The most common problem in the determination of the radius of curvature of trajectory of the thrown body in a given period of time. The trajectory in this case is described by the equations on the coordinate axes: x = f(t), y = f(t), where t is the time and date at which you want to find the radius. Evaluation will be based on the formula AP = V2/R. Here the radius R is detected from the ratio of the normal acceleration AP and the instantaneous speed V of the movement of the body. Knowing these values, you can easily find the desired component R.
Calculate the projection velocity of the body on the axes (OX, OY). The mathematical sense of speed is the first derivative of the equations of motion. So they are easily found by taking the derivative of the set of equations: Vx = x, Vy = y'. When considering geometric data display projection in a coordinate system shows that they are legs of a right triangle. And the hypotenuse in it – the desired velocity. Based on this, calculate the magnitude of the instantaneous velocity V by the Pythagorean theorem: V = √( Vx2 + Vy2). Substituting in the expression is a known value of time, find the index of V.
Module normal acceleration is also easy to identify, having another right triangle formed by the module full acceleration and tangential acceleration AK. And here the normal acceleration is the leg and calculated as: AP = √( A2 - ak2). To find the tangential acceleration differentiate in time equation instantaneous speed of motion: AK = |dV/dt|. Full acceleration calculate from its projections on the axis, similar to finding the instantaneous velocity. Just for this take from the set of equations of motion of second order derivatives: Ah = x," y = y". The module of the acceleration a = √( ах2 + ау2). Substituting all the found values, determine a numeric value for the normal acceleration AP = √( A2 - ak2).
Express from the formula AP = V2/R the desired variable radius of curvature of the trajectory: R = V2/ AP. Substitute the numerical values of speed and acceleration, and calculate the radius.
Advice 2: How to find tangential acceleration
Tangential acceleration is the bodies moving along a curved path. It is aimed in the direction of change of the velocity of the body at a tangent to the trajectory. Tangential acceleration happens to bodies moving uniformly in a circle, they have only centripetal acceleration.
You will need
- speedometer or the radar;
- - ruler or measuring tape;
- - stopwatch.
Find the tangential acceleration aτ, if you know the full acceleration of a particle moving along a curved trajectory a and its centripetal acceleration an. To do this, from the square full acceleration, subtract the square of the centripetal acceleration, and from this value, extract the square root aτ=√(a2-an2). If you do not know the centripetal acceleration, but there is a value of instantaneous speed, measure with a ruler or tape measure the radius of curvature of the trajectory and find its value by dividing the square of the instantaneous velocity v, which measure the speedometer or the radar on the radius of curvature of the trajectory R, an=v2/R.
Example. The body moves in a circle with a radius of 0.12 m. Its full acceleration is 5 m/S2, determine its tangential acceleration, at a time when its velocity is 0.6 m/s. First, find the centripetal acceleration of a body at a specified speed, it it square divide by the radius of the trajectory of an= v2/R= 0,62/0,12=3 m/S2. Find the tangential acceleration by the formula aτ=√(a2-an2)=√(52-32)=√(25-9)= √16=4 m/S2.
Find the magnitude of tangential acceleration through a change in the speed module. To do this using a speedometer or radar, determine the initial and final velocity of the body for a certain period of time, which you measure with a stopwatch. Find the tangential acceleration, v the ultimate subtracting from the initial value of speed v0 and dividing the time interval t during which that change occurred: aτ= (v-v0)/t. If the value of the tangential acceleration turned negative, so the body slows down, if positive accelerating.
Example. 4 with the velocity of a body moving along the circumference, decreased from 6 to 4 m/s. Determine the tangential acceleration. Calculated by applying the formula, obtain aτ= (v-v0)/t=(4-6)/4=-0,5 m/S2. This means that the body slows down with an acceleration of the absolute value of which is equal to 0.5 m/S2.