![]() ![]() Unless this object is undergoing a change of potential energy (unlikely if it’s still uniform circular motion), the kinetic energy and potential energy are constant. If an object is undergoing uniform circular motion, the speed is constant, and therefore the kinetic energy is also constant. Where \(v\) is in metres per second, \(ω\) is in radians per second, and \(r\) is the radius of the circular path. This can be calculated by determining the total change-in-angle (\(Δθ\)) the object undergoes in a certain amount of time (\(Δt\)) Ī special case is when you already know the period (\(T\)) of rotation – by definition, this is the time it takes for the object to rotate \(2π\) radians! If you know this value, you can determine the angular velocity using:Ĭan you see the similarities between this equation and our equation for linear velocity? From this, we can infer that angular velocity and linear velocity are related via the following: The angular velocity ω (in radians per second) of an object is the number of radians it rotates every second. The time taken for one revolution is known as the Period, \(T\).įrom this, we can derive the linear velocity of an object undergoing UCM as:Īnother way of thinking about the velocity of an object undergoing UCM is by considering the angular change. If the problem is one under gravity, you will be expected to know that \(a_y=g=-9.8 ms^\).įor an object undergoing UCM, the distance travelled in one revolution (trip around the circle) is the circumference \(\Delta d=2πr\). Of course, the equations of motion in the vertical direction are a little more complex: Setting \(a_x=0\) results in rather simple horizontal equations of motion: So, we can set the acceleration in the horizontal (\(x\)) direction to be zero. The simplest example of this is a projectile motion under gravity: since gravity only acts in the vertical (\(y\)) direction. In order to extend these equations to 2-dimensions, we are going to make use of the following assumption: the acceleration in one of the dimensions is zero. You’ll recall that these equations were valid only for constant acceleration (and zero is a constant, too!) In other words, there is no such thing as \(t_x\) or \(t_y\): the motion in each axis shares a common time \(t\).Ī key skill you developed in Year 11 was the use of the equations of motion that related the variables \(s\), \(u\), \(v\), \(a\) and \(t\): Whenever the motion is stopped in one dimension (for example, a ball hitting the ground after falling in the \(y\) direction), the motion in the other direction also stops. In fact, the only thing these two dimensions have in common is time. \(a_y\) – the acceleration in the \(y\)-axisĪn important thing to keep in mind is that the \(x\) and \(y\) axes are independent of one another – the position, velocity and acceleration of an object in the \(x\) direction have no impact on the position, velocity and acceleration of an object in the \(y\) direction, and vice versa.\(a_x\) – the acceleration in the \(x\)-axis. ![]()
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