With a quick flick of her wrist, an Ultimate Frisbee player can give a Frisbee an angular velocity of 4.00 revolutions per second. To do this, the player accelerates the Frisbee from rest through an angle of 55.0° before letting it go. (a) What is the Frisbee's angular velocity, in units of rad/s, when the Ultimate Frisbee player releases the Frisbee? rad/s (b) Through what angle, in radians, does the player rotate the Frisbee? rad (C) What is the Frisbee's angular acceleration while the Frisbee player is flicking her wrist? Assume the angular acceleration is constant. rad/s2 (d) How much time does this process take?

Question

Alonso Higgins

Physics

27 March, 18:45

With a quick flick of her wrist, an Ultimate Frisbee player can give a Frisbee an angular velocity of 4.00 revolutions per second. To do this, the player accelerates the Frisbee from rest through an angle of 55.0° before letting it go. (a) What is the Frisbee's angular velocity, in units of rad/s, when the Ultimate Frisbee player releases the Frisbee? rad/s (b) Through what angle, in radians, does the player rotate the Frisbee? rad (C) What is the Frisbee's angular acceleration while the Frisbee player is flicking her wrist? Assume the angular acceleration is constant. rad/s2 (d) How much time does this process take?

+1

Answers (1)

Know the Answer?

Quinn Copeland · Answer 1

a) w = 25.1 rad/s, b) θ = 0.9599 rad, c) α = 328.1 rad/s² d) t = 0.0765 s

Explanation: Let's work on this exercise with the equations of angular kinematics

a) The angular velocity is

w = 4.00 rev / s (2π rad / 1 rev)

w = 25.1 rad/s

b) let's reduce the angle of degrees to radians

θ = 55 ° (π rad / 180 °)

θ = 0.9599 rad

c) Let's use the initial angular velocity as the system part of the rest is zero

w² = w₀² + 2 α θ

α = w² / 2 θ

α = 25.1²/2 0.9599

α = 328.1 rad / s²

d)

w = w₀ + α t

t = w / α

t = 25.1 / 328.1

t = 0.0765 s