32.
A particle is moving on a circular path with constant speed v, The magnitude of the change in its velocity
after it has described an angle of 90° is :
(A)v
(B) root (2) v
(C) 0
(D) root (3) v
Answers
Explanation:
If a particle is moving along a circular path with constant speed, what is the change in velocity when the particle completes one fourth of a revolution?
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Let the constant speed of the particle be v . We will express this as vector. So assume the particle is going clockwise in the XY plane with the center of circular path at origin (0,0) and radius r .
Let the initial position of particle at our first observation be (0,−r) . At this instant the particle has a velocity directed towards negative X-axis.
v⃗ i=−vi^
After the quarter circle motion in the third quadrant, it comes to position (−r,0) . At this instance it's velocity is directed towards positive Y-axis.
v⃗ f=vj^
So the change in velocity =v⃗ f−v⃗ i
=(0i^+vj^)−(−vi^+0j^)
=vi^+vj^
This has a magnitude of 2–√v which is greater than both the velocities. That's how vectors work. You can check this from triangle law of vector addition. Initial and final velocity are represented by two sides of right isosceles triangle, each being of length v then the difference is the length of hypotenuse = 2–√v
Answer:
If a particle is moving along a circular path with constant speed, what is the change in velocity when the particle completes one fourth of a revolution?
Sale ends on 13th November!
Let the constant speed of the particle be v . We will express this as vector. So assume the particle is going clockwise in the XY plane with the center of circular path at origin (0,0) and radius r .
Let the initial position of particle at our first observation be (0,−r) . At this instant the particle has a velocity directed towards negative X-axis.
v⃗ i=−vi^
After the quarter circle motion in the third quadrant, it comes to position (−r,0) . At this instance it's velocity is directed towards positive Y-axis.
v⃗ f=vj^
So the change in velocity =v⃗ f−v⃗ i
=(0i^+vj^)−(−v\ha
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A particle is moving on a circular path of radius r=10 M with constant speed 10 M per second. The particle turns through 60 degree what is the magnitude of change in velocity of the particle?
Particle moving along a circular path with constant speed has different velocity same in magnitude but different in direction. Direction can be obtained by connecting particle to centre of circular path and then draw a tangent on That point.
Since particles completes 1/4 revolution so direction of velocity changes 90 degree to install direction of velocity. Calculation are attached .
Answer will √2v . However it is only magnitude direction will be 135 degree counter clockwise to initial velocity