An object is moving along a circular path of radius r. (5 marks)
(a) Show that its velocity vector is perpendicular to the radial rays at any moment. This confirms that the velocity vector is tangent to the path, as it should be.
(b) Show that its acceleration vector is perpendicular to its velocity vector if the object rotates with a fixed speed. This confirms that the acceleration vector will be parallel to the radial ray at any moment
(a) To show that the velocity vector is perpendicular to the radial rays at any moment, we need to examine the relationship between the velocity vector and the radius vector.
Let's denote the position vector of the object as r(t), where t is the time. The velocity vector v(t) is then given by the derivative of the position vector with respect to time:
v(t) = dr(t)/dt
Since the object is moving along a circular path of radius r, we can write the position vector as:
r(t) = r*cos(theta(t))*i + r*sin(theta(t))*j
where theta(t) is the angle between the position vector and the positive x-axis at time t, i is the unit vector in the x-direction, and j is the unit vector in the y-direction.
Taking the derivative of r(t) with respect to time, we get:
dr(t)/dt = -r*sin(theta(t))*d(theta(t))/dt * i + r*cos(theta(t))*d(theta(t))/dt * j
But d(theta(t))/dt is the angular velocity, denoted as w, since the object rotates with a fixed speed. Therefore, we have:
dr(t)/dt = -r*w*sin(theta(t))*i + r*w*cos(theta(t))*j
The velocity vector is given by v(t) = dr(t)/dt, so we can write:
v(t) = -r*w*sin(theta(t))*i + r*w*cos(theta(t))*j
To show that the velocity vector is perpendicular to the radial rays, we need to show that the dot product of the velocity vector and the radial vector is equal to zero.
The radial vector is given by r(t) = r*cos(theta(t))*i + r*sin(theta(t))*j.
Taking the dot product of v(t) and r(t), we have:
v(t) • r(t) = (-r*w*sin(theta(t))*i + r*w*cos(theta(t))*j) • (r*cos(theta(t))*i + r*sin(theta(t))*j)
Simplifying the dot product, we get:
v(t) • r(t) = -r*w*sin(theta(t))*r*cos(theta(t)) + r*w*cos(theta(t))*r*sin(theta(t))
v(t) • r(t) = 0
Since the dot product of the velocity vector and the radial vector is zero, we can conclude that the velocity vector is perpendicular to the radial rays at any moment. This confirms that the velocity vector is tangent to the circular path, as it should be.
(b) To show that the acceleration vector is perpendicular to the velocity vector if the object rotates with a fixed speed, we need to examine the relationship between the acceleration vector and the velocity vector.
The velocity vector v(t) can be written as:
v(t) = -r*w*sin(theta(t))*i + r*w*cos(theta(t))*j
The acceleration vector a(t) is defined as the derivative of the velocity vector with respect to time:
a(t) = dv(t)/dt
Taking the derivative of v(t) with respect to time, we get:
a(t) = -r*w*cos(theta(t))*d(theta(t))/dt * i - r*w*sin(theta(t))*d(theta(t))/dt * j
Since d(theta(t))/dt is the angular velocity, denoted as w, and the object rotates with a fixed speed, d(w)/dt = 0. Therefore, we have:
a(t) = -r*w*cos(theta(t))*w * i - r*w*sin(theta(t))*w * j
Simplifying the expression, we get:
a(t) = -r*w^2*cos(theta(t))*i - r*w^2*sin(theta(t))*j
To show that the acceleration vector is perpendicular to the velocity vector, we need to show that the dot product of the acceleration vector and the velocity vector is equal to zero.
Taking the dot product of a(t) and v(t), we have:
a(t) • v(t) = (-r*w^2*cos(theta(t))*i - r*w^2*sin(theta(t))*j) • (-r*w*sin(theta(t))*i + r*w*cos(theta(t))*j)
Simplifying the dot product, we get:
a(t) • v(t) = r*w^3*sin(theta(t))*cos(theta(t)) - r*w^3*sin(theta(t))*cos(theta(t))
a(t) • v(t) = 0
Since the dot product of the acceleration vector and the velocity vector is zero, we can conclude that the acceleration vector is perpendicular to the velocity vector at any moment. This confirms that the acceleration vector will be parallel to the radial ray at any moment.
To answer both parts (a) and (b), we will need to use some principles of circular motion.
(a) To show that the velocity vector is perpendicular to the radial rays at any moment, we can consider that the object's velocity vector always points tangent to its circular path.
To demonstrate this, we can use the concept of centripetal acceleration. The centripetal acceleration of an object moving along a circular path of radius r with speed v can be calculated using the formula:
a_c = v^2 / r
where a_c represents the centripetal acceleration.
Now, consider a point on the circular path. At that point, the radial ray is the line connecting the center of the circle to the object. The object's velocity vector points tangent to the circular path at that point.
Since the velocity vector is tangent to the circular path, and the centripetal acceleration is directed towards the center of the circle, we can conclude that the velocity vector is perpendicular to the radial ray at any moment.
(b) To show that the acceleration vector is perpendicular to the velocity vector when the object rotates with a fixed speed, we need to understand the relationship between centripetal acceleration, tangential acceleration, and total acceleration.
The total acceleration of an object moving along a circular path is the vector sum of its centripetal acceleration and tangential acceleration. In this case, as the object rotates with a fixed speed, the magnitude of the tangential acceleration is zero.
The centripetal acceleration for an object moving along a circular path with constant speed is given by the formula:
a_c = v^2 / r
where v represents the constant speed and r is the radius of the circular path.
Now, since the total acceleration is the vector sum of centripetal and tangential accelerations, and the tangential acceleration is zero in this scenario, the total acceleration vector is equal to the centripetal acceleration vector.
Since the centripetal acceleration is directed towards the center of the circle, and the velocity vector is tangent to the circular path, the total acceleration vector is perpendicular to the velocity vector at any moment. Therefore, we can conclude that the acceleration vector is also perpendicular to the velocity vector when the object rotates with a fixed speed.