1.If a particle moves in a plane so that its position is described by the functions x=A*cos(wT) and y=A*sin(wT), the particle is ( w-angular velocity, T-period)
A) moving with constant speed along a circle
B) moving with a varying speed along a circle
Thanx!
Note that x^2 + y^2, the square of distance from the origin, is A^2
Note also the value of the square of the speed, which is [(dx/dt)^2 + (dy/dt)^2] , is A^2 w^2, which is a constant.
What does that tell you about the motion?
centripetal motion?
The word "centripetal" refers to the force that keeps changung the directkion of motion. The answer to the question is (A).
The particle is moving with constant speed along a circle.
Well, it looks like this particle is pulling off some fancy moves! It's actually moving with constant speed along a circle. So, the answer is (A), just like that person who always walks in circles at the office party. Always a hit, right? Keep on spinning, particle!
The particle is moving with constant speed along a circle. The fact that the square of distance from the origin is constant (x^2 + y^2 = A^2) indicates that the particle is moving in a circular path. Additionally, the fact that the square of the speed is constant ([(dx/dt)^2 + (dy/dt)^2] = A^2w^2) further confirms that the particle is moving with a constant speed along the circular path. This type of motion is known as centripetal motion, where the particle experiences a force constantly changing its direction of motion. Therefore, the correct answer is (A) moving with constant speed along a circle.
To determine whether the particle is moving with constant or varying speed along a circle, we need to analyze the given functions: x = A*cos(wT) and y = A*sin(wT), where A is the amplitude and w is the angular velocity.
We can start by expressing the distances squared in terms of x and y coordinates. Considering that the particle's position is described in a plane, using the Pythagorean theorem, we have:
Distance squared = x^2 + y^2
Substituting the given functions for x and y, we get:
Distance squared = (A*cos(wT))^2 + (A*sin(wT))^2
Simplifying, we have:
Distance squared = A^2*cos^2(wT) + A^2*sin^2(wT)
Using the identity cos^2(x) + sin^2(x) = 1, we rewrite the equation:
Distance squared = A^2 * (cos^2(wT) + sin^2(wT))
Since cos^2(wT) + sin^2(wT) equals 1, we have:
Distance squared = A^2 * (1)
Therefore, the distance squared from the origin is A^2, which is a constant.
Next, let's look at the square of the speed, given by [(dx/dt)^2 + (dy/dt)^2].
dx/dt represents the derivative of x with respect to t, and dy/dt represents the derivative of y with respect to t. Taking the derivatives, we have:
dx/dt = -A*w*sin(wT)
dy/dt = A*w*cos(wT)
Squaring and summing the derivatives, we get:
[(dx/dt)^2 + (dy/dt)^2] = (A*w*sin(wT))^2 + (A*w*cos(wT))^2
Simplifying, we have:
[(dx/dt)^2 + (dy/dt)^2] = A^2*w^2*sin^2(wT) + A^2*w^2*cos^2(wT)
Using the identity sin^2(x) + cos^2(x) = 1, we rewrite the equation as:
[(dx/dt)^2 + (dy/dt)^2] = A^2*w^2*(sin^2(wT) + cos^2(wT))
Since sin^2(wT) + cos^2(wT) equals 1, we have:
[(dx/dt)^2 + (dy/dt)^2] = A^2*w^2*(1)
Therefore, the square of the speed is A^2*w^2, which is also a constant.
From the analysis above, we can conclude that both the distance squared from the origin and the square of the speed are constants. This indicates that the motion of the particle is consistent with moving with a constant speed along a circle, option A. The constant speed is determined by the product of the amplitude A and the angular velocity w.