In problems involving circular motion, you frequently decompose a force into a radial force, F_r, that points to the center of motion and a tangential force, F_t, that points perpendicular to F_r and tangential to the circular path. Two examples of these forces are those applied to objects pinned at a point and motion around a curve when friction is present.

## Object Pinned at a Point

Use the fact that if an object is pinned at a point and you apply a force F at a distance R from the pin at an angle θ relative to a line to the center, then F_r = R∙cos(θ) and F_t = F∙sin(θ).

Imagine that a mechanic is pushing on the end of a wrench with a force of 20 Newtons. From the position at which she is working, she must apply the force at an angle of 120 degrees relative to the wrench.

Calculate the tangential force. F_t = 20∙sin(120) = 17.3 Newtons.

## Torque

Use the fact that when you apply a force at a distance R from where an object is pinned, the torque is equal to τ= R∙F_t. You may know from experience that the farther out from the pin you push on a lever or wrench, the easier it is to make it rotate. Pushing at a greater distance from the pin means you are applying a larger torque.

Imagine that a mechanic is pushing on the end of a 0.3-meter-long torque wrench to apply 9 Newton-meters of torque.

Calculate the tangential force. F_t = τ/R = 9 Newton-meters/0.3 meters = 30 Newtons.

## Non-Uniform Circular Motion

Use the fact that the only force needed to keep an object in circular motion at a constant speed is a centripetal force, F_c, which points towards the center of the circle. But if the speed of the object is changing, then there must also be a force in the direction of motion, which is tangential to the path. An example of this is the force from the engine of a car causing it to speed up when going around a curve or the force of friction slowing it to stop.

Imagine that a driver takes his foot off of the accelerator and lets a 2,500 kilogram car coast to a stop beginning from a starting speed of 15 meters/second while steering it around a circular curve with a radius of 25 meters. The car coasts 30 meters and takes 45 seconds to stop.

Calculate the acceleration of the car. The formula incorporating the position, x(t), at time t as a function of the initial position, x(0), the initial velocity, v(0), and the acceleration, a, is x(t) – x(0) = v(0)∙t + 1/2∙a∙t^2. Plug in x(t) – x(0) = 30 meters, v(0) = 15 meters per second and t = 45 seconds and solve for the tangential acceleration: a_t = –0.637 meters per second squared.

Use Newton’s second law F = m∙a to find that friction must have applied a tangential force of F_t = m∙a_t = 2,500×(–0.637)= –1,593 Newtons.

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About the Author

Ariel Balter started out writing, editing and typesetting, changed gears for a stint in the building trades, then returned to school and earned a PhD in physics. Since that time, Balter has been a professional scientist and teacher. He has a vast area of expertise including cooking, organic gardening, green living, green building trades and many areas of science and technology.