Question

6 A particle of mass m moves in a circular path of radius r in a
horizontal plane with uniform angular velocity about the z-axis.
Write down the position vector of the particle at any instant and
hence derive expressions for velocity and acceleration of the particle.​

Answer 1

Answer: position vector are

Along x(Sx)=rcos(wt); Along y(Sy)=rsin(wt), where w is angular velocity

Net displacement=r from origin

Velocity along y(Vy)= wrcos(wt);along x(Vx)=-wrsin(wt)

Net v=wr

Acceleration along x(Ax)=-(w)^2rcos(wt); along y(Ay)=-(w)^2rsin(wt)

Net acceleration=-(w)^2r

Explanation:

Since angular velocity (w) is constant therefore angular displacement (theta)=wt

Then, position is given by the diagram where along x(Sx) = r cos(theta) and along y (Sy) = r sin(theta) therefore by substituting theta= wt

We get the position along x and y

(I.e)Along x(Sx)=rcos(wt); Along y(Sy)=rsin(wt)

And net displacement=√((Sx)^2+(Sy)^2)=r{√(sin(wt))^2+(cos(wt))^2}=r

And by differentiating it we get

velocity (i.e)Velocity along y(Vy)= wrcos(wt);along x(Vx)=-wrsin(wt)

Net (v) =√((Vx)^2+(Vy)^2)=wr{√(-sin(wt))^2+(cos(wt))^2}=wr

And by differentiating it again we get acceleration (i.e)

Acceleration along x(Ax)=-(w)^2rcos(wt); along y(Ay)=-(w)^2rsin(wt)

Net acceleration(a)=√((Ax)^2+(Ay)^2) =-w^2r{√(sin(wt))^2+(cos(wt))^2}=-w^2r