PhanthomJay said:This problem is a bit more complex than that. You have to look at the forces acting on the bead at angle beta
NO, the hoop is not rotating about its center point, it is rotating about the vertical axis, that is, out of and into the plane, as if you were spinning a top.cheater1 said:Will this be the direction of a_rad?
PhanthomJay said:NO, the hoop is not rotating about its center point, it is rotating about the vertical axis, that is, out of and into the plane, as if you were spinning a top.
Yes.cheater1 said:so that means the a_rad is pointing right from the bead?
PhanthomJay said:Yes.
No. Firstly, r is not 0.1; as you noted before, the radial acceleration points to the right from the bead, so that is the value of "r" to use, the distance from the bead to the vertical axis in the horizontal direction. Also, it is better to use the angular speed instead of the out of plane tangential speed, that is , a_radial = w^2(r), where r = R*(____??___), and w is 4 rev/s, or w = 8 pi rad/scheater1 said:Just to check is this right?
a_rad= 4pi^2(0.1)/.25= 15.79 m/s^2
PhanthomJay said:No. Firstly, r is not 0.1; as you noted before, the radial acceleration points to the right from the bead, so that is the value of "r" to use, the distance from the bead to the vertical axis in the horizontal direction. Also, it is better to use the angular speed instead of the out of plane tangential speed, that is , a_radial = w^2(r), where r = R*(____??___), and w is 4 rev/s, or w = 8 pi rad/s
where you get this?a_radial = w^2(r)
Yes, if the rotation speed was increased or decreased, then r would change with respect to the angle.cheater1 said:Ok, I got r=(sinB)R , since the hoop is spinning, the bead will be going up. As it goes up r will will be changing with respect to angle.
Since a_radial= v^2/r, and v=wr, then a_radial = w^r^2/r = w^2(r) (where w = omega, the angular speed in rad/sec)where you get (a_radial =w^2r)?
PhanthomJay said:Since a_radial= v^2/r, and v=wr, then a_radial = w^r^2/r = w^2(r) (where w = omega, the angular speed in rad/sec)
PhanthomJay said:No, w= rev/sec= 8(pi)radians/sec, then v=wr = 8(pi)r/ sec, where r = RsinB
PhanthomJay said:That's the biggest thanks I ever got,
You're Welcome!
Did you get parts b and c?
A radial acceleration problem is a physics problem that involves calculating the acceleration of an object moving in a circular path, also known as radial motion. This type of problem is common in fields such as astronomy, where objects orbit around a central mass, or in mechanics, where objects move along a curved path.
The formula for radial acceleration is a = v^2/r, where a is the acceleration, v is the velocity, and r is the radius of the circular path. This formula can be derived from Newton's second law of motion, F = ma, where the force (F) is equal to the mass (m) multiplied by the acceleration (a).
Radial acceleration is the acceleration of an object moving in a circular path, while tangential acceleration is the acceleration of an object moving in a straight line. In other words, radial acceleration is directed towards the center of the circle, while tangential acceleration is directed tangent to the circle.
Radial acceleration is a fundamental concept in many fields of science, including astronomy, mechanics, and engineering. It is used to understand the motion of objects in circular paths, such as planets orbiting around a star, cars navigating a curved road, or roller coasters moving along a track. It also has practical applications in designing and analyzing circular motion devices, such as centrifuges or particle accelerators.
One common mistake is confusing radial acceleration with tangential acceleration, as they have different formulas and directions. Another mistake is not accounting for the radius of the circular path, which is a crucial component in the formula for radial acceleration. It is also important to use consistent units and pay attention to the directions of vectors, as they can affect the final answer. Additionally, not considering factors such as friction or air resistance can lead to inaccurate solutions.