Calculate the acceleration of a car in m/s that can go from rest to 28 m/s in 10 seconds.
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Answer:
Toward the centre of the circular path
Explanation:
The can is moved in a circular path: this means that it is moving by circular motion (uniform circular motion if its tangential speed is constant).
In order to keep a circular motion, an object must have a force that pushes it towards the centre of the circular trajectory: this force is called centripetal force, and its magnitude is given by
where m is the mass of the object, v its tangential speed, r the radius of the trajectory. This force always points towards the centre of the circular path.
First, let's put 22 km/h in m/s:
Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by
The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:
Notice we don't use the kinetic coefficient even though the bike is moving. This is because when the tires meet the road they are momentarily stationary with the road surface. Otherwise the bike is skidding.
Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:
Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by
The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:
Notice we don't use the kinetic coefficient even though the bike is moving. This is because when the tires meet the road they are momentarily stationary with the road surface. Otherwise the bike is skidding.
Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve: