A horizontal rod rotates around its center with a rotational speed of 5.60 revolutions per second and experiences an angular acceleration of 6.70 revolutions per second squared. 1. Calculate how many revolutions the rod needs to complete to accelerate to a speed of 23.0 revolutions per second. 1.1b. Given that the rod has a mass of 0.560 kg and a length of 40.0 cm, determine the magnitude of the perpendicular forces, F, in newtons, that must be exerted on both ends of the rod to achieve the aforementioned acceleration. The diagram provided depicts a top view of the rod. The moment of inertia for the rod is given by (Irod = 1/12 ML²).
Angular speed, acceleration and displacement
Angular acceleration is defined as the rate of change in angular velocity with respect to time. It has both magnitude and direction. So, it is a vector quantity.
Angular Position
Before diving into angular position, one should understand the basics of position and its importance along with usage in day-to-day life. When one talks of position, it’s always relative with respect to some other object. For example, position of earth with respect to sun, position of school with respect to house, etc. Angular position is the rotational analogue of linear position.
A horizontal rod rotates around its center with a rotational speed of 5.60 revolutions per second and experiences an
1. Calculate how many revolutions the rod needs to complete to accelerate to a speed of 23.0 revolutions per second.
1.1b. Given that the rod has a mass of 0.560 kg and a length of 40.0 cm, determine the magnitude of the perpendicular forces, F, in newtons, that must be exerted on both ends of the rod to achieve the aforementioned acceleration. The diagram provided depicts a top view of the rod. The moment of inertia for the rod is given by (Irod = 1/12 ML²).
Step by step
Solved in 4 steps with 3 images
why did you use 1/2 ML^2 if the provied value is 1/12 ML^2?