet = Fnorm + Fgrav

Since Fnet = m•a = m•v2/R and Fgrav = m•g, substitutions can be made and the equation can be rewritten:

m•v2/R = Fnorm + m•g

Rearranging to solve for Fnorm yields:

Fnorm = m•v2/R - m•g = (62.6 kg) • (64.1 m/s)2 / (56.2 m) - (62.6 kg) • (9.8 m/s2)

Fnorm = 3.96 x 103 N

Bank Soal Circular Motion

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