This page supports the multimedia tutorial Circular Motion
The car drives at constant speed over a hill at whose summit the radius of curvature, in the vertical plane, is 30 m, as shown. Let's find the critical speed vcrit at which the car loses contact with the road.
Following the road at constant speed, the car is in uniform circular motion. At the top of the hill, its acceleration is therefore downwards and its magnitude is v2/r. What are the downwards forces on it? Its weight is one such force. In general, there is also a downwards component of the force that the air exerts on the car*. For our case, this second force is negligible in comparison with the weight. (See Air resistance to obtain an upper limit.)
So, at the top of the hill, the weight is the only downwards force, and so the maximum downwards acceleration is g, which, as we saw in the Projectile module, is 9.8 m.s−2. Now the downwards centripital acceleration when the car is at the top of the hill is v2/r, and this acceleration cannot be greater than g. So the critical acceleration and velocity are
* Racing cars often have a large foil whose effect, at high speed, is to provide a downwards force and thus allow greater frictional forces for cornering, braking and even acceleration. Some street cars have a miniature version of this foil. These have several possible functions. First, they identify the style of the driver, which may provide a warning to other drivers. Second, they increase drag and petrol consumption, which benefits oil companies (in the short term, at least). Do they have a significant effect in producing a downwards force? Taking the weight of the car as 20 kN, a 10% increase in downwards force would require the foil to exert 2 kN downwards. I do not recommend the following experiment, but if it can exert 2 kN, then it ought to support, elastically, the weight of a few people standing on it.
What is the acceleration of your room?
You will need to know some values: the period of the Earth's rotation is (slightly less than) 24 hours. The actual value depends on your latitude, but your distance from the Earth's axis is not greater than six thousand kilometres. So
You should do a similar calculation for the acceleration due to the Earth's orbit, which has a radius of 1.5 x 1011 m.
What if the circular motion is not uniform?
First, let's note that, in circular motion of any sort, the velocity is always tangential: in other words, v and r are at right angles. If v had a component in the r direction, then r would change in time, so it would not be circular motion. We can now derive some important results here using the scalar product.
We have just argued that v and r are at right angles to each other, so
a. r = - v. v
ar = - v2/r