Newton\u2019s first law (inertia) tells us that an object\u2019s velocity will only change if the object is acted upon by some force. Velocity is a vector<\/strong>\u2014both magnitude and direction are important. Therefore, your velocity<\/em> can be changing, even if your speed<\/em> is constant.<\/p>\n\n\n\n Objects traveling around a circular path do experience an acceleration, called centripetal (\u201ccenter-seeking\u201d) acceleration. Moving at a speed \\(v\\) around a path of radius \\(r\\), an object\u2019s acceleration is<\/p>\n\n\n \\[ There is no single force called \u201cthe<\/em> centripetal (or centrifugal) force\u201d Circular motion is an effect of the net<\/em> force\u2014the total of all forces\u2014acting on an object.<\/p>\n\n\n\n A 1200 kg car is driving at 20 m\/s around a curve that has a radius of 10 m. The curve is not banked; it is only the force of friction between the car’s wheels and the road that keeps it moving in a circle (think about it: if the road were icy, the car would just go straight off the road). What is the magnitude of the friction?<\/p>\n\n\n\n Let’s say that “towards the center” of the curve is the \\(x\\) direction. The only force acting in this direction is friction. (If there were no friction, the car would slide out from the curve; therefore, friction points in.) We know the acceleration in this direction is centripetal acceleration since the car is moving in a circle. Newton’s second law gives us:<\/p>\n\n\n \\[ A 10 kg bucket of water is tied to a 1 m long rope and swung in a vertical circle with a speed of 4 m\/s. Find the tension in the rope when the bucket is at the very top and the very bottom of the circle.<\/p>\n\n\n\n When the bucket is at the top of the circle, tension and gravity are both acting in the \\(-y\\) direction. The direction towards the center of the circle is also the \\(-y\\) direction. Applying Newton’s second law gives:<\/p>\n\n\n \\[ When the bucket is at the bottom of the circle, gravity is acting in the \\(-y\\) direction as before, but now tension is acting in the \\(+y\\) direction. The direction towards the center of the circle is also the \\(+y\\) direction. Applying Newton’s second law, we have:<\/p>\n\n\n \\[
\na_c = \\frac{v^2}{r} \\tag{9.7}
\n\\]<\/p>\n\n\n\nExample 9.10<\/h4>\n\n\n\n
\n \\begin{align*}
\n F_\\textit{net,x} &= ma_c \\\\
\n F^f &= m\\frac{v^2}{r} \\\\
\n &= (1200\\ \\textrm{kg})\\frac{(20\\ \\textrm{m\/s})^2}{10\\ \\textrm{m}} \\\\
\n &= 48\\thinspace000\\ \\textrm{N}
\n \\end{align*}
\n\\]<\/p>\n\n\n\nExample 9.11<\/h4>\n\n\n\n
Top of the circle<\/h5>\n\n\n\n
\n \\begin{align*}
\n F_\\textit{net,y} &= ma_y \\\\
\n -F^T – F^G &= -ma_c \\\\
\n F^T + mg &= m\\frac{v^2}{r} \\\\
\n F^T &= m\\left(\\frac{v^2}{r} – g\\right) \\\\
\n &= (10\\ \\textrm{kg})\\left(\\frac{(4\\ \\textrm{m\/s})^2}{1\\ \\textrm{m}} – \\left(9.81\\ \\textrm{m\/s}^2\\right)\\right) \\\\
\n &= 62\\ \\textrm{N}
\n \\end{align*}
\n\\]<\/p>\n\n\n\nBottom of the circle<\/h5>\n\n\n\n
\n \\begin{align*}
\n F_\\textit{net,y} &= ma_y \\\\
\n F^T – F^G &= ma_c \\\\
\n F^T – mg &= m\\frac{v^2}{r} \\\\
\n F^T &= m\\left(\\frac{v^2}{r} + g\\right) \\\\
\n &= (10\\ \\textrm{kg})\\left(\\frac{(4\\ \\textrm{m\/s})^2}{1\\ \\textrm{m}} + \\left(9.81\\ \\textrm{m\/s}^2\\right)\\right) \\\\
\n &= 258\\ \\textrm{N}
\n \\end{align*}
\n\\]<\/p>\n\n\n\nPractice 9.8<\/h4>\n\n\n\n\n