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.

## Sciencing Video Vault

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.