erm just wanna clarify this.If one were to move in a circular motion,and u complete one revolution thus meaning you come back to the same point to where you started off,is your displacement equals to zero??
Treat "s" as the distance along the circumference measured from the starting point. All the usual kinematic formulas for uniformly accelerated motion apply (assuming the tangential acceleration is constant).SVG84R said:what my tutor did and set me puzzling was that he used kinematic equation:
v^2=u^2+2as,to solve for the linear accleration.Hence s,which is the displacement,turns out to be 2pi*r which is the circumference of the circle.How can this equation be used if we conclude that the displacment is zero for one revolution???
In addition, one of them pointed out that when the particle comes back to the same point in a circular motion,its tangential velocity is pointing in the same direction as it first started off.Now what he claimed was that when one walked in a straight line,in order to move back to where you started off,you have to change direction,hence your velocity must point in the opposite direction.
Assuming the tangential acceleration is constant, the usual kinematic equations apply.SVG84R said:hmm so if we want to find the linear acceleration,we can apply kinematics equation solve for it?
The tangential velocity at any time is just [itex]v_t = r \omega[/itex]; thus the tangential acceleration is [itex]a_t = r \alpha[/itex].is there any other way than using the kinematics equation to solve for the linear accleration?? say can we differentiate the tangential velocity wrt to time to get the linear acceleration??