Adisk rotates around an axis through its center that is perpendicular to the plane of the disk. the disk has a line drawn on it that extends from the axis of the disk to the rim. at t = 0 this line lies along the x-axis and the disk is rotating with positive angular velocity ω0z. the disk has constant positive angular acceleration αz. at what time after t = 0 has the line on the disk rotated through an angle θ?

express your answer in terms of the variables ω0z, αz, and θ.

Explanation:

The rotated angle is given by:

Since this is a quadratic equation it can be solved using:

Rewriting our equation:

Since we discard the negative solution.

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft