circular motion

linear velocity

$$v=\frac{2\pi r}{T}=2\pi rf$$

this changes with radius

linear displacement

$$s=\frac{2\pi rt}{T}=\theta r$$

angular displacement

• the angle in radians passed in a time t

• T is the time for the full rotation

$$\theta=\frac{2\pi t}{T}=2\pi tf$$

angular velocity

$$\omega=\frac{\theta}{t}=\frac{2\pi}{T}=2\pi f$$

$$rads^{-1}$$

angular velocity and linear velocity

$$v=\omega r$$

• if both objects have the same angular speed, the object with greater radius has the greater linear speed

centripetal force and acceleration

• net constant force that always acts towards the centre and causes an object to move in a circular path

• perpendicular to tangential velocity

• velocity changes but speed is constant - acceleration changes the direction

$F=\frac{mv^2}{r}$ $a=\frac{v^2}{r}$

speed of an object travelling in a circular path remains constant despite always being acted on by a resultant force

• the centripetal force always acts perpendicular to the tangential / linear velocity

• W = Fdcostheta so no work is done

• kinetic energy of the object and therefore speed does not change

• velocity changes due to changing direction only

vertical circular motion

• at the very top, the normal contact force / tension is least

• T1 = mv²/r - mg

• at the very bottom, the normal contact force / tension is greatest

• T2 = mv² / r +mg

conical pendulum

• if there a two masses and one of them now has a much greater mass than the other, the tension will be greater

• the horizontal component which provides the centripetal force, will be greater

• the vertical component of tension increases, so there is a greater compressive force on the pole

banked road

• the centripetal force is provided by the horizontal component of the normal contact force

• friction acts up or down the slope

• F (centripetal) = Rsintheta

• no side slipping so no change in elevation - maintaining circular motion - no resultant vertical force

• W = Rcostheta

• tan theta = v²/gr

• this is the optimal speed at which there is no sideways friction

horizontal circular road

• centripetal force provide by the sideways friction between the road and types

• velocity must be less than a maximum value, or the centripetal force required to maintain circular motion will be greater than the sideways friction that can be provided, causing slipping

circular hill

• the centripetal force is provided by the resultant force of the car’s weight and the normal force of the road

• the normal contact force is less than the weight

• s = mg - mv²/r

• if the speed increases, the centripetal force required to keep the object in circular motion increases - support force decreases

• at a maximum speed, support force becomes 0 - beyond this speed mg = V²/r , object will not stay in contact with hill and will not stay in circular motion