7. Newton’s second law of motion states that the centripetal force, fc , acting on an automobile of mass, m, while it is going around a curve of radius, r, is affected by its velocity, v. More specifically the force is directly proportional to the square of the velocity and inversely proportional to the radius keeping the mass constant. a) Write this statement as an equation b) Compute the centripetal force, fc, on a car that weighs 3200 Lbf and is traveling at a velocity of 45 mi/hr around a curve with a radius of 50 feet. Pay attention to the units of measure. mass measured in Lbm = weight measured in Lbf / 32 ft/sec" 1 mile = 5,280 feet 1 hour = 3600 sec c) Compute the radius of the curve if the centripetal force is reduced to 7,000 Lbf to account for bad road conditions. Assume the car has the same mass and travels at the same speed.
Added by Ann M.
Close
Step 1
Newton's second law of motion states that the centripetal force is directly proportional to the square of the velocity and inversely proportional to the radius. So, we can write this as: $$ f_c \propto \frac{v^2}{r} $$ Show more…
Show all steps
Your feedback will help us improve your experience
Madhur L and 80 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A car with a mass of 1300 kg travels around a banked curve with a constant speed of 25 m/s. The radius of curvature of the curve is 35 m. Calculate the centripetal acceleration of the car. (You must provide an answer before moving to the next part.) (Round the final answer to four decimal places.) The centripetal acceleration of the car is ______m/s2. What is the magnitude of the horizontal component of the normal force that would be required to produce this centripetal acceleration in the absence of any friction? (Round the final answer to four decimal places.) The magnitude of the force is ______kN.
Bhushan K.
A car is driving around a curve that can be approximated as being circular. In which direction does the centripetal force point? towards the center of the circle in the direction of motion away from the center of the circle tangential to the circle perpendicular to the plane of the circle The centripetal force (FC) of an object can be calculated using the equation FC=mv2/r where m is the object's mass, v is the object's velocity, and r is the radius of the circle. If the radius of the circle changes by a factor of 0.5, the force changes by a factor of 0.8 3 0.3 0.5 2 1 If the velocity changes by a factor of 4, the force changes by a factor of 5 4 6 1 16 12
Ivan K.
Centripetal Force An object of mass $m$ moves at a constant speed $v$ in a circular path of radius $r .$ The force required to produce the centripetal component of acceleration is called the centripetal force and is given by $F=m v^{2} / r .$ Newton's Law of Universal Gravitation is given by $F=G M m / d^{2}$, where $d$ is the distance between the centers of the two bodies of masses $M$ and $m$, and $G$ is a gravitational constant. Use this law to show that the speed required for circular motion is $v=\sqrt{G M / r}$
Vector-Valued Functions
Tangent Vectors and Normal Vectors
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD