1 Answer | Add Yours
The figure is below attached.
a) In a circular motion the speed is tangent to the trajectory in each point of the curve. Since in the problem the absolute value of the speed is unchanged, to draw the speed vectors at points 1, 2, 3 and 4 it is convenient to draw first the tangents to the trajectory at these points, then to measure on the tangents the same amount having the direction towards the motion of the point.
b) Since the absolute value of the velocity is the same at all points of the trajectory, only its direction is changing. To change only the direction of a vector one needs to apply a perpedicular acceleration to it. Thus the direction of the acceleration at points 1,2,3,4 is towards the instantaneous center of rotation. If the motion would be in a circle with constant speed the absolute value of the acceleration would be
Since here the instantaneous center of rotation is changing also the radius of trajectory is changing. The absoulte value of the acceleration at points 1,2,3,4 will be inversely proportional to the radius of trajectory. Thus
`R_4 >R_1>R_2>R_3 => a_4< a_1<a_2<a_3`
Newton second law is always valid. Thus at all points of the trajectory the force on the car is `F=m*a` . Since mass `m` is the same, the forces at points 1,2,3,4 will just be proportinal to the acceleration vectors at these points. The schetch is showing the forces for a value of mass just under 1 kg.
We’ve answered 319,632 questions. We can answer yours, too.Ask a question