<< Chapter < Page | Chapter >> Page > |
Of course, a net external force is needed to cause any acceleration, just as Newton proposed in his second law of motion. So a net external force is needed to cause a centripetal acceleration. In Centripetal Force , we will consider the forces involved in circular motion.
Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration, and see how the vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze the behavior.
Can centripetal acceleration change the speed of circular motion? Explain.
A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration 1.50 times that due to gravity?
12.9 rev/min
A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 30 m. If he completes the 200 m dash in 23.2 s and runs at constant speed throughout the race, what is his centripetal acceleration as he runs the curved portion of the track?
Taking the age of Earth to be about $4\times {\text{10}}^{9}$ years and assuming its orbital radius of $\mathrm{1.5\; \times}{\text{10}}^{11}$ has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun).
$4\times {\text{10}}^{\text{21}}\phantom{\rule{0.25em}{0ex}}\text{m}$
The propeller of a World War II fighter plane is 2.30 m in diameter.
(a) What is its angular velocity in radians per second if it spins at 1200 rev/min?
(b) What is the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac?
(c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of $g$ .
An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min.
(a) Calculate the centripetal acceleration at its edge in meters per second squared and convert it to multiples of $g$ .
(b) What is the linear speed of a point on its edge?
a) $3.\text{47}\times {\text{10}}^{\text{4}}\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{2}$ , $3.\text{55}\times {\text{10}}^{\text{3}}\phantom{\rule{0.25em}{0ex}}g$
b) $51.\text{1}\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{}$
Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip.
(a) Calculate the centripetal acceleration at the tip of a 4.00 m long helicopter blade that rotates at 300 rev/min.
(b) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s).
Olympic ice skaters are able to spin at about 5 rev/s.
(a) What is their angular velocity in radians per second?
(b) What is the centripetal acceleration of the skater’s nose if it is 0.120 m from the axis of rotation?
(c) An exceptional skater named Dick Button was able to spin much faster in the 1950s than anyone since—at about 9 rev/s. What was the centripetal acceleration of the tip of his nose, assuming it is at 0.120 m radius?
(d) Comment on the magnitudes of the accelerations found. It is reputed that Button ruptured small blood vessels during his spins.
a) $\text{31.4 rad/s}$
b) $\text{118 m/s}$
c) $\text{384 m/s}$
d)The centripetal acceleration felt by Olympic skaters is 12 times larger than the acceleration due to gravity. That’s quite a lot of acceleration in itself. The centripetal acceleration felt by Button’s nose was 39.2 times larger than the acceleration due to gravity. It is no wonder that he ruptured small blood vessels in his spins.
What percentage of the acceleration at Earth’s surface is the acceleration due to gravity at the position of a satellite located 300 km above Earth?
Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:
(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.
(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).
a) 0.524 km/s
b) 29.7 km/s
A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of $9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$ at the rim?
At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m.
(a) At how many rev/min are the tires rotating?
(b) What is the centripetal acceleration at the edge of the tire?
(c) With what force must a determined $1\text{.}\text{00}\times {\text{10}}^{-\text{15}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ bacterium cling to the rim?
(d) Take the ratio of this force to the bacterium’s weight.
(a) $\text{1.35}\times {\text{10}}^{\text{3}}\phantom{\rule{0.25em}{0ex}}\text{rpm}$
(b) $\text{8.47}\times {\text{10}}^{\text{3}}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{\text{2}}$
(c) $\text{8.47}\times {\text{10}}^{\text{\u201312}}\phantom{\rule{0.25em}{0ex}}\text{N}$
(d) $\text{865}$
Integrated Concepts
Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity.
(a) What is the centripetal acceleration at the bottom of the arc?
(b) Draw a free body diagram of the forces acting on a rider at the bottom of the arc.
(c) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.
(d) Discuss whether the answer seems reasonable.
(a) $\text{16.6}\phantom{\rule{0.25em}{0ex}}\text{m/s}$
(b) $\text{19.6}\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{2}$
(c)
(d) $\text{1}.\text{76}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{N or 3}.\text{00}\phantom{\rule{0.25em}{0ex}}w$ , that is, the normal force (upward) is three times her weight.
(e) This answer seems reasonable, since she feels like she’s being forced into the chair MUCH stronger than just by gravity.
Unreasonable Results
A mother pushes her child on a swing so that his speed is 9.00 m/s at the lowest point of his path. The swing is suspended 2.00 m above the child’s center of mass.
(a) What is the centripetal acceleration of the child at the low point?
(b) What force does the child exert on the seat if his mass is 18.0 kg?
(c) What is unreasonable about these results?
(d) Which premises are unreasonable or inconsistent?
a) $\text{40}.5\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{2}$
b) 905 N
c) The force in part (b) is very large. The acceleration in part (a) is too much, about 4 g.
d) The speed of the swing is too large. At the given velocity at the bottom of the swing, there is enough kinetic energy to send the child all the way over the top, ignoring friction.
Notification Switch
Would you like to follow the 'Introduction to applied math and physics' conversation and receive update notifications?